2015南京大学考博真题泛函分析
2015年南京大学考博英语真题及详解【圣才出品】
2015年南京大学考博英语真题及详解Part I Vocabulary and Structure (20%)Directions: There are 20 incomplete sentences in this part. For each sentence there are four choices marked A, B, C and D respectively. Choose the ONE thatbest complete the sentences. Then blacken your answer in thecorresponding letter on your Answer Sheet with a single line through thecenter.1. The ambassador was accused of having _____ on domestic affairs.A. trespassedB. encroachedC. entrenchedD. invaded【答案】A【解析】句意：大使被指控干涉国内事务。
该题为近义词辨析，选项中的四个词均有侵犯的意思。
trespass为不及物动词，一般与介词on搭配使用，意思为“擅自进入；侵犯，侵害；打扰”，强调非法侵入，符合题意。
encroach意思为“蚕食；侵占”，强调侵入并占领。
entrench意思为“用壕沟围绕或保护…；牢固地确立…”，强调在某处站稳脚跟。
invade一般用作及物动词，指“侵入，攻占；侵袭”。
2. The goal is to use crops, weeds, and even animal waste _____ the petroleum that fuels much of American manufacturing.A. in terms ofB. in favor ofC. in spite ofD. in place of【答案】D【解析】句意：目标是使用农作物、杂草甚至动物粪便来代替石油为美国制造业提供能源。
泛函分析答案
泛函分析答案泛函分析题1_3列紧集p191.3.1 在完备的度量空间中，求证：为了子集A是列紧的，其充分必要条件是对?ε > 0，存在A的列紧的ε网．证明：(1) 若子集A是列紧的，由Hausdorff定理，ε > 0，存在A的有限ε网N．而有限集是列紧的，故存在A的列紧的ε网N．(2) 若?ε > 0，存在A的列紧的ε/2网B．因B列紧，由Hausdorff定理，存在B的有限ε/2网C．因C ?B ?A，故C为A的有限ε网．因空间是完备的，再用Hausdorff定理，知A是列紧的．1.3.2 在度量空间中，求证：紧集上的连续函数必是有界的，并且能达到它的上、下确界．证明：设(X, ρ)是度量空间，D是紧子集，f : D→ 是连续函数．(1) 若f无上界，则?n∈ +，存在x n∈D，使得f (x n) > 1/n．因D是紧集，故D是自列紧的．所以{x n}存在收敛子列x n(k) →x0∈D (k→∞)．由f的连续性，f (x n(k))→f (x0) (k→∞)．但由f (x n) > 1/n知f (x n)→ +∞(n→∞)，所以f (x n(k))→ +∞ (k→∞)，矛盾．故f有上界．同理，故f有下界．(2) 设M = sup x∈D f(x)，则?n∈ +，存在y n∈D，使得f (y n) > M- 1/n．{y n}存在子列y n(k) →y0∈D (k→∞)．因此f ( y0 ) ≥M．而根据M的定义，又有f ( y0 ) ≤M．所以f ( y0 ) = M．因此f能达到它的上确界．同理，f能达到它的下确界．1.3.3 在度量空间中，求证：完全有界的集合是有界的，并通过考虑l 2的子集E = {e k }k≥ 1，其中e k = { 0, 0, ..., 1, 0, ... } (只是第k 个坐标为1，其余都是0 )，来说明一个集合可以是有界的但不完全有界的．证明：(1) 若A是度量空间(X, ρ)中的完全有界集．则存在A的有限1-网N = { x0, x1, x2, ..., x n }．令R = ∑1 ≤j≤nρ(x0, x j) + 1．则?x∈A，存在某个j使得0 ≤j≤n，且ρ(x, x j) < 1．因此，ρ(x, x0) ≤ρ(x, x j) + ρ(x j, x0) ≤ 1 + ∑1 ≤j≤nρ(x0, x j) = R．所以A是度量空间(X, ρ)中的有界集．(2) 注意到ρ(e k , e j) = 21/2 ( ?k ≠ j )，故E中任意点列都不是Cauchy列．所以，E中任意点列都没有收敛子列(否则，该收敛子列就是Cauchy列，矛盾)．因此，E不是列紧集．由l 2是完备的，以及Hausdorff定理，知E不是全有界集．但E显然是有界集．1.3.4 设(X, ρ)是度量空间，F1, F2是它的两个紧子集，求证：?x i ∈F i( i = 1, 2)，使得ρ(F1, F2) = ρ(x1, x2)．其中ρ(F1, F2) = inf {ρ(x, y) | x∈F1, y∈F2 }证明：由ρ(F1, F2)的定义，?n∈ +，?x i(n)∈F i( i = 1, 2)，使得ρ(x1(n), x2(n)) < ρ(F1, F2) + 1/n．因F1, F2紧，故不妨假设{x1(n)}, {x2(n)}都是收敛列．设它们的极限分别为x1, x2，则ρ(x1, x2) ≤ρ(F1, F2)．因此ρ(F1, F2) = ρ(x1, x2)．1.3.5 设M是C[a, b]中的有界集，求证集合{F(x) =?[a, x]f(t) dt | f∈M }是列紧集．证明：设A = {F(x) =?[a, x]f(t) dt | f∈M }．由M有界，故存在K > 0，使得?f∈M，ρ( f, 0) ≤K．先证明A是一致有界的和等度连续的．F∈A，存在f∈M，使得F(x) =?[a, x]f(t) dt．由于ρ(F, 0) = max x∈[a,b] | F(x) | = max x∈[a, b] | ?[a, x]f(t) dt |≤ max x∈[a, b] | f(t) | · (b -a ) = ρ( f, 0) · (b -a ) ≤K (b -a )．故A是一致有界的．ε > 0，?s, t∈[a, b]，当| s-t| < ε/K时，F∈A，存在f∈M，使得F(x) =?[a, x]f(u) du．| F(s) -F(t) | = | ?[s, t]f(u) du | ≤ max u∈[a, b] | f(u) | · | s -t |= ρ( f, 0) · | s -t | ≤K · (ε/K) = ε．故A是等度连续的．由Arzela-Ascoli定理，A是列紧集．1.3.6 设E = {sin nt}n≥ 1，求证：E在C[0, π]中不是列紧的．证明：显然E是一致有界的．根据Arzela-Ascoli定理，我们只要证明E不是等度连续的即可．我们的想法是找一个E中的点列f n，以及[0, π]中的两个点列s n 和t n，使得| s n -t n | → 0，但| f n(s n)-f n(t n)|不收敛于0．事实上，这是可以做到的，只要令f n (u) = sin (2n u)，s n = (π/2)(1 + 1/(2n))，t n = (π/2)(1 - 1/(2n))．则s n + t n = π；s n -t n = π/(2n)→ 0(n→∞)．因此，| f n(s n)-f n(t n)| = 2 | sin (2n s n) - sin (2n t n) |= 2 | sin (n (s n -t n)) cos (n (s n + t n)) |= 2 | sin (π/2) cos (n π) | = 2．所以，E不是等度连续的．进而，E在C[0, π]中不是列紧的．1.3.7 求证S空间的子集A是列紧的充要条件是：?n∈ +，?C n> 0，使得x = (ξ1, ξ2, ..., ξn, ...)∈A，都有| ξn | ≤C n( n = 1, 2, ...)．证明：(?) 设x k = (ξ1(k), ξ2(k), ..., ξn(k), ...) ( k = 1, 2, ... )是A中的点列．存在{x k}的子列{x1, k}使得其第1个坐标ξ1(1, k)收敛；存在{x1, k}的子列{x2, k}使得其第2个坐标ξ2(2, k)收敛；如此下去，得到一个{x k}的子列的序列，第( j +1)个子列是第j个子列的子列，且第j个子列的第j个坐标是收敛的．选取对角线构成的点列{x j, j}，则{x j, j}是{x k}的子列，且每个坐标都收敛．根据习题1.2.1的证明可知，S空间的点列收敛的充要条件是坐标收敛．故{x j, j}是收敛点列．所以，A是列紧的．(?) 我们只要证明，?n∈ +，A中的点的第n个坐标所构成的集合是有界集．若不然，设A中的点的第N个坐标所构成的集合是无界的．则存在A中的点列x k = (ξ1(k), ξ2(k), ..., ξn(k), ...) ( k = 1, 2, ... )，使得| ξN(k) | > k．显然，{ ξN(k) }无收敛子列，故{ x k }也无收敛子列，这与A列紧相矛盾．这样就完成了必要性的证明．1.3.8 设(X, ρ)是度量空间，M是X中的列紧集，映射f : X →M满足ρ( f (x1), f (x2)) < ρ( x1, x2 )(?x1, x2∈M, x1≠x2)．求证：f在X中存在唯一的不动点．证明：(1) 首先证明cl(M)是紧集．为此只要证明cl(M)列紧即可．设{ x n }是cl(M)中的点列，则存在M中的点列{ y n }使得ρ( x n, y n) < 1/n．因M列紧，故{ y n }有收敛子列{ y n(k)}，设y n(k) →u∈cl(M)．显然{ x n(k)}也是收敛的，并且也收敛于u∈cl(M)．所以cl(M)是自列紧的，因而是紧集．(2) 令g(x) = ρ( x, f (x))，则g是X上的连续函数．事实上，由ρ( f (x1),f (x2)) < ρ( x1, x2 )可知f : X →M是连续的，因而g也连续．由习题1.3.2知存在x0∈cl(M)，使得g(x0) = inf {ρ( x, f (x)) | x∈cl(M) }．(3) 若g(x0) > 0，则ρ( x0, f (x0)) > 0，即x0≠f (x0)．故ρ( x0, f (x0)) = g(x0) ≤g( f (x0)) = ρ( f (x0), f ( f (x0))) < ρ( x0, f (x0))，矛盾．所以，必有g(x0) = 0，即ρ( x0, f (x0)) = 0，因此x0就是f的不动点．1.3.9 设(M, ρ)是一个紧距离空间，又E?C(M)，E中的函数一致有界并且满足下列的H?lder条件：| x(t1) -x(t2) | ≤Cρ(t1, t2)α(?x∈E，?t1, t2∈M )，其中0 < α≤ 1，C > 0．求证：E在C(M)中是列紧集．证明：由H?lder条件易知E是等度连续的．又E中的函数一致有界，由Arzela-Ascoli定理知E是C(M)中的列紧集．[第3节完] 泛函分析题1_4线性赋范空间p391.4.1 在2维空间 2中，对每一点z = (x, y)，令|| z ||1 = | x | + | y |；|| z ||2 = ( x 2 + y 2 )1/2；|| z ||3 = max(| x |, | y |)；|| z ||4 = ( x 4 + y 4 )1/4；(1) 求证|| · ||i( i = 1, 2, 3, 4 )都是 2的范数．(2) 画出( 2, || · ||i )( i = 1, 2, 3, 4 )各空间中单位球面图形．(3) 在 2中取定三点O = (0, 0)，A = (1, 0)，B= (0, 1)．试在上述四种不同的范数下求出?OAB三边的长度．证明：(1) 正定性和齐次性都是明显的，我们只证明三角不等式．设z = (x, y), w = (u, v)∈ 2，s = z + w= (x + u, y + v )，|| z||1 + || w||1 = (| x | + | y |) + (| u | + | v |) = (| x | + | u |) + (| y | + | v |)≥ | x + u | + | y + v | = || z+ w||1．( || z||2 + || w||2 )2 = ( ( x 2 + y 2 )1/2 + ( u 2 + v 2 )1/2 )2= ( x 2 + y 2 ) + ( u 2 + v 2 ) + 2(( x 2 + y 2 )( u 2 + v 2 ))1/2 ≥ ( x 2 + u 2 ) + ( y 2 + v 2 ) + 2( x u+ y v )= ( x + u )2 + ( y + v)2 = ( || z+ w||2 )2．故|| z||2 + || w||2 ≥ || z+ w||2．|| z||3 + || w||3 = max(| x |, | y |) + max(| u |, | v |)≥ max(| x | + | u |, | y | + | v |) ≥ max(| x + u |, | y + v |) = || z+ w||3．|| ·||4我没辙了，没找到简单的办法验证，权且用我们以前学的Minkowski不等式(离散的情况，用H?lder不等式的离散情况来证明)，可直接得到．(2) 不画图了，大家自己画吧．(3) OA = (1, 0)，OB = (0, 1)，AB = (- 1, 1)，直接计算它们的范数：|| OA||1 = 1，|| OB||1 = 1，|| AB||1 = 2；|| OA||2 = 1，|| OB||2 = 1，|| AB||2 = 21/2；|| OA||3 = 1，|| OB||3 = 1，|| AB||3 = 1；|| OA||4 = 1，|| OB||4 = 1，|| AB||4 = 21/4．1.4.2 设c[0, 1]表示(0, 1]上连续且有界的函数x(t)全体．?x∈c[0, 1]，令|| x || = sup{| x(t) | | 0 < t≤ 1}．求证：(1) || ·||是c[0, 1]空间上的范数．(2) l∞与c[0, 1]的一个子空间是等距同构的．证明：(1) 正定性和齐次性都是明显的，我们只证明三角不等式．|| x || = sup{| x(t) | | 0 < t≤ 1}．|| x || + || y || = sup{| x(t) | | 0 < t≤ 1} + sup{| y(t) | | 0 < t≤ 1}≥ sup{| x(t) + y(t) | 0 < t≤ 1} = || x + y ||．所以|| ·||是c[0, 1]空间上的范数．(2) 任意取定(0, 1]中的一个单调递减列{a k }，满足(i) a1 = 1；(ii) lim k→∞a k = 0．显然，在每个[a k + 1, a k]上为线性函数的f∈c[0, 1]是存在的．设X = { f∈c[0, 1] | f在每个[a k + 1, a k]上为线性函数}．容易验证X是c[0, 1]的子空间．定义? : X →l∞，f #? ( f ) = ( f (a1), f (a2), ...)．则? : X →l∞是线性双射，且|| ? ( f ) ||∞= sup k ≥ 1 | f (a k) | = sup0 < t≤ 1 { | f (t ) | } = || f ||．所以，? : X →l∞是等距同构．因此，l∞与c[0, 1]的一个子空间是等距同构的．1.4.3 在C1[a, b]中，令|| f ||1 = (?[a, b] ( | f(x) |2 + | f’(x) |2) dx )1/2 (?f∈C1[a, b])．(1) 求证：|| · ||1是C1[a, b]上的范数．(2) 问(C1[a, b], || · ||1)是否完备？证明：(1) 正定性和齐次性都是明显的，和前面的习题一样，只验证三角不等式．我们先来证明一个比较一般的结果：若线性空间X上的非负实值函数p, q都满足三角不等式：p(x) + p(y) ≥p(x +y)，q(x) + q(y) ≥q(x +y)，?x, y∈X；则函数h = ( p2 + q2 )1/2也满足三角不等式．事实上，?x, y∈X，由Minkowski不等式，我们有h(x) + h(y) = ( p(x)2 + q(x)2 )1/2 + ( p(y)2 + q(y)2 )1/2≥ (( p(x)+ p(y))2 + ( q(x) + q(y))2 )1/2 ≥ ( p(x + y)2 + q(x + y)2 )1/2 = h(x + y)．回到本题：若令p( f ) = (?[a, b] | f(x) |2dx )1/2，q( f ) = (?[a, b] | f’(x) |2dx )1/2，则( p( f ) + p( g ))2 = ((?[a, b] | f(x) |2dx )1/2 + (?[a, b] | g(x) |2dx )1/2)2= ?[a, b] | f(x) |2dx + 2(?[a, b] | f(x) |2dx )1/2 · (?[a, b] | g(x)|2dx )1/2 + ?[a, b] | g(x) |2dx≥?[a, b] | f(x)|2dx + 2 ?[a, b] | f(x) | · | g(x)| dx + ?[a, b] | g(x)|2dx = ?[a, b] ( | f(x) | + | g(x)| )2dx ≥?[a, b] ( | f(x) + g(x)| )2dx = ( p( f + g ))2．所以有p( f ) + p( g ) ≥p( f + g )．特别地，p( f’) + p( g’) ≥p( f’+ g’)，即q( f ) + q( g ) ≥q( f + g )．因此，线性空间C1[a, b]上的非负实值函数p, q都满足三角不等式．根据开始证明的结论，|| · ||1也满足三角不等式．所以，|| · ||1是C1[a, b]上的范数．(2) 在C1[- 1, 1]中，令f n(x) = (x2 + 1/n2 )1/2 ( ?x∈[- 1, 1] )．则f’n(x) = 2x (x2 + 1/n2 )-1/2 ( ?x∈[- 1, 1] )．显然，f n(x)几乎处处收敛于| x |，f’n(x)几乎处处收敛于2sign( x )．因此，f n(x)依测度收敛于| x |，f’n(x)依测度收敛于2sign( x )．则f’n(x) = 2x (x2 + 1/n2 )-1/2 ( ?x∈[- 1, 1] )．显然，f n(x)几乎处处收敛于| x |，f’n(x)几乎处处收敛于2sign( x )．因此，f n(x)依测度收敛于| x |，f’n(x)依测度收敛于2sign( x )．故在L2[- 1, 1]中，f n(x) → | x |，f’n(x) → 2sign( x )．因此，它们都是L2[- 1, 1]中的基本列，故[- 1, 1] | f n(x) -f m(x) |2 dx → 0(m, n→∞)；[- 1, 1] | f’n(x) -f m’(x) |2 dx → 0(m, n→∞)．故|| f n-f m ||1 = (?[- 1, 1] ( | f n(x) -f m(x) |2 + | f’n(x) -f m’(x) |2 ) dx )1/2→ 0 (m, n→∞)．即{ f n }是C1[- 1, 1]中的基本列．下面我们证明{ f n }不是C1[- 1, 1]中的收敛列．若不然，设{ f n }在C1[- 1, 1]中的收敛于f∈C1[- 1, 1]．因|| f n-f ||1 = (?[- 1, 1] ( | f n(x) -f(x) |2 + | f’n(x) -f’(x) |2 ) dx )1/2≥ (?[- 1, 1] | f n(x) -f(x) |2dx )1/2，故在L2[- 1, 1]中，f n(x) →f．而在前面已说明L2[- 1, 1]中，f n(x) → | x |；由L2[- 1, 1]中极限的唯一性以及f的连续性，知f(x) = | x |．这样就得到f?C1[- 1, 1]，矛盾．所以，{ f n }不是C1[- 1, 1]中的收敛列．这说明C1[- 1, 1]不是完备的．对一般的C1[a, b]，只要令f n(x) = (x - (a + b )/2)2 + 1/n2 )1/2( ?x∈[a, b] )就可以做同样的讨论，就可以证明C1[a, b]不是完备空间．1.4.4 在C[0, 1]中，对每个f∈C[0, 1]，令|| f ||1 = (?[0, 1] | f(x) |2dx )1/2，|| f ||2 = (?[0, 1] ( 1 + x) | f(x) |2dx )1/2．求证：|| · ||1和|| · ||2是C[0, 1]中的两个等价范数．证明：(1) 在习题1.4.3的证明中已经包含了|| · ||1是C[0, 1]中的范数的证明．下面我们证明|| · ||2是C[0, 1]中的范数，我们仍然只要验证三角不等式．|| f ||2 + || g ||2 = (?[0, 1] ( 1 + x) | f(x) |2dx )1/2 + (?[0, 1] ( 1 + x) | g(x) |2dx )1/2= || (1 + x)1/2f(x) ||1 + || (1 + x)1/2g(x) ||1≥ || (1 + x)1/2f(x) + (1 + x)1/2g(x) ||1= || (1 + x)1/2 ( f(x) + g(x) ) ||1≥ (?[0, 1] (1 + x) | f(x) + g(x) |2dx )1/2= || f + g ||2．所以，|| · ||2也是C[0, 1]中的范数．(2) 我们来证明两个范数的等价性．?f∈C[0, 1]|| f ||1 = (?[0, 1] | f(x) |2dx )1/2 ≤ (?[0, 1] ( 1 + x) | f(x) |2dx )1/2 = || f ||2，|| f ||2 = (?[0, 1] ( 1 + x) | f(x) |2dx )1/2 ≤ 2 (?[0, 1] | f(x) |2dx )1/2 = 2 || f ||1．因此两个范数等价．1.4.5 设BC[0, ∞)表示[0, ∞)上连续且有界的函数f(x)全体，对每个f ∈BC[0, ∞)及a > 0，定义|| f ||a = (?[0, ∞) e-ax | f(x) |2dx )1/2．(1) 求证|| ·||a是BC[0, ∞)上的范数．(2) 若a, b > 0，a≠b，求证|| ·||a与|| ·||b作为BC[0, ∞)上的范数是不等价的．证明：(1) 依然只验证三角不等式．|| f ||a + || g ||a = (?[0, ∞) e-ax | f(x) |2dx )1/2 + (?[0, ∞) e-ax | g(x) |2dx )1/2= || e-ax/2f(x)||L2 + || e-ax/2g(x)||L2≤ || e-ax/2f(x)+ e-ax/2g(x)||L2= || e-ax/2 ( f(x)+ g(x))||L2= (?[0, ∞) e-ax | f(x)+ g(x) |2dx )1/2= || f + g ||a，所以|| ·||a是BC[0, ∞)上的范数．(2) 设f n(x)为[n, +∞)上的特征函数．则f n∈BC[0, ∞)，且|| f n||a = (?[0, ∞) e-ax | f n(x) |2dx )1/2 = (?[n, ∞) e-ax dx )1/2 = ((1/a)e-an)1/2．同理，|| f n||b = ((1/b)e-bn)1/2．故若a < b，则|| f n||a/|| f n||b = (b/a)1/2e-(b -a)n/2→ +∞ (n→+∞)．因此|| ·||a与|| ·||b作为BC[0, ∞)上的范数是不等价的．1.4.6 设X1, X2是两个B*空间，x1∈X1和x2∈X2的序对(x1, x2)全体构成空间X = X1?X2，并赋予范数|| x || = max{ || x1 ||1, || x2 ||2 }，其中x = (x1, x2)，x1∈X1，x2∈X2，|| · ||1和|| ·||2分别是X1和X2的范数．求证：如果X1, X2是B空间，那么X也是B空间．证明：(1) 先验证|| · ||的三角不等式．设x = (x1, x2), y = (y1, y2)∈X1?X2，则|| x + y || = || (x1 + y1, x2 + y2) || = max{ || x1 + y1 ||1, || x2 + y2 ||2 }≤ max{ || x1 ||1 + || y1 ||1, || x2 ||2 + || y2 ||2 }≤ max{ || x1 ||1, || x2 ||2 } + max{ || y1 ||1, || y2 ||2 }= || (x1, x2) || + || (y1, y2) ||= || x || + || y ||，而|| · ||的正定性和齐次性是显然的，所以，|| · ||是X1?X2的范数．(2) 设X1, X2是B空间，我们来证明X也是B空间．设x(n) = (x1(n), x2(n))是X = X1?X2中的基本列，则|| x(n) -x(m) || = max{ || x1(n) -x1(m) ||1, || x2(n) -x2(m)||2 } ≥ || x1(n) -x1(m) ||1，故{x1(n)}是X1中的基本列，同理，{x2(n)}是X2中的基本列．因X1, X2是B空间，故{x1(n)}和{x2(n)}分别是X1, X2中的收敛列．设x1(n) →x1∈X1，x2(n) →x2∈X2，令x = (x1, x2)．则|| x(n) -x || = max{ || x1(n) -x1 ||1, || x2(n) -x2 ||2 }≤ || x1(n) -x1 ||1 + || x2(n) -x2 ||2→ 0 (n→∞)．所以，|| x(n) -x ||→ 0 (n→∞)．即{ x(n) }为X = X1?X2中的收敛列．所以X = X1?X2也是B空间．1.4.7 设X是B*空间．求证：X是B空间，必须且只须对?{x n}?X，∑n≥ 1 || x n || < +∞?∑n≥ 1x n 收敛．证明：(?) ?{x n}?X，记S n = ∑1 ≤j≤n x j，B n = ∑1 ≤j≤n || x n ||，则|| S n + p-S n || = || ∑1 ≤j≤n + p x j -∑1 ≤j≤n x j ||= || ∑n +1 ≤j≤n + p x j ||≤∑n +1 ≤j≤n + p || x j ||= B n + p-B n → 0，(n→∞)．故{ S n }为X中的Cauchy列．由X完备，故{ S n }为X中的收敛列，即∑n≥ 1x n 收敛．(?) 反证法．若(X, ρ)不完备，设(Y, d )为(X, ρ)的一个完备化．不妨设(X, ρ)是(Y, d )的子空间，则存在y∈Y \ X．因cl( X ) = Y，故?n∈ +，存在x n∈X，使得d(x n, y) < 1/2n．则ρ(x n, x m) = d(x n, x m) ≤d(x n, y) + d(x m, y) ≤ 1/2n+ 1/2m → 0，因此{x n}是X中的Cauchy列，但不是收敛列．令z n = x n+1-x n，S n = ∑1 ≤j≤n z j；则z n, S n∈X．因|| z n || = || x n+1-x n || = ρ(x n+1, x n) ≤d(x n+1, y) + d(x n+1, y) ≤ 1/2n+1+ 1/2n < 1/2n - 1，故∑n≥ 1 || z n || < +∞．而S n = ∑1 ≤j≤n z j = ∑1 ≤j≤n ( x j+1-x j ) = x n+1-x1；故∑n≥ 1z n 在中不收敛．矛盾．1.4.8 记[a, b]上次数不超过n的多项式全体为n．求证：?f(x)∈C[a, b]，存在P0(x)∈ n，使得max a ≤x≤b| f(x) –P0(x) | = min{ max a ≤x≤b| f(x) –P(x) | | P∈ n }．证明：注意到 n是B*空间C[a, b]中的n+1维子空间．{1, x, x2, ..., x n}是 n中的一个向量组，把它看成C[a, b]中的一个有限向量组．根据定理p35, 1.4.23，对任意?f(x)∈C[a, b]，存在最佳逼近系数{λ0, λ1, ..., λn}，使得|| f(x) –∑0 ≤j≤n λj x j || = min{ || f(x) –∑0 ≤j≤n a j x j || | (a0, a1, ..., a n)∈ n+1}．令P0(x) = ∑0 ≤j≤n λj x j 就得到要证明的结论．1.4.9 在 2中，对?x = (x1, x2)∈ 2，定义范数|| x || = max(| x1 |, | x2 |)，并设|| x0–λ e1 ||．e1 = (1, 0)，x0 = (0, 1)．求a∈ 适合|| x0–a e1 || = minλ∈并问这样的a是否唯一？请对结果作出几何解释．解：g(λ) = || x0–λ e1 || = || (0, 1) –λ(1, 0)|| = || (–λ, 1)|| = max(| λ |, 1) ≥ 1，故g(λ) 当| λ| ≤ 1时取得最小值1．所以a = 0满足要求．显然满足要求的a不是唯一的．从几何上看就是某线段上的点到某定点的距离都是1．1.4.10 求证范数的严格凸性等价于下列条件：|| x + y || = || x || + || y || ( ?x≠θ, y≠θ) ?x = c y ( c > 0)．证明：(?) 设范数是严格凸的，若x, y ≠θ满足|| x + y || = || x || + || y ||，事实上，我们总有|| (x/|| x ||) || = || (y/|| y ||) || = 1．因x, y ≠θ，故|| x || + || y || > 0，所以|| x + y || ≠ 0．于是|| x ||/|| x + y || + || y ||/|| x + y || = 1．假若x/|| x || ≠y/|| y ||，由严格凸性，得到|| (|| x ||/|| x + y ||)(x/|| x ||) + (|| y ||/|| x + y ||)(y/|| y ||) || < 1，即|| (( x + y )/|| x + y ||) || < 1，矛盾．因此必然有x/|| x || = y/|| y ||，即x = (|| x ||/|| y ||) y．(?) 设?x, y ≠θ，|| x + y || = || x || + || y ||蕴涵x = c y ( c > 0)．下面证明范数是严格凸的．设x≠y，且|| x || = || y || = 1，又设α, β∈(0, 1)，且α + β= 1．我们知道|| α x + β y || ≤ || α x || + || β y || = α || x || + β|| y || = α + β= 1．假若|| α x + β y || = 1，根据我们的条件，就得到α x = c (β y)，其中c > 0．那么，就有|| α x || = || c (β y) ||，而|| x || = || y || = 1，所以α= c β；故x = y，这就与x≠y相矛盾．所以必然有|| α x + β y || < 1，即范数是严格凸的．1.4.11 设X是线性赋范空间，函数? : X → 1称为凸的，如果不等式( λ x + (1 -λ) y ) ≤λ?( x ) + (1 -λ)?( y ) ( ? 0 ≤λ≤ 1)成立．求证凸函数的局部极小值必然是全空间的最小值．证明：设x0是凸函数?的一个局部极小点．如果存在x∈X，使得?( x ) < ?( x0)，则? t ∈(0, 1)，( t x + (1 -t ) x0) ≤t ?( x ) + (1 -t )?( x0) < t ?( x0) + (1 -t )?( x0) = ?( x0)．而对x0的任意邻域U，都存在t ∈(0, 1)，使得t x + (1 -t ) x0∈U．这就与x0是局部极小点相矛盾．因此?x∈X，都有?( x0) ≤?( x )，即x0是?的最小点．1.4.12 设(X, || · ||)是一线性赋范空间，M是X的有限维子空间，{e1, e2, ..., e n}是M的一组基，给定g∈X，引进函数F : n → 1．对?c = (c1, c2, ..., c n)∈ n，规定F(c) = F(c1, c2, ..., c n) = || ∑1 ≤i≤n c i e i-g ||．(1) 求证F是一个凸函数；(2) 若F的最小值点是c = (c1, c2, ..., c n)，求证f = ∑1 ≤i≤n c ie i给出g在M中的最佳逼近元．证明：(1) 设c = (c1, c2, ..., c n), d = (d1, d2, ..., d n)∈ n, λ∈[0, 1]，则F(λ c + ( 1 -λ) d ) = || ∑1 ≤i≤n ( λ c i + ( 1 -λ) d i ) e i-g || = || λ∑1 ≤i≤n c i e i + ( 1 -λ) ∑1 ≤i≤n d i e i- (λ g+ ( 1 -λ)g )|| = || λ(∑1 ≤i≤n c i e i -g) + ( 1 -λ) ( ∑1 ≤i≤n d i e i-g )||≤λ|| ∑1 ≤i≤n c i e i -g || + ( 1 -λ) || ∑1 ≤i≤n d i e i-g ||= λ F(c)+ ( 1 -λ)F(d)，故F是一个凸函数．(2) 因为{e1, e2, ..., e n}是M的一组基，故M中的每个元h都可表示为h = ∑1 ≤i≤n d i e i，其中d = (d1, d2, ..., d n)∈ n．因为F(c) ≤F(d)，故|| f-g || = F(c) ≤F(d) = || h-g ||．那么f就是g在M中的最佳逼近元．1.4.13 设X是B*空间，X0是X的线性子空间，假定?c∈(0, 1)使得?y∈X，有inf { || y–x || | x ∈X0 } ≤c || y ||．求证：X0在X中稠密．证明：设y∈X，?ε > 0，x1∈X0，s.t. || y–x1 || < c || y || + ε /4．x2∈X0，s.t. || (y–x1) –x2 || < c || y–x1 || + ε /8．x3∈X0，s.t. || (y–x1 –x2 ) –x3 || < c || y–x1 –x2 || + ε /16．如此下去，可得到一个X0中的点列{ x n }，满足|| y–∑1 ≤j≤n +1x j|| < c || y–∑1 ≤j≤n x j|| + ε /2n + 2(?n∈ +)．那么，我们可以用数学归纳法证明|| y–∑1 ≤j≤n x j|| < c n || y || + ε (∑1 ≤j≤n 1/2j + 1)．当n = 1时，|| y–x1 || < c || y || + ε /4．结论成立．当n = 2时，|| (y–x1) –x2 || < c || y–x1 || + ε /8< c (c || y || + ε /4) + ε /8 < c 2 || y || + ε (1/4 + 1/8)，结论成立．当n≥ 3时，若|| y–∑1 ≤j≤n x j|| < c n || y || + ε (∑1 ≤j≤n 1/2j + 1)成立，则|| y–∑1 ≤j≤n +1x j|| < c || y–∑1 ≤j≤n x j|| + ε /2n + 2< c (c n || y || + ε (∑1 ≤j≤n 1/2j + 1)) + ε /2n + 2< c n+1 || y || + ε (∑1 ≤j≤n 1/2j + 1)) + ε /2n + 2< c n+1 || y || + ε (∑1 ≤j≤n+ 11/2j + 1))，因此结论也成立．由数学归纳法原理，?n∈ +，|| y–∑1 ≤j≤n x j|| < c n || y || + ε (∑1 ≤j≤n 1/2j + 1)．因为c∈(0, 1)，故存在N∈ +，使得c N || y || < ε /2．令x = ∑1 ≤j≤N x j，则x∈X0．且|| y–x || < ε /2 + ε (∑1 ≤j≤N 1/2j + 1) < ε．所以，X0在X中稠密．[张峰同学的证明] 反证法．若不然，则cl(X0)是X的真闭线性子空间．用Riesz引理，存在y∈X，使得|| y || = 1，且inf { || y–x || | x ∈ cl(X0)} > c．故对此y∈X，有inf { || y–x || | x ∈X0 } > c || y ||，矛盾．1.4.14 设C0表示以0为极限的实数全体，并在C0中赋以范数|| x || = max n≥1| ξn |，( ?x = (ξ1, ξ2, ..., ξn, ...)∈C0 )．又设M = {x = (ξ1, ξ2, ..., ξn, ...)∈C0 | ∑n ≥1 ξn/2n = 0}．(1) 求证：M是C0的闭线性子空间．(2) 设x0= (2, 0, 0, ...)，求证：inf z ∈M || x0–z || = 1，但?y∈M，有|| x0–y || > 1．证明：(1) 显然M ≠?，容易直接验证M是C0的线性子空间．若x k = (ξ1(k), ξ2(k), ..., ξn(k), ...)为M中的点列，且x k→x = (ξ1, ξ2, ..., ξn, ...)∈C0．则?ε > 0，存在N∈ +，使得?k > N，|| x k -x || < ε．此时，?n∈ +，有|ξn -ξn(k)| ≤ max n≥1| ξn -ξn(k) | = || x k -x || < ε．| ∑n ≥1 ξn/2n | = | ∑n ≥1 ξn/2n-∑n ≥1 ξn(k)/2n | = | ∑n ≥1 (ξn -ξn(k))/2n |≤∑n ≥1 |ξn -ξn(k)|/2n≤∑n ≥1 ε/2n = ε．所以，∑n ≥1 ξn/2n = 0，即x = (ξ1, ξ2, ..., ξn, ...)∈M．所以M是C0的闭线性子空间．(2) x0= (2, 0, 0, ...)，?z = (ξ1, ξ2, ..., ξn, ...)∈M，|| x0–z || = max{| 2 -ξ1 |, | ξ2 |, | ξ3 |, ... }．如果| 2 -ξ1 | > 1，则|| x0–z || > 1．如果| 2 -ξ1 | ≤ 1，则| ξ1 | ≥ 1，我们断言{| ξ2 |, | ξ3 |, ... }中至少有一个大于1者．否则，假若它们都不超1，因为ξn → 0 (n→∞)，故它们不能全为1．由∑n ≥1 ξn/2n = 0知| ξ1 |/2 = | ∑n ≥2 ξn/2n | ≤∑n ≥2 | ξn | /2n < ∑n ≥2 1/2n = 1/2，这样得到| ξ1 | < 1，矛盾．故{| ξ2 |, | ξ3 |, ... }中至少有一个大于1者．因此也有|| x0–z || > 1．综上所述，但?y∈M，有|| x0–y || > 1．由此，立即知道inf z ∈M || x0–z || ≥ 1．下面证明inf z ∈M || x0–z || ≤ 1．n∈ +，令z n= (1 - 1/2n, -1, -1, ..., -1, 0, 0, ...)．( z n从第2个坐标开始有连续的n个-1，后面全部是0 )，则(1 - 1/2n)/2 - 1/4 - 1/8 - ... - 1/2n + 1 = 0，因此z n∈M．此时，|| x0–z n || = max{| 1 + 1/2n|, | 1/4|, | 1/8|, ... } = 1 + 1/2n．故inf z ∈M || x0–z || ≥ inf n || x0–z n || = inf n (1 + 1/2n ) = 1．所以，inf z ∈M || x0–z || = 1．1.4.15 设X是B*空间，M是X的有限维真子空间，求证：?y∈X，|| y|| = 1，使得|| y–x || ≥ 1 ( ?x ∈M )．证明：取定z∈X \ M，令Y = span{z} + M．记S = { y∈Y | || y || = 1 }．则M是Y的真闭子空间，而S是Y中的单位球面．由Riesz引理，?n∈ +，存在y n∈S，使得d( y n, M ) ≥ 1 - 1/n．因为Y也是有限维的，故其中的单位球面为自列紧集．存在{y n}的收敛子列．不妨设y n(k) →y∈S．则d( y n(k), M ) ≥ 1 - 1/n(k)，故有d( y, M ) ≥ 1．即|| y–x || ≥ 1 ( ?x ∈M )．1.4.16 若f是定义在区间[0, 1]上的复值函数，定义ωδ( f ) = sup{| f (x) –f (y) | | ?x, y∈[0, 1], | x–y | ≤δ}．如果0< α≤ 1对应的Lipschitz空间Lipα，由满足|| f || = | f(0) | + supδ > 0{δ–αωδ( f )} < +∞的一切f组成，并且以|| f ||为模．又设lipα = { f∈Lipα| lim δ→ 0 δ–αωδ( f ) = 0}．求证Lipα是B空间，而且lipα是Lipα的闭子空间．证明：(1) 显然，C1[0, 1]?Lipα，因此Lipα不空．对区间[0, 1]上的复值函数f, g，?λ∈ ，我们有ωδ( f + g ) = sup{| f (x) + g (x) – f (y) –g (y) | | ?x, y∈[0, 1], | x–y | ≤δ}≤ sup{| f (x) – f (y) | + | g (x) –g (y) | | ?x, y∈[0, 1], | x–y | ≤δ}≤ωδ( f ) + ωδ( g )．ωδ( λ f ) = sup{|λ f (x) –λ f (y) | | ?x, y∈[0, 1], | x–y | ≤δ}= | λ| sup{| f (x) –f (y) | | ?x, y∈[0, 1], | x–y | ≤δ}= | λ| ·ωδ( f )．若f, g∈Lipα，λ∈ ，则|| f + g || = | f(0) + g(0) | + supδ > 0{δ–αωδ( f + g ) }≤ | f(0) | + | g(0) | + supδ > 0{δ–α(ωδ( f ) + ωδ( g )) }= | f(0) | + | g(0) | + supδ > 0{δ–αωδ( f ) + δ–αωδ( g ) }≤ | f(0) | + | g(0) | + supδ > 0{δ–αωδ( f ) }+ supδ > 0{ δ–αωδ( g ) }= || f || + || g || < +∞．|| λ f || = | λ f(0) | + supδ > 0{δ–αωδ( λ f )}= | λ| · | f(0) | + | λ| · supδ > 0{δ–αωδ( f )}= | λ| · || f || < +∞．因此，f + g, λ f∈Lipα，且上述两个不等式表明|| · ||有齐次性和三角不等式．显然，|| f || ≥ 0．当|| f || = 0时，| f(0) | + supδ > 0{δ–αωδ( f )} = 0，意味着f(0) = 0，且ωδ( f ) = 0(?δ> 0)．而ωδ( f ) = 0(?δ> 0)则意味着f为常值．所以，f = 0．即|| · ||有正定性．综上所述，Lipα是B*空间．(2) 我们首先证明集合Lipα?C[0, 1]．f∈Lipα，?x, y∈[0, 1]，x ≠y，记δ = | x -y |．则| f (x) –f (y) | ≤ωδ( f )．而δ–αωδ( f ) ≤ supδ > 0{δ–αωδ( f n-f m) } ≤ || f ||，所以，| f (x) – f (y) | ≤ || f || δα= || f || · | x -y |α，故f∈C[0, 1]．我们再证明，?f∈Lipα，|| f ||C≤ || f ||，其中|| ·||C是C[0, 1]范数．事实上，?x∈[0, 1]，| f (x) | ≤ | f (0) | + | f (x) – f (0) |，故|| f ||C = max x∈[0, 1] | f (x) | ≤ | f (0) | + max x∈[0, 1] | f (x) –f (0) |≤ | f (0) | + sup x∈(0, 1] | f (x) –f (0) |/| x |α≤ | f (0) | + sup x∈(0, 1] { δ–αωδ( f ) } ≤ || f ||．这说明，如果{ f n }是Lipα中的基本列，则它也必是C[0, 1]中的基本列．而C[0, 1]是完备的，故存在f∈C[0, 1]，使得{ f n }一致收敛于f．而{ f n }作为Lipα中的基本列，有|| f n-f m || = | f n(0) -f m(0) | + supδ > 0{δ–αωδ( f n-f m) } → 0 (n, m→∞)，因此?ε > 0，?N∈ +，使得?n, m > N，有| f n(0) -f m(0) | + supδ > 0{δ–αωδ( f n-f m) } < ε．因此supδ > 0{δ–αωδ( f n-f m) } < ε．故?δ > 0，ωδ( f n-f m) < εδα．即?x, y∈[0, 1]，| x -y | ≤δ，都有| ( f n(x) -f m(x)) - ( f n(y) -f m(y)) | < εδα．令m→∞，得到| ( f n(x) -f(x)) - ( f n(y) -f(y)) | ≤εδα．因此，sup {| ( f n(x) -f(x)) - ( f n(y) -f(y)) | | x, y∈[0, 1]，| x -y | ≤δ}≤εδα．即?δ > 0，ωδ( f n-f ) ≤εδα．故supδ > 0{δ–αωδ( f n-f ) } ≤ε．同样地，对不等式| f n(0) -f m(0) | < ε令m→∞，就得到| f n(0) -f(0) | ≤ε．所以，| f n(0) -f(0) | + supδ > 0{δ–αωδ( f n-f ) } ≤ 2ε．这说明f n-f∈Lipα．而f n∈Lipα，故f = ( f -f n ) + f n∈Lipα．而前面的式子也表明|| f -f n || ≤ 2ε．因此|| f n-f || → 0 (n→∞)，即{ f n }为Lipα中的收敛列．所以，Lipα是Banach空间．(3) 记lipα = { f∈Lipα| lim δ→ 0 δ–αωδ( f ) = 0 }．f, g∈lipα，?λ∈ ，我们有δ–αωδ( f + g ) ≤δ–α(ωδ( f ) + ωδ( g ) ) = δ–αωδ( f ) + δ–αωδ( g ) → 0 (δ→ 0)．δ–αωδ( λ f ) = | λ| ·δ–αωδ( f ) → 0 (δ→ 0)．故f + g, λ f∈lipα，因此，lipα是Lipα的线性子空间．设{ f n }是lipα中的序列，且f n→f∈Lipα(n→∞)．则{ f n }一致收敛于f．ε > 0，存在N∈ +，使得|| f N →f || < ε /2．故有supδ > 0{δ–αωδ( f N-f ) } < ε /2．因为lim δ→ 0 δ–αωδ( f N) = 0，所以，?? > 0，使得?δ∈(0, ?)，有δ–αωδ( f N) < ε /2．此时我们有δ–αωδ( f ) ≤δ–α(ωδ( f N) + ωδ( f -f N))= δ–αωδ( f N) + δ–αωδ( f -f N)< ε /2 + supδ > 0{δ–αωδ( f N-f ) } < ε．所以，lim δ→ 0 δ–αωδ( f ) = 0，即f∈lipα．所以lipα是Lipα的闭子空间．1.4.17 (商空间) 设X是线性赋范空间，X0是X的闭线性子空间，将X中的向量分类，凡是适合x’-x’’∈X0的两个向量x’, x’’归于同一类，称其为等价类，把一个等价类看成一个新的向量，这种向量的全体组成的集合为X/X0表示，并称其为商空间．下列是关于商空间的命题．(1) 设[ y ]∈X/X0，x∈X，求证：x∈[ y ]的充分必要条件是[ y ] = x + X0．证明：设x’, x’’∈X，若它们归于同一类，则记为x’～x’’．我们用[ x ]表示x所在的等价类（大家注意，题目形式已经作了相应的修改）．(?) 若x∈[ y ]，则x～y．u ∈[ y ]，u～y，故u～x，即u –x∈X0．因此u ∈x + X0．所以[ y ] ?x + X0．反过来，?u ∈x + X0，则u～x，故u～y．因此u ∈[ y ]．所以x + X0 ? [ y ]．所以[ y ] = x + X0．(?) 若[ y ] = x + X0，则y –x∈X0，即y～x．从而x∈[ y ]．(2) 在X/X0中定义加法与数乘如下：[ x ] + [ y ] = x + y + X0(?[ x ], [ y ] ∈X/X0 )λ[ x ] = λ x + X0(?[ x ]∈X/X0 , ?λ∈ )其中x和y分别表示属于等价类[ x ]和[ y ]的任一元素．又规定范数|| [ x ] ||0 = inf z∈[ x ] || z || ( ?[ x ]∈X/X0 )求证：(X/X0, || · ||0)是一个B*空间．证明：第(1)部分说明了[ x ] = x + X0．容易看出加法与乘法的定义是合理的．进一步可以证明X/X0 构成数域上的线性空间，且其零元为[ θ] = X0．下面证明|| · ||0是X/X0 上的范数．显然，?[ x ]∈X/X0，|| [ x ] ||0≥ 0．若[ x ] = [ θ] = X0，则|| [ x ] ||0 = 0．若|| [ x ] ||0 = 0，则inf z∈[ x ] || z || = 0．存在z n∈[ x ]使得|| z n || → 0，即z n→θ (n→∞)．那么，x-z n∈X0，x-z n→x (n→∞)，而X0是闭集，故x∈X0．所以x～θ，即[ x ] = X0．因此|| · ||0有正定性．[ x ]∈X/X0，?λ∈ ，|| λ[ x ]||0 = || [ λ x ] ||0 = inf y∈[ x ] || λ y || = inf y∈[ x ] | λ| · || y ||= | λ| · inf y∈[ x ] || y || = | λ| · ||[ x ]||0．因此|| · ||0有齐次性．[ x ], [ y ]∈X/X0，|| [ x ] + [ y ] ||0 = inf z∈[ x ] + [ y ] || z || = inf u∈[ x ], v∈[ y ] || u + v ||≤ inf u∈[ x ], v∈[ y ] { || u || + || v || } ≤ inf u∈[ x ] { inf v∈[ y ] { || u || + || v ||} }≤ inf u∈[ x ] { inf v∈[ y ] { || u || + || v ||} } = inf u∈[ x ] { || u || + inf v∈[ y ] || v || }= inf u∈[ x ] || u || + inf v∈[ y ] || v || = || [ x ] ||0 + || [ y ] ||0．因此|| · ||0的三角不等式成立．所以，(X/X0, || · ||0)是一个B*空间．(3) 设[ x ]∈X/X0, 求证对?y∈[ x ]有inf { || y -z || | z∈X0 } = ||[ x ] ||0．证明：|| [ x ] ||0 = inf u∈[ x ] || u || = inf u∈[ y ] || u || = inf { || u || | u∈y + X0 }= inf { || y + v || | v∈X0 } = inf { || y -z || | z∈X0 }．(4) 定义映射? : X →X/X0为? (x) = [ x ] = x + X0(?x∈X )．求证?是线性连续映射．证明：?x, y∈X，?α, β∈ ，( α x + β y ) = [α x + β y ] = [α x ] + [ β y ] = α [ x ] + β[ y ] = α? (x) + β? (y)．|| ? (x) -? (y) ||0 = || [ x ] - [ y ] ||0 = || [ x-y ] ||0 = in f z∈[ x-y ] || z || ≤ || x-y ||．所以，?是线性连续映射．(5) ?[ x ]∈X/X0，求证?y∈X，使得? (y) = [ x ]，且|| y || ≤ 2|| [ x ] ||0．证明：因为|| [ x ] ||0 = inf z∈[ x ] || z ||，若|| [ x ] ||0 = 0，则由|| · ||0的正定性，知[ x ] = X0，取y = θ即满足要求．若|| [ x ] ||0≠ 0，则inf z∈[ x ] || z || = || [ x ] ||0 < 2 || [ x ] ||0，存在?y∈[ x ]，使得|| y || ≤ 2|| [ x ] ||0．此时显然有? (y) = [ x ] = [ y ]．(6) 设(X, || · ||)完备，求证(X/X0, || · ||0)也是完备的．证明：设{ [ x ]n }是X/X0中的基本列．为证明它是收敛列，只需证明它存在收敛子列．由基本列性质，可选出子列{ [ x ]n(k)}使得|| [ x ]n(k) - [ x ]n(k+1) ||0 ≤ 1/2k．故∑k ≥ 1 || [ x ]n(k) - [ x ]n(k+1) ||0 收敛．根据(5)，?k∈ +，?y k∈[ x ]n(k+1) - [ x ]n(k)，使得|| y k || ≤ 2|| [ x ]n(k+1) - [ x ]n(k) ||0．那么，∑k ≥ 1|| y k ||收敛．由X的完备性，s k = ∑ 1 ≤j ≤k y j是X中的收敛列．设其极限为s．由(5)中?的连续性，在X/X0中，?(s k) →?(s) ( k→∞ )．而?(s k) = ?( ∑ 1 ≤j ≤k y j ) = ∑ 1 ≤j ≤k ?( y j )= ∑ 1 ≤j ≤k ( [ x ]n(j+1) - [ x ]n(j)) = [ x ]n(k+1) - [ x ]n(1)．故{[ x ]n(k+1) - [ x ]n(1)}收敛，因而{[ x ]n(k)}是收敛列．因此X/X0中的基本列{ [ x ]n }存在收敛子列{[ x ]n(k)}，所以，{ [ x ]n }是X/X0中的收敛列．因此，(X/X0, || · ||0)是完备的．(7) 设X = C[0, 1]，X0 = { f∈X | f (0) = 0 }，求证：X/X0 ? ，其中记号“?”表示等距同构．证明：显然，X0是C[0, 1]中的线性子空间．记X0所确定的等价关系为～，则f～g ? f (0) = g (0)．定义Φ : X/X0 → ，Φ([ f ]) = f (0)．显然定义是合理的．f, g∈X，?α, β∈ ，Φ(α[ f ] + β[ g ]) = Φ([αf + β g ]) = (αf + β g )(0)= αf (0)+ β g (0) = αΦ([ f ])+ βΦ([ g ])．因此Φ是线性映射．因Φ(X0) = 0，故Φ是单射．而?c∈ ，若记所对应的常值函数为h c∈C[0, 1]，则Φ( [ h c] ) = c．故Φ是满射．综上所述，Φ : X/X0 → 是线性同构．f∈X，|| [ f ]||0 = inf g∈[ f ] { || g || } ≥ inf g∈[ f ] { | g (0) | }= inf g∈[ f ] { | f (0) | } = | f (0) | = | Φ([ f ]) |．另一方面，因为常值函数h f (0)∈[ f ]，故|| [ f ]||0 = inf g∈[ f ] { || g || } ≤ || h f (0) || = | f (0) | = | Φ([ f ]) |．所以，?f∈X，都有|| [ f ]||0 = | Φ([ f ]) |，因此Φ : X/X0 → 是等距同构．[第4节完] 泛函分析题1_5凸集与不动点p521.5.1 设X是B*空间，E是以θ为内点的真凸子集，P是由E产生的Minkowski 泛函，求证：(1) x∈int(E) ?P(x) < 1；(2) cl(int(E)) = cl(E)．证明：(1) (?) 若x∈int(E)，存在δ > 0，使得Bδ(x) ?E．注意到x + x/n→x ( n→∞ )，故存在N ∈ +，使得x + x/N ∈Bδ(x) ?E．即x/( N/( 1 + N ) ) ∈E．因此P(x) ≤N/( 1 + N ) < 1．(?) 若P(x) < 1．则存在a > 1，使得y = a x∈E．因θ∈int(E)，故存在δ > 0，使得Bδ(θ) ?E．令η = δ(a - 1)/a，?z∈Bη(x)，令w = (a z-y )/(a - 1)，则|| w || = || (a z-y )/(a - 1) || = || a z-y ||/(a - 1)= || a z-a x ||/(a - 1) = a || z-x ||/(a - 1) < aη/(a - 1) = δ．故w∈Bδ(θ) ?E．故z = ((a - 1)w + y )/a ∈E，因此，Bη(x) ?E．所以x∈int(E)．(2) 因int(E) = E，故有cl(int(E)) ? cl(E)．下面证明相反的包含关系．若x∈cl(E)，则?ε > 0，存在y∈E，使得|| x -y || < ε/2．因ny/(n + 1) →y ( n →∞ )．故存在N ∈ +，使得|| Ny/(N + 1) -y || < ε/2．令z = Ny/(N + 1)，则z∈E，且P(z) ≤N/(N + 1) < 1，由(1)知z∈int(E)．而|| z -x || ≤ || z -y || + || y -x || < ε/2 + ε/2 = ε．故x∈cl(int(E))，因此cl(E) ? cl(int(E))所以cl(int(E)) = cl(E)．1.5.2 求证在B空间中，列紧集的凸包是列紧集．证明：设A是B空间X中的列紧集，?ε > 0，存在A的有限ε /3网B．设B = {b1, b2, ..., b n}，M = max j{ || b j || }，取δ > 0，使得n δ M < ε /3．设[0, 1]分划D为0 = t0 < t1 < t2 < ... < t m = 1，使得max 1 ≤j ≤m {| t j–t j–1|} < δ．设?x∈co(A)，设x= λ1 a1 + λ2 a2+ ... + λ k a k，其中a j∈A，λ j > 0，∑ j λ j = 1．对每个j ≤k，存在b i( j )∈B使得|| a j-b i( j ) || < ε /3；令y= λ1 b i(1) + λ2 b i(2)+ ... + λ k b i(k)，则|| x - y || = || λ1 (a1 -b i(1)) + λ2 (a2 -b i(2))+ ... + λ k (a k-b i(k))||，≤λ1 · || a1 -b i(1) || + λ2 · || a2 -b i(2) || + ... + λ k · || a k-b i(k) ||≤ ( λ1 + λ2 + ... + λ k ) · (ε /2) = ε /3．将y= λ1 b i(1) + λ2 b i(2)+ ... + λ k b i(k)中的那些含有相同b j 的项合并起来，于是，y可表示为y= μ1 b1 + μ2 b2+ ... + μ n b n，其中μj ≥ 0，且∑ j μj = 1．对每个l ≤n，存在t s( l )∈D，使得|| μl-t s( l ) || < δ；令z= t s(1) b1 + t s(2) b2+ ... + t s(n) b n，则|| y - z || = || (μ1 -t s(1))b1 + (μ2 -t s(2))b2+ ... + (μn -t s(n))b n ||≤∑ l | μl-t s( l ) | · max j{ || b j || } ≤n δ M < ε /3；令C = {t s(1) b1 + t s(2) b2+ ... + t s(n) b n | t s(i)∈D，1 ≤i≤n}，则C是有限集，且C是co(A)的有限ε网．因空间是完备的，故co(A)是列紧集．1.5.3 设C是B*空间X中的一个紧凸集，映射T : C →C连续，求证T在C上有一个不动点．证明：因为C是紧集，所以C是闭集．因为C是紧集，故C的任意子集都列紧．而T(C) ?C，故T(C)列紧．于是，由Schauder不动点定理，T在C上有一个不动点．。
2015年全国医学博士入学统一考试英语真题及答案解析
2015年全国医学博士入学统一考试英语真题及答案解析Part I: Listening comprehension(略)Part II: Vocabulary(10%)Section ADirection: In this section, all the sentences are incomplete. Four word or phrases marked A,B,C and D are given beneath each of them. You are to choose the word the word or phrase that best completes the sentence, then mark your answer on the ANSWER SHEET.31. Despite his doctor’s note of caution, he never____from dring and smorking.A. retainedB. dissuadedC. alleviatedD. abstained32. people with a history of recurrent infections are warned that the use of personal stereos with headsets is likely to____their hearing.A. rehabilitateB. jeopardizeC. tranquilizeD. supplement33. impartial observers had to acknowledge that lack of formal education did not seem to____larry in any way in his success.A. refuteB. ratifyC. facilitateD. impede34. when the supporting finds were reduced, they should have revised their plan______.A. accordinglyB. alternativelyC. considerablyD. relatively35. it is increasingly believed among the expectant parents that prenatal education of classical music can_____ future adults with appreciation of music.A. acquaintB. familiarizedC. endowD. amuse36. if the gain of profit is solely due to rising energy prices, then inflation should be subsided when energy prices_____A. level outB. stand outC. come offD. wear off37. heat stroke is a medical emergency that demands immediate_____ from qualified medical personnel.A. prescriptionB. palpationC. interventionD. interposition38. asbestos exposure results in Mesothelioma, asbestosis and internal organ cancers, and_____ of these diseases is often decades after the initial exposure.A. offsetB. intakeC. outletD. onset39. ebola, which spreads through body fluid or secretions such as urine,______ and semen, can kill up to 90% of those infected.A. salineB. salivaC. scabiesD. scrabs40. the newly designed system is ____ to genetic transfections, and enables an incubation period for studying various genes.A. comparableB. transmissibleC. translatableD. amenable Section BDirections: each of the following sentences has a word or phrase underlined. There are four words or phrases beneath each sentence. Choose the word or phase which can best keep the meaning of the original sentence if it issubstituted for the underlined part. Mark your answer on the ANSWER SHEET.41. every year more than 1000 patients in Britain die on transplant waiting lists, prompting scientists to consider other ways to produce organs.A. propellingB. prolongingC. puzzlingD. promising42. improved treatment has changed the outlook of HIV patients, but there is still a serious stigma attached to AIDS.A. disgraceB. discriminationC. harassmentD. segregation43. surviviors of the shipwreck were finally rescued after their courage of persistence lowered to zero by their physical lassitude.A. depletionB. dehydrationC. exhaustionD. handicap44. scientists have invented a 3D scan technology to read the otherwise illegible wood-carved stone, a method that may apply to other areas such as medicine.A. negativeB. confusingC. eloquentD. indistinct45. top athletes scrutinize both success and failure with their coach to extract lessons from them, but they are never distracted from long-term goals.A. anticipateB. clarifyC. examineD. verify46. his imperative tone of voice reveals his arrogance and arbitrariness.A. challengingB. solemnC. hostileD. demanding47. the discussion on the economic collaboration between the United States and the European Union may be eclipsed by the recent growing trade friction.A. erasedB. triggeredC. shadowedD. suspended48. faster increases in prices foster the belief that the future increases will be also stronger, so that higher prices fuel demand rather than quench it.A. nurtureB. eliminateC. assimilateD. puncture49. some recent developments in photography allow animals to be studied in previously inaccessible places and in unprecedented detail.A. unpredictableB. unconventionalC. unparalleledD. unexpected50. a veteran negotiation specialist should be skillful at manipulating touchy situation.A. estimatingB. handlingC. rectifyingD. anticipatingPart III Cloze(10%)Direction: in this section there is a passage with ten numbered blanks. For each blank, there are four choices marked A, B, C, and D on the right side. Choose the best answer and mark the letter of your choice on the ANSWER SHEET.A mother who is suffering from cancer can pass on the disease to her unborn child in extremely rare cases 51 a new case report published in PNAS this week.According to researchers in Japan and at the Institute for Cancer Research in Sutton, UK, a Japanese mother had been diagnosed with leukemia a few weeks after giving birth 52 tumors were discovered in her daughter’s cheek and lung when she was 11 months old. Genetic analysis showed that the baby’s cancer cells had the same mutation as the cancer cellsof the mother. But the cancer cells contained no DNA whatsoever from the father 53 would be expected if she had inherited the cancer from conception. That suggests the cancer cell made it into the unborn child’s body across the placental barrier.The Guardian claimed this to be the fires 54 case of cells crossing the placental barrier. But this is not the case----microchimerism 55 cells are exchanged between a mother and her unborn child, is thought to be quite common, with some cells thought to pass from fetus to mother in about 50 to 70 percent of cases and to go the other way about half,56.As the BBC pointed out, the greater 57 in cancer transmission from mother to fetus had been how cancer cells that have slipped through the placental barrier could survive in the fetus without being killed by its immune system. The answer, in this case at least, lies in a second mutation of the cancer cells, which led to the 58 of the specific features that would have allowed the fetal immune system to detect the cells as foreign. As a result, no attack against the invaders was launched.59, according to the researchers there is little reason for concern of “cancer danger”. Only 17 probable cases have been reported worldwide and the combined 60 of cancer cells both passing the placental barrier and having the right mutation to evade the baby’s immune system is extremely low.51. A. suggests B. suggesting C. having suggested D. suggested52. A. since B. although C. whereas D. when53. A. what B. whom C. who D. as54. A. predicted B. notorious C. proven D. detailed55. A. where B. when C. if D. whatever56. A. as many B. as much C. as well D. as often57. A. threat B. puzzle C. obstacle D. dilemma58. A. detection B. deletion C. amplification D. addition59. A. therefore B. furthermore C. nevertheless D. conclusively60. A. likelihood B. function C. influence D. flexibilityPart IV Reading Comprehension(30%)Directions: in this part there are six passages, each of which is followed by five questions. For each question there are four possible answers marked A, B, C, and D. choose the best answer and mark the letter of your choice on the ANSWER SHEET.Passage OneThe American Society of Clinical Oncology wrapped its annual conference this week, going through the usual motions of presenting a lot of drugs that offer some added quality or extension of life to those suffering from a variety of as-yet incurable diseases. But buried deep in an AP story are a couple of promising headlines that seems worthy of more thorough review, including one treatment study where 100 percent of patients saw their cancer diminish byhalf.First of all, it seems pharmaceutical companies are moving away from the main cost-effective one-size-first-all approach to drug development and embracing the long cancer treatments, engineering drugs that only work for a small percentage of patients but work very effectively within that group.Pfizer announced that one such drug it’s pushing into late-stage testing is target for 4% of lung cancer patients. But more than 90% of that tiny cohort responded to the drug initial tests, and 9 out of ten is getting pretty close to the ideal ten out of ten. By gearing toward more boutique treatments rather than broad umbrella pharmaceuticals that try to fit for everyone it seems cancer researchers are making some headway. But how can we close the gap on that remaining ten percent?Ask Takeda Pharmaceutical and Celgene, two drug makers who put aside competitive interests to test a novel combination of their treatments. In a test of 66 patients with the blood disease multiple myeloma, a full 100 percent response to a cancer drug(or in this case a drug cocktail) is more or less unheard of. Moreover, this combination never would’ve been two competing companies hadn’t sat down and put their heads together.Are there more potentially effective drug combos out there separated by competitive interest and proprietary information? Who’s to say, but it seems like with the amount of money and research being pumped into cancer drug development, the outcome pretty good. And if researchers can start pushing more of their response numbers toward 100 percent, we can more easily start talking about oncology’s favorite four-letter word: cure.61. which of the following can be the best title for the passage?A. Competition and CooperationB.Two Competing Pharmaceutical CompaniesC. The promising Future of PharmaceuticalsD. Encouraging News: a 100% Response to a Cancer Drug62. in cancer drug development, according to the passage, the pharmaceuticals now ____A. are adopting the cost-effective one-size-fits-all approachB. are moving towards individualized and targeted treatmentsC. are investing the lion’s shares of their moneyD. care only about their profits63. from the encouraging advance by the two companies, we can infer that____A. the development can be ascribed to their joint efforts and collaborationB. it was their competition that resulted in the accomplishmentC. other pharmaceuticals will join them in the researchD. the future cancer treatment can be nothing but cocktail therapy64. from the last paragraph it can be inferred that the answer to the question___A. is nowhere to be foundB. can drive one crazyC. can be multipleD. is conditional65. the tone of the author of this passage seems to be_____A. neutralB. criticalC. negativeD. potimistPassage TwoLiver disease is the 12th leading cause of death in the US, chiefly because once it’s determined that a patient needs a new liver it’s difficult to get one. Even in case where a suitable donor match is found, there’s guarantee a transplant will be successful. But researchers Massachusetts General Hospital have taken a huge step toward building functioning livers in the lab, successfully transplanting culture-grown livers into rats.The livers aren’t grown from scratch, but rather within the infrastructure of a donor liver. The liver cells in the donor organ are washed out with a detergent that gently strips away the liver cells, leaving behind a biological scaffold of proteins and extracellular architecture that is very hard to duplicate synthetically.With all of that complicated infrastructure already in place, the researchers then seeded the scaffold(支架) with liver cells isolated from health livers, as well as some special endothelial cells to line the bold vessels. Once repopulated with healthy cells, these livers lived in culture for 10 days.The team also translated some two-day-old recellularized livers back into rats, where they continued to thrive for eight hours while connected into the rat’s vascular systems. However, the current method isn’t perfect and can not seem to repopulate the blood vessels quite densely enough and the transplanted livers can’t keep functioning for more than about 24 hours(hence the eight-hour maximum for the rat thansplant).But the initial successes are promising, and the team thinks they can overcome the blood vessel problem and get fully functioning livers into rats within two years. It still might be a decade before the tech hits the clinic, but if nothing goes horribly wrong—and especially if stem-cell research established a reliable way to create health liver cells from the every patients who need transplants-lab-generated livers that are perfect matches for their recipients could become a reality.66. it can be inferred from the passage that the animal model was mainly intended to____A. investigate the possibility of growing blood vessels in the labB. explore the unknown functions of the human liverC. reduce the incidence of liver disease in the US.D. address the source of liver transplants67. what does the author mean when he says that the livers aren’t grown from scratch?A. the making of a biological scaffold of proteins and extracellular architectureB. a huge step toward building functioning livers in the labC. the building of the infrastructure of a donor liverD. growing liver cells in the donor organ68. the biological scaffold was not put into the culture in the lab until____A. duplicated syntheticallyB. isolated from the healthy liverC. repopulated with the healthy cellsD. the addition of some man-made blood vessels69. what seems to be the problem in the planted liver?A. the rats as wrong recipientsB. the time point of the transplantationC. the short period of the recellularizationD. the insufficient repopulation of the blood vessels70. the research team holds high hopes of_____A. creating lab-generated livers for patients within two yearsB. the timetable for generating human livers in the labC. stem-cell research as the future of medicineD. building a fully functioning liver into ratsPassage ThreePatients whose eyes have suffered heat or chemical bums typically experience severe damage to the cornea—the thin, transparent front of the eye that refracts light and contributes most of the eye’s focusing ability. In a long-term study, Italian researchers use stem cells taken from the limbus, the border between the cornea and the white of the eye, to cultivate a graft of healthy cells in a lab to help restore vision in eyes. During the 10-years study, the researchers implanted the healthy stem cells into the damaged cornea in 113 eyes of 112 patients. The treatment was fully successful in more than 75 percent of the patients, and partially successful in 13 percent. Moreover, the restored vision remained stable over 10 years. Success was defined as an absence of all symptoms and permanent restoration of the cornea.Treatment outcome was initially assessed at one year, with up to 10 years of follow-up evaluations. The procedure was even successful on several patients whose bum injuries had occurred years earlier and who had already undergone surgery.Current treatment for burned eyes involves taking stem cells from a patient’s healthy eye, or from the eyes of another person, and transferring them to the burned eye. The new procedure, however, stimulates the limbal stem cells from the patient’s own eye to reproduce in a lab culture. Several types of treatments using stem cells have proven successful in restoring blindness, but the long-term effectiveness shown here is significant. The treatment is only for blindness caused by damage to the cornea; it is not effective for repairing damaged retinas or optic nerves.Chemical eye burns often occur in the workplace, but can also happen due to mishaps involving household cleaning products and automobile batteries.The result of the study, based at Italy’s University of Modena and Reggio Emilia, were published in the June 23 online issue of the New England Journalof Medicine.71. what is the main idea of this passage?A. stem cells can help restore vision in the eyes blinded by bums.B. the vision in the eyes blinded by bums for 10 years can be restoredC. the restored vision of the burned eyes treated with stem cells can last for10 yearsD. the burned eyes can only be treated with stem cells from other healthy persons72. the Italian technique reported in this passage_____A. can repair damaged retinasB. is able to treat damaged optic nervesC. is especially effective for burn injuries in the eyes already treated surgicallyD. shows a long-term effectiveness for blindness in vision caused by damage to cornea73. which of the following is NOT mentioned about eye bums?A. the places in which people workB. the accidents that involve using household cleaning productsC. the mishaps that involved vehicles batteriesD. the disasters caused by battery explosion at home74. what is one of the requirements for the current approach?A. the stem cells taken from a healthy eyeB. the patient physically healthyC. the damaged eye with partial visionD. the blindness due to damaged optic nerves75. which of the following words can best describe the author’s attitude towards the new method?A. sarcasticB. indifferentC. criticalD. positivePassage FourHere is a charming statistic: divide the us by race, sex and county of residence, and differences in average life expectancy across the various groups can exceed 30 years. The most disadvantaged look like denizens of a poor African country: a boy born on a Native American reservation in Jackson County, South Dakota, for example, will be lucky to reach his 60th birthday, a typical child in Senegal can expect to live longer than that.America is not alone in this respect. While the picture is extreme in other rich nations, health inequalities based on race, sex and class exist in most societies—and are only party explained by access to healthcare.But fresh insights and solutions may soon be at hand. An innovative project in Chicago to unite sociology and biology is blazing the trail(开创), after discovering that social isolation and fear of crime can help to explain the alarmingly high death rate from breast cancer among the city’s black women. Living in these conditions seems to make tumors more aggressive by changing gene activity, so that cancer cells can use nutrients more effectively.We are already familiar with the lethal effect of stress on people clinging to the bottom rungs of the societal ladder, thanks to pioneering studies of British civil servants conducted by Michael Marmot of University College London. What’s exciting about the Chicago project is that it both probes the mechanisms involved in a specific disease and suggests precise remedies that it both probes the mechanisms invlilved in a specific disease and suggests precise remedies. There are drugs that may stave tumors of nutrients and community coordinators could be employed to help reduce social isolation. Encouraged by the US National Institutes of Health , similar projects are springing up to study other pockets of poor health, in populations ranging from urban black men to while poor women in rural Appalachia.To realize the full potential of such projects, biologists and sociologists will have to start treating one other with a new respect and learn how to collaborate outside their comfort zones. Too many biomedical researchers still take the arrogant view that sociology is a “soft science” with little that’s serious to say about health. And too many sociologists reject any biological angle—fearing that their expertise will be swept aside and that this approach will be used to bolster discredited theories of eugenics, or crude race-based medicine.It’s time to drop these outdated attitudes and work together for the good of society’s most deprived members. More important, it’s time to use this fusion of biology and sociology to inform public policy. This endeavor has huge implications, not least in cutting the wide health gaps between blacks and whites, rich and poor.76. as shown in the 1st paragraph, the shaming statistic reflects______.A. injustice everywhereB. racial discriminationC. a growing life spanD. health inequalities77. which of the following can have a negative impact on health according to the Chicago-based project?A. where to liveB. which race to belong toC. how to adjust environmentallyD. what medical problem to suffer78. the Chicago-based project focuses its management on_____A. a particular medical problem and its related social issueB. racial discrimination and its related social problemsC. the social ladder and its related medical conditionsD. a specific disease and its medical treatment79. which of the following can most probably neglected by sociologists?A. the racial perspectiveB. the environmental aspectC. the biological dimensionD. the psychological angel80. the author is a big fan of______A. the combination of a traditional and new way of thinking in promoting healthB. the integration of biologists and sociologists to reduce health inequalitiesC. the mutual understanding and respect between racesD. public education and health promotionPassage FiveAmerican researchers are working on three antibodies that many mark a new step on the path toward an HIV vaccine, according to a report published online Thursday, July 8,2010, in the journal Science.One of the antibodies suppresses 91 percent of HIV strains, more than any AIDS antibody ever discovered, according to a report on the findings published in the Wall Street Journal. The antibodies were discovered in the cells of a 60-year-old African-American gay man whose body produced them naturally. One antibody in particular is substantially different from its precursors, the Science study says.The antibodies could be tried as a treatment for people already infected with HIV, the WSJ reports. At the very least, they might boost the efficacy of current antiretroviral drugs.It is welcome news for the 33 million people the United Nations estimated were living with AIDS at the end of 2008.The WSJ outlines the painstaking method the team used to find the antibody amid the cells of the African—American man, known as Donor 45. First they designed a probe that looks just like a spot on a particular molecule on the cells that HIV infects. They used the probe to attract only the antibodies that efficiently attack that spot. They screened 25 million of Donor 45’s cell to find just 12 cells that produced the antibodies.Scientists have already discovered plenty of antibodies that either don’t work at all or only work on a couple of HIV strains. Last year marked the first time that researchers found ”broadly neutralizing antibodies”, which knock out many HIV strains. But none of those antibodies neutralized more than about 40 percent of them, the WSJ says. The newest antibody, at 91 percent neutralization , is a marked improvement.Still, more work needs to be done to ensure the antibodies would activate the immune system to produce natural defenses against AIDS, the study authors say. They suggest there test methods that blend the three new antibodies together—in raw form to prevent transmission of the virus, such as from mother to child; in a microbicide gel that women or gay men could use before sex to prevent infection; or as a treatment for HIV/AIDS, combined with antiretroviral drug.If the scientists can find the right way to stimulate production of the antibodies, they think most people could produce then, the WSJ says.81. we can learn from the beginning of the passage that_______A. a newly discovered antibody defeats 91% of the HIV strainsB. a new antiretroviral drug has just come on the marketC. American researchers have developed a new vaccine for HIVD. the African—American gay man was cured of this HIV infection82. what is the implication of the antibodies discovered in the cells of the African—American gay man?A. they can cure the 33 million AIDS patients in the worldB. they may strengthen the effects of the existing antiretroviral drugsC. they will kill all the HIV virusesD. they will help make a quick diagnosis of an HIV infection83. the newest antibody found in Donor 45 reflects a dramatic advance in terms of_____.A. pathologyB. pharmacologyC. HIV neutralizationD. HIV epidemiology84. according to the study authors, the three test methods are intended to____.A. advance the technology in condom production to prevent HIV infectionB. facilitate the natural immune defense against AIDSC. develop more effective antiretroviral drugs85. the passage is most likely_____.A. a news reportB. a paper in ScienceC. an excerpt from an Immunology TextbookD. an episode in a science fiction novel.Passage SixWhitening the world's roofs would offset the emissions of the world's cars for 20 years, according to a new study from Lawrence Berkeley National Laboratory.Overall, installing lighter-colored roofs and pavement can cancel the heat effect of two years of global carbon dioxide emissions, Berkeley Lab says. It's the first roof-cooling study to use a global model to examine the issue.Lightening-up roofs and pavement can offset 57 billion metric tons of carbon dioxide, about double the amount the world emitted in 2006, the study found. It was published in the journalEnvironmental Research Letters.Researchers used a conservative estimate of increased albedo, or solar reflection, suggesting that purely white roofs would be even better. They increased the albedo of all roofs by 0.25 and pavement by 0.15. That means a black roof, which has an albedo of zero, would only need to be replaced by a roof of a cooler color -- which might be more feasible to implement than a snowy white roof, Berkeley Lab says.The researchers extrapolated a roof's CO2 offset over its average lifespan. If all roofs were converted to white or cool colors, they would offset about 24 gigatons (24 billion metric tons) of CO2, but only once. But assuming roofs last about 20 years, the researchers came up with 1.2 gigatons per year. That equates to offsetting the emissions of roughly 300 million cars, all the cars in the world, for 20 years.Pavement and roofs cover 50 to 65 percent of urban areas, and cause a heat-island effect because they absorb so much heat. That's why cities aresignificantly warmer than their surrounding rural areas. This effect makes it harder -- and therefore more expensive -- to keep buildings cool in the summer. Winds also move the heat into the atmosphere, causing a regional warming effect.Energy Secretary Steven Chu, a Nobel laureate in physics (and former Berkeley Lab director), has advocated white roofs for years. He put his words into action Monday by directing all Energy Department offices to install white roofs. All newly installed roofs will be white, and black roofs might be replaced when it is cost-effective over the lifetime of the roof."Cool roofs are one of the quickest and lowest-cost ways we can reduce our global carbon emissions and begin the hard work of slowing climate change," he said in a statement.86. which of the following can be the best title for the passage?A. a Decline in Car EmissionsB. white Roofs or Black PavementsC. the Effect of Linghting-up RoofsD. climate Change and Extreme Weathers87. a indicated by the passage, black roofs______A. are better than snowy white onesB. reflect not heat from the sunC. are more expensive to build in the urban areasD. are supposed to be placed by snowy white ones88. if they are converted to white or cooler colors, all roofs in the world in their lifetime_____A. can absorb 1.2 gigattons of CO2 a yearB. could serve as 300 million cars in terms of emissionC. would offset the emissions from 300 million carsD. would offset about 24 gigatons of CO2 as emitted from the cars89. according to the passage, it is hard and expensive to keep the urban buildings cool because of______A. the heat-island effectB. the lack of seasonal windsC. the local unique weatherD. the fast urban shrinkage90. energy Secretary Steven Chu implies that_____A. nothing could be more effective in cooling global warming than method he has advocatedB. the method in question still needs to be justified in the futureC. our global carbon emissions can be reduced by half if cool roofs are installedD. weather change and global warming can be addressed in no timePart V Writing(20%)Directions: in this part there is an essay in Chinese. Read it carefully and then write a summary of 200 words in English on the ANSWER SHEET. Make sure that your summary covers the major points of the passage.什么是健康？人的健康包括身体健康和心理健康两个方面。
2015考博真题
一、单选1×50
上下尖牙区别
monson球面的半径
下颌神经前支中的感觉神经
前牙切割运动的杠杆运动形式
单囊性成釉细胞瘤处理方式
腺淋巴瘤病理特点
舌下腺结构
放射性骨髓炎病理表现
翼下颌间隙内容
下颌运动特点
下颌体骨化中心
颞下颌关节手术时切口方式
牙受垂直向力时牙龈主纤维中不受力的是
（以后想起来再补充）
二、名解2×10
近唇线角
pterygoid process
Terra dentition index
mento-cervical angle
taste threshold
alveolar bone proper
candidiasis
chronic gingivitis
branchial cleft cyst
lymphoepithelial carcinoma
三、简答5×6
解剖
1.根管系统在根部侧面开口的系统名称，并从解剖角度解释牙周病和牙髓病的相互影响。
2.口颌系统肌链的组成与功能？
3.临床上面神经的解剖方法，面神经主干的解剖标志点？
病理
1.口腔黏膜鳞癌有很多亚型，请举3例口腔黏膜鳞癌亚型，并叙述其镜下特点及生物学行为？
2.根据牙骨质组织结构学特性，叙述牙骨质龋特点？
3.肌上皮细胞来源的唾液腺良恶性肿瘤各举两例，及其镜下鉴别要点。
泛函分析试卷
泛函分析期末考试试卷（总分100分） 一、选择题（每个3分，共15分）1、设X 是赋范线性空间，X y x ∈,，T 是X 到X 中的压缩映射，则下列哪个式子成立（ ）.A ．10<<-≤-αα， y x Ty Tx B.1≥-≤-αα， y x Ty Tx C.10<<-≥-αα， y x Ty Tx D.1≥-≥-αα， y x Ty Tx 2、设X 是线性空间，X y x ∈,，实数x 称为x 的范数,下列哪个条件不是应满足的条件：（ ）.A. 0等价于0且,0==≥x x xB.()数复为任意实,αααx x =C. y x y x +≤+D. y x xy +≤ 3、下列关于度量空间中的点列的说法哪个是错误的（ ）. A ．收敛点列的极限是唯一的 B. 基本点列是收敛点列 C ．基本点列是有界点列 D.收敛点列是有界点列4、巴拿赫空间X 的子集空间Y 为完备的充要条件是（ ）. A ．集X 是开的 B.集Y 是开的 C.集X 是闭的 D.集Y 是闭的5、设(1)p l p <<+∞的共轭空间为q l ，则有11p q+的值为（ ）.A. 1-B.12 C. 1 D. 12- 二、填空题（每个3分，共15分）1、度量空间中的每一个收敛点列都是（ ）。
2、任何赋范线性空间的共轭空间是（ ）。
3、1l 的共轭空间是（ ）。
4、设X按内积空间<x,y>成为内积空间，则对于X中任意向量x,y 成立不等式（）当且仅当x与y线性相关时不等式等号成立。
5、设T为复希尔伯特空间X上有界线性算子，则T为自伴算子的充要条件是（）。
三、判断题（每个3分，共15分）1、设X是线性赋范空间，X中的单位球是列紧集，则X必为有限维。
( )2、距离空间中的列紧集都是可分的。
( )3、若范数满足平行四边形法则，范数可以诱导内积。
( )4、任何一个Hilbert空间都有正交基。
2015南京师范大学考博英语真题阅读理解精练
2015南京师范大学考博英语真题阅读理解精练Every living thing has an inner biological clock that controls behavior.The clock works all the time;even when there are no outside signs to mark the passing of time.The biological clock tells plants when to form flowers and when the flowers should open.It tells insects when to leave the protective cocoon and fly a way.And it tells animals when to eat，sleep and wake.It controls body temperature，the release of some hormones and even dreams.These natural daily events are circadian rhythms.Man has known about them for thousands of years.But the first scientific observation of circadian rhythms was not made until1729. In that year French astronomer，Jean-Jacques d“Ortous de Mairan，noted that one of his plants opened it s leaves at the same time every morning，and closed them at the same time every night.The plant did this even when he kept it in a dark place all the ter scientists wondered about circadian rhythms in humans.They learned that man”s biological clock actually keeps time with a day of a little less than 25hours instead of the24hours on a man-made clock.About four years ago an American doctor，Eliot Weitzman，established a laboratory to study how our biological clock works.The people in his experiments are shut off from the outside world.They are free to listen to and live by their circadian rhythms.Dr.Weitzman hopes his research will lead to effective treatments for common sleep problems and sleep disorders caused by ageing and mental illness.The laboratory is inthe Monteflore Hospital in New York City.It has two living areas with three small rooms in each.The windows are covered，so no sunlight o r moonlight comes in.There are no radios or television receivers. There is a control room between the living areas.It contains computers，one-way cameras and other electronic devices for observing the person in the living area.The instruments measure heartbeat，body temperature，hormones in the blood，other substances in the urine and brain waves during sleep.A doctor or medical technician is on duty in the control room24hours a day during an experiment.They do not work the same time each day and are not permitted to wear watches，so the person in the experiment has no idea what time it is.In the first four years of research，Dr Weitzman and his assistant have observed16men between the ages of21and80. The men remained in the laboratory for as long as six st month，a science reporter for“The New York Times”newspaper，Dava Sobol，became the first woman to take part in the experiment.She entered the laboratory on June13th and stayed for25days.Miss Sobol wrote reports about the experiment during that time，which were published in the newspaper.（PS:The way to contact yumingkaobo TEL:si ling ling-liu liu ba-l iu jiu qi ba QQ:si jiu san san qi yi liu er liu）1、The biological clock is believed to play an essential role in ____.A、the regulation of body temperatureB、the secretion of hormonesC、animal reproductionD、many aspects of plant and animal physiology2、In his observation，the French scientist noticed that the leaves of a certain plant maintained its opening-and-closing cycles ____.A、even when it was kept in a murky place all dayB、even if it was placed in the moonlightC、even when he was observing it from a dark placeD、even during the night time3、The sentence“They are free to listen to and live by their circadian rhythms.”(In Paragraph4)probably means____.A、They can lead their daily lives according to their biological clocks，without referring to a man-made clock.B、They can listen to the wonderful rhythms of the biological clock and live close to them.C、They can live by regulating their own circadian rhythms.D、They are free from the annoying rhythms of everyday life.4、In the experiment conducted by Mr.Weitzman，the doctor who is on duty does not work the same time each day____.A、insgroupsto observe the abnormal behavior of the people at different timesB、so as not to be recognized by the peopleC、so as to avoid indicating to the people what time it is whenhe starts workD、so as to leave the people“s circadian rhythms in disorder5、Miss Sobol left the laboratory____.A、on June13thB、on June25thC、at the end of JuneD、on July7thKeys to PassageD A A C D本文由“育明考博”整理编辑。
【2024版】2015年南京大学新闻传播学院620传播史论考研真题及详解【圣才出品】
可编辑修改精选全文完整版2015年南京大学新闻传播学院620传播史论考研真题及详解【圣才出品】2015年南京大学新闻传播学院620传播史论考研真题及详解科目一：传播史论一、名词解释（每题5分，共20分）1．卡尔·霍夫兰2．社会顺从理论3．KDKA4．时间偏向的媒介二、简答题（每题10分，共50分）1．简述议程设置理论的作用机制。
2．简述芝加哥学派和哥伦比亚学派观点的异同。
3．简述书面文化的社会影响。
4．经验性研究方法的主要特征和原则。
5．简述法兰克福学派的主要观点。
三、论述题（每题20分，共80分）1．谈谈第三人效果的学术意义及现实意义。
2．有人认为，是广播而不是报纸和电视开启了大众传播时代并促成了大众传播学的诞生。
谈谈你的看法。
3．试论全球化与身份认同之间的关系，以及如何理解与媒介的关系。
4．20世纪70年代，受斯图亚特·霍尔的影响，英国的文化研究学派理论出现转型，你如何理解这一转型，哪些理论促使此转型。
参考答案：一、名词解释（每题5分，共20分）1．卡尔·霍夫兰答：卡尔·霍夫兰是美国实验心理学家，传播学四大先驱之一。
霍夫兰于二战期间担任美国陆战心理实验室主任，研究宣传电影对士兵士气的影响，通过对传播者、传播技巧、传播内容、受传者等进行实验研究，提出影响传播效果的一系列因素。
二战期间和战后，霍夫兰和一批心理学家进行了大量实验，对态度与说服进行了细致研究，形成了具有影响力的“耶鲁学派”。
其传播研究的成果集结于著作《传播与说服》一书中。
霍夫兰对传播学的贡献主要表现在三个方面：①他在彼得森、瑟斯顿等学者的基础上，首次较完善地把心理学控制试验的方法用于传播效果的研究。
②他对军事教育电影的研究不仅从一个方面证明了坎特里尔的观点，而且为打破“魔弹论”的神话提供了更有价值的证据。
③他注意到影响说服效果的多种因素，尤其是说服者及其发出的信息者两个因素，并提出了改善说服效果的一系列有价值的建议，为“可说服性”这个当代传播学的重要课题奠定了基础。
泛函分析9§1-5,习题选讲与答案
第九章 内积空间和希尔伯特空间例题选讲例1． Hilbert 是X 可分的充分必要条件X 存在一个可数的完全规范正交系{}n e证明：若X 是可分的，设{}n x 是X 的一个可数稠密子集。
不妨设{}n x 是线性无关的。
用Gram Schmidt -方法，存在可数的完全规范正交系{}n e ，使span{}1,,n e e L = {}1,,n span x x L 。
这样。
因此{}n e 是完全的。
反之，若{}n e 是X 的一个完全规范正交系，则span {}n e 在X 中稠密。
()01,,1,2,3,n k k k k k k X a ib e a b Q N =⎧⎫=+∈=⎨⎬⎩⎭∑L 是X 中的可数稠密子集，因此 X 是可分的。
证毕例2．求证：P 是Hilbert 空间X 上的投影算子的充分必要条件是：2P P =且*P P =证明：设P 是X 中相对应与闭线性子空间Y 的投影算子。
对任意x ∈X ，存在1x Y∈，2x ∈Y ⊥，使12x x x =+，1Px x =。
对于1x ，1x =10x +，其中1x Y ∈，0Y ⊥∈。
因此11Px x =，即21P x Px Px ==，因此2P P = 设,x y X ∈，12x x x =+，12y y y =+。
其中11,x y Y ∈，22,x y Y ⊥∈。
这样()()()()()1121112,,,,,Px y x y y x y x x Py x Py =+==+=。
这就证明了*P P =。
反之，若P 满足*P P =，*P P =。
令{}Y x Px x ==，则Y 是X 中的线性子空间。
Y 还是闭的。
事实上，若n x Y ∈，0lim n n x x →∞=，则00lim lim n n n n Px Px x x →∞→∞===。
故0x Y ∈，因此Y 是闭的线性子空间，我们要证明P 是Y 上的投影算子。
设x X ∈，则()x Px x Px =+-。
泛函分析习题及参考答案
， 令n → ∞， 可得 m( E ( x n − x ≥ σ ) → 0 。 即 x n (t )
由 x(t ) 的积分绝对连续性可知，对任何 ε > 0 ，存在 δ 1 > 0 ，使得 e ⊂ E ，me < δ 1 时，
( ∫ x(t ) dt ) <
p e
1 p
ε
, ξi , ) ∈ l p ，则 n → ∞
；
p⎞ ⎛ ∞ (n) (n) 时，d ( xn , x) = ⎜ ∑ ξi − ξ i ⎟ → 0 的充要条件为 (1) n → ∞ 时，ξi → ξ i ，i = 1, 2, ⎝ i =1 ⎠
(2) ∀ε > 0 ，存在 N > 0 ，使得
i = N +1
泛函分析习题及参考答案
一、在 R 中定义如下三种距离： x = ( x1 , x2 ), y = ( y1 , y2 ) ∈ R ，
2
2
d1 ( x, y ) = ( x1 − y1 ) 2 + ( x2 − y2 ) 2 ， d 2 ( x, y ) = max{ x1 − y1 , x2 − y2 } ，
2
。 对上述 ε > 0 ， 存在 N > 0 ， 使得 n > N 时， （ x n (t ) − x(t ) dt） <
E
∫
p
1 p
ε
2
，
从而
∫
e n
x n (t ) dt ) ≤ ( ∫ x n − x dt ) + ( ∫ x dt ) ≤ ( ∫ x n − x dt ) + ( ∫ x dt ) p < ε ，
南京大学历年社会学理论与方法考博试题
南京大学历年社会学理论与方法考博试题
一、简述题（每题8分，共40分）
1、简述马克斯.韦伯的理想类型
2、简述达伦多夫社会冲突理论
3、简述社会交换理论在人类学、经济学和心理学中的理论来源
4、简述布尔迪厄德“文化资本”理论
5、有关政府预算的一项社会调查想知道某市市民对政府在教育经费投入、交通建设经费投入方面的意见。
由于调查经费有限，甲研究人员把问卷发往全市各小学，然后通过学生转到家长手中；乙研究人员从全市18岁以上成人中随机抽出调查对象，再用邮寄方式发放问卷。
请比较两种不同的调查方法，指出各自的长短。
二、论述题（每题15分，共45分）
1、论述唯名论和唯实论两种社会观
2、论述宏观社会学理论和微观社会学理论的区别
3、社会学的中国化应该如何借鉴西方社会学发展的历史教训？
三、计算题（每题15分，共15分）
在一次对研究生考生的社会调查中，某地区所有考生的资料显示，该地区考生5门课程的总分成绩近似服从正态分布。
每个考生的平均分为320分，标准差为25.38，试求：
（1）、从该地区中随机抽取一名考生，其总分超过340分的概率；
（2）、如果抽取一个容量为9的随机样本，求其平均总分超过340分的概率
（3）、比较（1）和（2）的结果，请做出统计意义的解释；另外，如果总体不是正态，那么（2）的答案将是什么？
（附：查表可知，Ф（Z<=0.79）=0.7851
Ф（Z<=0.78）=0.7823 Ф（Z<=2.36）=0.981）。
泛函分析答案
泛函分析答案：1、 所有元素均为0的n ×n 矩阵2、 设E 为一线性空间，L 是E 中的一个子集，若对任意的x,y ∈L ，以及变数λ和μ均有λx ＋μy ∈L ，则L 称为线性空间E 的一个子空间。
子空间心室包含零元素，因为当λ和μ均为0时，λx ＋μy ＝0∈L ，则L 必定含零元素。
3、 设L 是线性空间E 的子空间，x 0∈E\L,则集合x 0+L={x 0+l,l ∈L}称为E 中一个线性流形。
4、 设M 是线性空间E 中一个集合，如果对任何x,y ∈M ，以及λ＋μ＝1，λ≥0，μ≥0的λ和μ，都有λx ＋μy ∈M ，则称M 为E 中的凸集。
5、 设x,y 是线性空间E 中的两个元素，d(x,y)为其之间的距离，它必须满足以下条件：（1） 非负性：d(x,y)＞0，且d(x,y)＝0＜―――＞x=y （2） d(x,y)=d(y,x)（3） 三角不等式：d(x,y)≤d(x,z)+d(y,z) for every x,y,z ∈E n 维欧几里德空间常用距离定义：设x={x 1,x 2,…x n }T ,y={y 1y 2,…y n }Td 2(x,y)=(21||niii x y=-∑)1/2d 1(x,y)=1||ni i i x y =-∑d p (x,y) = (1||np iii x y=-∑ )1/p d ∞(x,y)=1max ||i i i nx y ≤≤-6、距离空间(x,d)中的点列{x n }收敛到x 0是指d(x n ,x 0)→0(n →∞)，这时记作0lim nn xx -->∞=，或简单地记作x n →x 07、设||x||是线性空间E 中的任何一个元素x 的范数，其须满足以下条件： （1）||x||≥0,且||x||＝0 iff x=0 （2）||λx||=λ||x||,λ为常数（3）||x+y||≤||x||+||y||,for every x,y ∈E8、设E 为线性赋范空间，｛x n ｝∞n=1是其中的一个无穷列，如果对于任何ε>0,总存在自然数N ，使得当n>N,m>N 时，均有|x m -x n |<ε,则称序列{x n }是E 中的基本列。
2015年南京大学考博考研生物化学重点试题及答案
2015年南京大学考博考研生物化学重点试题及答案第一章糖类化学一、填空题1.纤维素是由_ D-葡萄糖_组成,它们之间通过_____β-1，4____糖苷键相连。
2.常用定量测定还原糖的试剂为___Fehling___试剂和__Benedict__试剂。
3.人血液中含量最丰富的糖是__葡萄糖_,肝脏中含量最丰富的糖是__糖原_,肌肉中含量最丰富的糖是__糖原_。
4.乳糖是由一分子_D-半乳糖_和一分子_D-葡萄糖_ 组成,它们之间通过_β-1，4__糖苷键相连。
5.鉴别糖的普通方法为___Molisch__试验。
6.蛋白聚糖是由__糖胺聚糖和___蛋白质_共价结合形成的复合物。
7.糖苷是指糖的____半缩醛（半缩酮）羟基___和醇、酚等化合物失水而形成的缩醛(或缩酮)等形式的化合物。
8.判断一个糖的D-型和L-型是以_离羰基最远的一个不对称__碳原子上羟基的位置作依据。
9.多糖的构象大致可分为_螺旋、带状__、皱折_和_无规卷曲_四种类型,决定其构象的主要因素是_糖链的一级结构_。
二、是非题1.[✗]果糖是左旋的，因此它属于L-构型。
2.[ ✗]从热力学上讲,葡萄糖的船式构象比椅式构象更稳定。
3.[ ✗]糖原、淀粉和纤维素分子中都有一个还原端,所以它们都有还原性。
4.[ ✗]同一种单糖的α-型和β-型是对映体。
5.[ ✗]糖的变旋现象是指糖溶液放置后,旋光方向从右旋变成左旋或从左旋变成右旋。
6.[ ✗]D-葡萄糖的对映体为L-葡萄糖,后者存在于自然界。
7.[√]D-葡萄糖,D-甘露糖和D-果糖生成同一种糖脎。
8.[√]糖链的合成无模板,糖基的顺序由基因编码的转移酶决定。
9.[√]醛式葡萄糖变成环状后无还原性。
10.[√]肽聚糖分子中不仅有L-型氨基酸,而且还有D-型氨基酸。
三、选择题1.[ ]下列哪种糖无还原性?A.麦芽糖B.蔗糖C.阿拉伯糖D.木糖E.果糖2.[ ]环状结构的己醛糖其立体异构体的数目为A.4B.3C.18D.32E.643.[ ]下列物质中哪种不是糖胺聚糖?A.果胶B.硫酸软骨素C.透明质酸D.肝素E.硫酸粘液素4.[ ]下图的结构式代表哪种糖？A.α-D-葡萄糖B.β-D-葡萄糖C.α-D-半乳糖D.β-D-半乳糖E.α-D-果糖5.[ ]下列有关葡萄糖的叙述,哪个是错的?A.显示还原性B.在强酸中脱水形成5-羟甲基糠醛C.莫利希(Molisch)试验阴性D.与苯肼反应生成脎E.新配制的葡萄糖水溶液其比旋光度随时间而改变6.[ ]糖胺聚糖中不含硫的是A.透明质酸B.硫酸软骨素C.硫酸皮肤素D.硫酸角质素E.肝素7.[ ]下列哪种糖不能生成糖脎?A.葡萄糖B.果糖C.蔗糖D.乳糖E.麦芽糖8.[ ]下列四种情况中,哪些尿能和班乃德(Benedict)试剂呈阳性反应?(1).血中过高浓度的半乳糖溢入尿中(半乳糖血症)(2).正常膳食的人由于饮过量的含戊醛糖的混合酒造成尿中出现戊糖(戊糖尿)(3).尿中有过量的果糖(果糖尿)(4).实验室的技术员错把蔗糖加到尿的样液中A.1,2,3B.1,3C.2,4D.4E.1,2,3,49.[ ]α-淀粉酶水解支链淀粉的结果是(1).完全水解成葡萄糖和麦芽糖(2).主要产物为糊精(3).使α-1,6糖苷键水解(4).在淀粉-1,6-葡萄糖苷酶存在时,完全水解成葡萄糖和麦芽糖A.1,2,3B.1,3C.2,4D.4E.1,2,3,410.[ ]有关糖原结构的下列叙述哪些是正确的?(1).有α-1,4糖苷键(2).有α-1,6糖苷键(3).糖原由α-D-葡萄糖组成(4).糖原是没有分支的分子A.1,2,3B.1,3C.2,4D.4E.1,2,3,4四、问答与计算1.大肠杆菌糖原的样品25mg,用2ml 1mol/L H2SO4水解。
南京大学考博英语真题2015年_真题(含答案与解析)-交互
南京大学考博英语真题2015年(总分100, 做题时间180分钟)Part Ⅰ Vocabulary and Structure1.The ambassador was accused of having ______ on domestic affairs.•**•**•****SSS_SIMPLE_SINA B C D该题您未回答:х该问题分值: 1答案:B近义词词义辨析。
encroach，invade，trespass这三个词均指损害他人权利，侵占其财产或侵犯别国的领土。
encroach通常指蚕食，即逐渐地、不声不响或偷偷摸摸地进入别国的领土，或攫取别人的财物，侵犯他人的权利，常与介词on或upon连用。
invade暗示着明目张胆、凶残与暴行，常用来指一国武装侵略另一国。
也可用来指疾病、虫害的侵袭。
trespass是个法律用语，指未经许可进入私人土地，或非法侵入，常与on或upon连用。
句中提及侵犯他国内政，encroach更加符合语境。
entrench与on搭配使用时表示挖掘壕沟，与题意不符。
故答案为B。
2.The goal is to use crops, weeds and even animal waste ______ the petroleum that fuels much of American manufacturing.•** terms of•** favor of•** spite of** place ofSSS_SIMPLE_SINA B C D该题您未回答:х该问题分值: 1答案:D介词词组辨析。
句中crops，weeds，animal waste与petroleum“石油”对比可知，这两组是性质不同的能源，由此可知空格处的词组应为“取代”的意思，选项中只有D表示“取代，替换”的意思。
故答案为D。
3.As computer security systems become even more advanced, ______ the methods of those who try to break into them illegally.•** much do•** too do•** much as** well asSSS_SIMPLE_SINA B C D该题您未回答:х该问题分值: 1答案:A语法知识。
2016年南京大学物理学院博士生量子力学入学考试真题
2015年南京大学物理学院博士生“申请-考核”制入学
专业课程笔试试题
考试科目： 量子力学 考试时间：三小时
本试卷共计五大题
一、基本概念题
简述量子力学的基本原理。
二、设一个质量为m 的粒子处于区域为(0, a )的一维无限深势阱中， 其状态波函数为2=sin cos x
x
a a ππψ ，试求：
1)、一维无限深势阱的本征值问题；
2)、测量到粒子处于不同能量本征态的几率。
三、设两个算子ˆA
与ˆB 满足交换关系式：ˆˆˆˆˆˆ[,]1A B AB BA =-=，试求： 1)、n 为正整数， ˆˆ[,]n A
B ； 2)、()f x 为解析函数，ˆˆ[,()]A
f B 。
四、 已知两个算子ˆa 与ˆa +满足ˆˆˆˆ1a a aa ++=-，令ˆˆˆN a a +=，且有ˆN
n n n =， 求证：n 为实数。
五、量子力学中的韦尔(Weyl)波动方程式为：
(,)(,)i r t c r t t i ψσψ∂
=⋅∇∂
，
其中=x x y y z z e e e σσσσ++
为泡利矩阵所组成的矢量，
(,)r t ψ 为泡利二 分量波函数，其它为量子力学标准符号。
求
1)、该系统的韦尔定态方程式与力学量完全集；
2)、该系统的能量本征值并说明其物理意义；
3)、该系统的本征波函数。
泛函分析习题参考答案
泛函分析习题参考答案一、设),(y x d 为空间X 上的距离，试证：),(1),(),(~x y d x y d x y d +=也是X 上的距离。
证明：显然,0),(~≥y x d 并且y x y x d y x d =⇔=⇔=0),(0),(~。
再者，),(~),(1),(),(1),(),(~y x d y x d y x d x y d x y d x y d =+=+=；最后，由tt t +-=+1111的单调增加性及),(),(),(y z d z x d y x d +≤，可得 ),(),(1),(),(),(1),(),(),(1),(),(),(1),(),(~y z d z x d y z d y z d z x d z x d y z d z x d y z d z x d y x d y x d y x d +++++=+++≤+=),(~),(~),(1),(),(1),(y z d z x d y z d y z d z x d z x d +=+++≤。
二、设1p ≥，1()()(,,,)i n n p n x l ξξ=∈， ,2,1=n ，1(,,,)pi x l ξξ=∈，则n →∞时，1()1(,)0pp n n i i i d x x ξξ∞=⎛⎫=-→ ⎪⎝⎭∑的充要条件为)1(n →∞时，()n i i ξξ→，1,2,i =；)2(0ε∀>，存在0N >，使得()1pn i i N ξε∞=+<∑对任何自然数n 成立。
必要性证明：由1()1(,)0ppn n i i i d x x ξξ∞=⎛⎫=-→ ⎪⎝⎭∑可知，()n i i ξξ→，1,2,i =。
由1(,,,)pi x lξξ=∈可知，ε∀>，存在10N >，使得11(2ppi i N εξ∞=+<∑，并且1n N >时，()1(2p n p i i i εξξ∞=-<∑。
2015年博士生入学考试外语真题
2015年博士生入学考试外语真题中国社会科学院研究生院2015年攻读博士学位研究生入学考试试卷英语2015年3月14 日8:30 – 11:30PART I: Vocabulary and GrammarSection A (10 points)Directions: Choose the answer that best fills in the blank.1. Even the president is not really the CEO. No one is. Power in a corporation is concentrated and vertically structured. Power in Washington is ______________ and horizontally spread out.a. prudentb. reversiblec. diffused. mandatory2. In describing the Indians of the various sections of the United States at different stages in their history, some of the factors which account for their similarity amid difference can be readily accounted for, others are difficult to _______________.a. refineb. discernc. embedd. cluster3. The partial transfer of legislative powers from Westminster, implemented by T ony Blair, wasdesigned to give the other members of the club a bigger ______________ and to counter centrifugal forces that seemed to threaten the very idea of the union.a. sayb. transmissionc. decayd. contention4. It can hardly be denied the proliferation of so-called dirty books and films has, to date, reached almost a saturation point. People do not acknowledge the _______________ fact that children are bound to be exposed to ―dirty words‖in a myriad of ways other than through the public airwaves.a.i rrefutableb. concretec. inevitabled. haphazard5. A condition is an essential term of the contract. If a contract is not performed, it may constitute a substantial breach of contract and allow the other party to _______________ the contract, that is, treat the contract as discharged or terminated.a. repudiateb. spurnc. declined. halt6. Each of us shares with the community in which we live a store of words as well as agreed conventions ______________ these words should be arranged to convey a particular message. a. as the way by which b. by the way in whichc. as to the way in whichd. in the way of which7. Rarely ______________ a technological development _______________ an impact on many aspects of social, economic, and cultural development as greatly as the growth of electronics.a. has… hadb. had…hadc. has…hasd. have…had8. If early humans ______________ as much as they did, they probably ______________ to evolve into different species.a. did not move and intermingle…would continueb. would not move and intermingle…had continuedc. had not moved and intermingled…would have continuedd. were not to move and intermingle…could have continued9. It was ______________ the last time around the track ______________ I really kicked itin--passing the gossiping girlfriends, blocking out the whistles of boys who had already completed their run and now were hanging out on the grassy hill, I ran--pushing hard, breathing shallowly, knowing full well that I was going to have to hear about it from my disapproving friends for the next few days.a. not until…whenb. not until…thatc. until…whend. until…that10．One impediment ______________ the general use of a standard in pronunciation is the fact ______________ pronunciation is learnt naturally and unconsciously, while orthography is learnt deliberately and consciously.a. in…whichb. of …in whichc. on…thatd. to…thatSection B (5 points)Directions: Choose the word that is the closest in meaning with the underlined word.11. It is some 15 million Hindus, Sikhs and Muslims swept up in a tumultuous shuffle of citizensbetween India and Pakistan after the partition of the subcontinent in 1947.a. divisionb. turmoilc. fusiond. consolidation12. Concerning speculation, philosophy looks upon things from the broadest possible perspective;for criticism, it has the twofold role of questioning and judging everything that pertains either to the foundations or to the superstructure of human thinking.a. inebriatesb. forsakesc. relatesd. emaciates13. Meeting is, in fact, a necessary though not necessarily productive psychological side show.Perhaps it is our civilized way to moderating，if not preventing, change.a. promotingb. impedingc. temperingd. arresting14. The truth about alliances and their merit probably lies somewhere between the travel utopiapresented by the players and the evil empires portrayed bytheir critics.a. collaborationb. worthc. triumphd. defect15. But Naifeh and Smith reveal a keen intellect, an avid reader and a passionate observer of otherartists’ work who progressed from labored figure studies to inspir ed outbursts of creative energy.Far from an artistic flash in the pan, he pursued his calling with dogged determination against nearly insurmountable odds.a. insuperableb. unsurpassablec. uncountabled. invaluableSection C (5 points)Directions: Choose the letter that indicates the error in the sentence.16.One of the most important non-legislative functions of the U.S. Congress is the power toinvestigate, which is usually delegated to committees—either standing committees, specialA Bcommittees set for a specific purpose, or joint committees consisting of members of bothC Dhouses.17.One of the important corollaries to the investigative power is the power to publicizeinvestigations and their results. Most committee hearings areopen to public and are reportedA Bwidely in the mass media. Congressional investigations thus represent one important toolCavailable to lawmakers to inform the citizenry and to arouse public interest in national issues.D18.It is not a voice we recognize at once, whereas our own handwriting is something which weA Balmost always know. We begin the natural learning of pronunciation long before we start Clearning to read or write, and in our early years we go on unconsciously imitating andDpracticing the pronunciation of those around us for many hours everyday.19. It had happened too often that the farmers sold their wheat soon after harvest when farm debtsAwere coming due, only to see prices rising and speculators getting rich. On various occasions,Bproducer groups, asked firmer control, but the government had no wish to become involved, atCleast not until wartime when wheat prices threatened to run wild.D20.Detailed studies of the tribe by the food scientists at the University of London showed thatAgathering is a more productive source of food than is hunting. An hour of hunting yields onB Caverage about 100 edible calories as an hour of gathering produces 240.DPART II: Reading comprehension (30 points)Directions: Choose the best answers based on the information in the passages below. Passage 1Plato’s Republic has been the source of great consternation, especially in literary circles, for itsattack on the poets. Socrates in fact asserts that they should have no place in the ideal state. Eric Havelock suggests that there are several misunderstandings in this regard, and in his Preface to Plato he identifies the issues, explains the historical context.Havelock opens his discussion by suggesting that the very title of the Republic is the source of much confusion. The book is commonly understood to be a treatise on the ideal political entity, but even a casual analysis will show that only one-third of the text is concerned with statecraft. The other two-thirds cover a variety of su bjects, but the thrust of Plato’s argument amounts to an attack on the traditional Greek approach to education.The educational methods still in use in the 4th century BC had their origins in what has been called the Greek Dark Age beginning around 1200 BC when the Mycenaean era collapsed. Very little is known about the whys and wherefores of this collapse, but it wasn’t until around 700 BC that the Phoenicianalphabet began to be adapted and used in the Greek-speaking world. During the intervening centuries, all knowledge concerning Greek history, culture, mores and laws were orally transmitted down through the generations. The most effective device in aid of memorizing vast amounts of information was rhyme. The epic form we see in Homer’s Iliad grew out of the need to preserve the Greek cultural memory. Havelock takes the reader through Book 1 of The Iliad and dissects it in detail to show how this cultural, historical and ethical heritage was conveyed. The Iliad takes on new and significant meaning to the reader of this minute examination.The Iliad and presumably other poetic vehicles were taught to children from an early age. The whole of the Greek-speaking world was immersed in the project of memorizing, and out of the masses arose those individuals with superior memories and theatrical skills who became the next generation of minstrels and teachers. Education was thus comprised of memorization and rote learning, and the people enjoyed constant reminders through public readings and festivals.Plato’s focus in the Republic and elsewhere is on Homer and Hesiod and to some extent the dramatists which at the time were the centerpieces of the educational regime. Their works presented gods and heroes as fundamentally immoral and thus bad examples for youth. The overall result is that the Greek adolescent is continually conditioned to an attitude which at bottom is cynical. It is more important to keep up appearances than to practice the reality. Decorum and decent behavior are not obviously violated, but the inner principle of morality is. Once the Republic is viewed as a critique of the educational regime, Havelock says that the logic of its total organization becomesclear.What Pl ato was railing against was an ―oral state of mind‖ which seems to have persisted even though the alphabet and written documentation had been in use for three centuries. Illiteracy was thus stil l a widespread problem in Plato’s time, and the poetic state of mind was the main obstacle to scientific rationalism and analysis. This is why Plato regarded the poetic or oral state of mind as the arch-enemy. In his teachings he did the opposite. He ask ed his students to ―think about what they were sa ying instead of just saying it.‖The epic had become, in Plato’s view, not ―an act of creation bu t an act of reminder and recall‖ and cont ributed to what Havelock terms ―the Homeric state of mind.‖It was So crates’project (and by extension Plato’s) to reform Greek education to encourage thinking and analysis. Thus all the ranting and railing about the ―poets‖ in Plato’s Republic was limited basically to Homer and Hesiod because of what he viewed as a wholly inadequate approach to education of which these particular poets were an integral part.Unfortunately, Western culture has misconstrued wh at Plato and Socrates meant by ―the poets.‖And because we view poetry as a highly creative and elevated form of expression, our critics have failed to recognize that Plato’s diatribe had a very specific and limited target which had nothing to do with high-minded creativity, of which there is plenty, by the way, in the proscribed poet s. It wasn’t really the poets who were the problem; it was the use of them that was deemed unacceptable.Post-Havelock, we can now read the Republic with the scales lifted from our eyes and see it for what it really was: an indictment of an antiquated educational regime which had no place in a democratic society.Comprehension Questions:21. The mistaken understanding of Plato's Republic consists in the widespread belief that it consistsof _______________.a.literary criticismb. a treatise on the ideal polityc. a critique of rationalismd. an indictment of an obsolete pedagogy22. According to Havelock, Plato’s anger with the poets arose from:I: Their representation of gods and heroes as fundamentally immoral and thus bad examples for youth.II: Their transmission of culture, mores and laws.a. I.b. II.c. Both I and II.d. Neither I nor II.23. Prior to the 4th century BC, recitation was considered the best educational method because______________.a.poetry was seen as a highly creative and elevated form of expressionb.rhyme was the most effective device in aid of memorizing vast amounts of informationc.there was no writing systemd.the people enjoyed constant reminders through public readings and festivals24. In Plato's diatribe the poetic or oral state of mind is the arch-enemy of _______________.a.democratic societyb. the Mycenaean Republicc .the Phoenicians d. literacy25. A common critique of the present-day Chinese educational system resembles the educationalsystem that Plato fulminated against in that it often _______________.a.asks students to think about what they were saying instead of just saying it/doc/8e18884558.htmlprises of memorization and rote learningc.has a very specific and limited targetd.encourages thinking and analysisPassage 2To govern is to choose how the revenue raised from taxes is spent. So far so good, or bad. But some people earn more money than others. Should they pay proportionately more money to the government than those who earn less? And if they do pay more money are they entitled to more services than those who pay less or those who pay nothing at all? And should those who pay nothing at all because they have nothing get anything? These matters are of irritable concern to ourrulers, and of some poignancy to the rest.Although the equality of each citizen before the law is the rock upon which the American Constitution rests, economic equality has never been an American ideal. In fact, it is the one unmentionable subject in our politics, as the senator from South Dakota recently discovered when he came up with a few quasi-egalitarian tax reforms. The furious and enduring terror of Communism in America is not entirely the work of those early cold warriors Truman and Acheson.A dislike of economic equality is something deep-grained in the American Protestant character. After all, given a rich empty continent for vigorous Europeans to exploit (the Indians were simply a disagreeable part of the emptiness, like chiggers), any man of gumption could make himself a good living. With extra hard work, any man could make himself a fortune, proving that he was a better man than the rest. Long before Darwin the American ethos was Darwinian.The vision of the rich empty continent is still a part of the American unconscious in spite of the Great Crowding and its attendant miseries; and this lingering belief in the heaven any man can make for himself through hard work and clean living is a key to the majority’s prevailing and apparently unalterable hatred of the poor, kept out of sight at home, out of mind abroad.Yet there has been, from the beginning, a significant division in our ruling class. The early Thomas Jefferson had a dream: a society of honest yeomen, engaged in agricultural pursuits, without large cities, heavy industry, banks, military pretensions. The early (and the late) Alexander Hamilton wanted industry, banks, cities, and a military force capable of making itself felt in world politics. It is a nice irony that so many of toda y’s laissez-faire conservatives think that they descend from Hamilton, the proponent of a strong federal government, and that so many liberals believe themselves to be the heirs of the early Jefferson, who wanted little more than a police force and a judiciary. Always practical, Jefferson knew that certain men would rise through their own good efforts while, sadly, others would fall. Government would do no more than observe this Darwinian spectacle benignly, and provide no succor.In 1800 the Hamiltonian view was rejected by the people andtheir new President Thomas Jefferson. Four years later, the Hamiltonian view had prevailed and was endorsed by the reelected Jefferson. Between 1800 and 1805 Jefferson had seen to it that an empire in posse had become an empire in esse. The difference between Jefferson I and Jefferson II is reflected in the two inaugural addresses.It is significant that nothing more elevated than greed changed the Dr. Jekyll of Jefferson I into the Mr. Hyde of Jefferson II. Like his less thoughtful countrymen, Jefferson could not resist a deal. Subverting the Constitution he had helped create, Jefferson bought Louisiana from Napoleon, acquiring its citizens without their consents. The author of the Declaration of Independence was quite able to forget the unalienable rights of anyone whose property he thought should be joined to our empire—a word which crops up frequently and unselfconsciously in his correspondence.In the course of land-grabbing, Jefferson II managed to get himself into hot water with France, England, and Spain simultaneously, a fairly astonishing thing to do considering the state of politics in Napoleonic Europe.Comprehension Questions:26. The author believes that Americans ________________.a. still believe America to be largely unpopulatedb. largely believe in lower taxationc. are in favor of taxation without representationd. should reconsider the Louisiana purchase27. From the passage, we may assume that the senator from South Dakota _______________.a. opposed tax reformb. was Thomas Jeffersonc. failed in his attempt to reform tax lawd. was Alexander Hamilton28. Jefferson made it possible for ________________.a. a potential empire to become a real oneb. tax laws to reflect the will of the peoplec. France, England, and Spain to simultaneously vacillate upon their mutual feelings towardsthe United States.d. Darwinian social theories to be accepted without question29. Jefferson’s early political writings espoused what would today b e called _______________.a. collectivismb. libertarianismc. socialismd. liberalism30. The author holds that Jefferson’s purchase of the Louisiana territories _______________.a. may be seen as a hypocritical actb. rigorously held with his previous views of inalienable rightsc. cannot be seen as an act of empire-expansiond. was an act meant to lower taxes and improve the wealth of the nationPassage 3If, besides the accomplishments of being witty and ill-natured, a man is vicious into the bargain, he is one of the most mischievous creatures that can enter into a civil society. His satire will then chiefly fall upon those who ought to be the most exempt from it. Virtue, merit, and everything that is praiseworthy, will be made the subject of ridicule and buffoonery. It is impossible to enumerate the evils which arise from these arrowsthat fly in the dark; and I know no other excuse that is or can be made for them, than that the wounds they give are only imaginary, and produce nothing more than a secret shame or sorrow in the mind of the suffering person. It must indeed be confessed that a lampoon or a satire do not carry in them robbery or murder; but at the same time, how many are there that would not rather lose a considerable sum of money, or even life itself, than be set up as a mark of infamy and derision? And in this case a man should consider that an injury is not to be measured by the notions of him that gives, but of him that receives it. Those who can put the best countenance upon the outrages of this nature which are offered them, are not without their secret anguish. I have often observed a passage in Socrates’ behavio r at his death in a light wherein none of the critics have considered it. That excellent man entertaining his friends a little before he drank the bowl of poison, with a discourse on the immortality of the soul, at his entering upon it says that he does not believe any the most comic genius can censure him for talking upon such a subject at such at a time. This passage, I think, evidently glances upon Aristophanes, who write a comedy on purpose to ridicule the discourses of that divine philosopher. It has been observed by many writers that Socrates was so little moved at this piece of buffoonery, that he was several times present at its being acted upon the stage, and never expressed the least resentment of it. But, with submission, I think the remark I have here made shows us that this unworthy treatment made an impression uponhis mind, though he had been too wise to discover it. When Julius Caesar was lampooned by Catullus, he invited him to a supper, and treated him with such a generous civility, that hemade the poet his friend ever after. Cardinal Mazarine gave the same kind of treatment to the learned Quillet, who had reflected upon his eminence in a famous Latin poem. The cardinal sent for him, and, after some kind expostulations upon what he had written, assured him of his esteem, and dismissed him with a promise of the next good abbey that should fall, which he accordingly conferred upon him in a few months after. This had so good an effect upon the author, that he dedicated the second edition of his book to the cardinal, after having expunged the passages which had given him offence. Though in the various examples which I have here drawn together, these several great men behaved themselves very differently towards the wits of the age who had reproached them, they all of them plainly showed that they were very sensible of their reproaches, and consequently that they received them as very great injuries. For my own part, I would never trust a man that I thought was capable of giving these secret wounds; and cannot but think that he would hurt the person, whose reputation he thus assaults, in his body or in his fortune, could he do it with the same security. There is indeed something very barbarous and inhuman in the ordinary scribblers of lampoons. I have indeed heard of heedless, inconsiderate writers that, without any malice, have sacrificed the reputation of their friends and acquaintance to a certain levity of temper, and a silly ambition of distinguishing themselves by a spirit of raillery and satire; as if it were not infinitely more honourable to be a good-natured man than a wit. Where there is this little petulant humor in an author, he is often very mischievous without designing to be so.Comprehension Questions:31. According to the author, those who want to trivializesatire tend to suggest that_______________.a. the damage is immaterialb. the effect is mere buffooneryc. wit is a streak of geniusd. the mischief must be taken in a spirit of raillery32. What would be the best strategy for the object of satire to adopt, according to the author?a. To take no heed.b. To placate the author.c. To take offence.d. To suffer the consequences.33. The main purpose of this article is ________________.a. the derision of the perpetrators of satireb. a warning against mischievous scribblersc. creating understanding of the genred. reproaching fellow satirists34. When the author speaks of ―this little petulant humor‖it is evident that he means________________.a. good-natured witb. the choleric temperc. a silly ambitiond. submission35. In view of the opinion of the author, it is unlikely that the author is a ________________.a. man of lettersb. satiristc. witd. a good-natured man Passage 4Alexander the Great’s conquests in the Eastern Mediterranean initiated a series of profound cultural transformations in the ancient centers of urban civilization of the Fertile Crescent. The final destruction of native rule and the imposition of an alien elite culture instigated a cultural discourse—Hellenism—which irrevocably marked all participants, both conquerors and conquered. This discourse was particularly characterized by a transformation of indigenous cultural traditions, necessitated by their need to negotiate their place in a new social order. As Bowerstock has argued, the process of Hellenization did not accomplish the wholesale replacement of indigenous cultural traditions with Greek civilization. Instead, it provided a new cultural vocabulary through which much pre-existing cultural tradition was often able to find new expression. This phenomenon is especially intriguing as it relates to language and literacy. The ancient civilizations of the Syro-Mesopotamian and Egyptian cultural spheres were, of course, literate, possessing indigenous literary traditions already of great antiquity at the time of the Macedonian conquests. The disenfranchisement of traditional elites by the imposition of Greek rule had the related effect of displacing many of the traditional social structures where in indigenous literacy functioned and was taught—in particular, the institutions of the palace and the temple. A new language of power, Greek, replaced the traditional language of these institutions. This had the unavoidable effect of displacing the traditional writing systems associated with these indigenous languages. Traditional literacy’s longstanding association with the centers of social and political authority began to be eroded.Naturally, the eclipse of traditional, indigenous literacy did not occur overnight. The decline of Cuneiform and Hieroglyphicliteracies was a lengthy process. Nor was the nature of their respective declines identical. Akkadian, the ancient language of Mesopotamian court and temple culture, vanished forever, along with cuneiform writing, in the first century CE. Egyptian lived on beyond the disappearance of hieroglyphic in the fourth century CE in the guise of Coptic, to succumb as a living, spoken language of daily social intercourse only after the Islamic conquest of Egypt. Even then, Coptic survives to this day as the liturgical language of the Coptic Orthodox Church. This latter point draws attention to an aspect of the decline of these indigenous literacies worthy of note: it is in the sphere of religion that these literacies are often preserved longest, after they have been superseded in palace circles—the last dated cuneiform text we have is an astrological text; the last dated hieroglyphic text a votive graffito. This should cause little surprise. The sphere of religion is generally one of the most conservative of cultural subsystems. The local need to negotiate the necessities of daily life and individual and collective identity embodied in traditional religious structures is slow to change and exists in ongoing dialogue with the more readily changeable royal and/or state ideologies that bind various locales together in an institutional framework.The process of ―Hellenization‖ of the an cient cultures of the Eastern Mediterranean provides us, then, with an opportunity to observe the on-going effect on traditional, indigenous literacy of the imposition of a new status language possessed of its own distinct writing system. The cultural politics of written and spoken language-use in such contexts has been much discussed and it is clear that the processes leading to the adoption of a new language—in written form, or spoken form, or both—in some cultural spheres and the retention of traditional languages inothers are complex. Factors including the imposition of a new language from above, adoption of a new language of social prestige from below, as well as preservation of older idioms of traditional statusin core cultural institutions, must have affected different sectors of a conquered society in different fashions and at different rates.Comprehension Questions:36. The languages that have to some extent managed to survive Hellenization did so in what area?a. In palace circles.b. In governmental institutions.c. In the religious sphere.d. In philological circles.37. Which aspect of society, according to the passage, is one of the most resistant to change?a. Monarchical institutions.b. Religious institutions.c. Linguistic norms.d. State ideologies.38. In the first paragraph, you saw the underlined word disenfranchisement. Choose, among thefollowing expressions, the closest in similar meaning.a. the removal of power, right and/or privilegeb. a strong sense of disappointmentc. the prohibition of the right to conduct businessd. the loss of social position39. Who was the leader of the Macedonian Conquest?a. King Philip of Macedon.b. Pericles of Athens.。
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f ( x ) g ( x )dx M
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1 1 1 ，证明 f Lp () 并且 ‖f‖ M. Lp ( ) p q
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试题编号
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5． （15 分）设 X 为紧的度量空间，证明在 X 上存在 Borel 测度 使得对 X 上的任何非负连 续函数 f ，并且 f 0 ，有
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2． （20 分）设 (, ) 为正测度空间， () ，再设 { f n } L (, ), 1 p ，满 足如下条件： （i）存在 上的可测函数 f 使得 { f n } 在 上几乎处处收敛于 f ； （ii）对任意 0 ，存在 M 0 ，使得对任何 n 有
南京大学 2015 年博士学 数 学 满分： 分
注意：①所有答案必须写在答题纸或答题卡上，写在本试题纸或草稿纸上均无效； ②本科目不允许使用计算器。 1． （15 分）求
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n