张恭庆 泛函分析资料报告上册问题详解

1.1.51.1.61.1.71.2.21.2.31.2.41.3.31.3.41.3.51.3.71.3.81.3.91.4.11.4.5-61.4.91.4.111.4.121.4.131.4.141.4.151.4.171.5.1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)．1.5.3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．[Schauder定理：B*空间中闭凸集C上使T(C)列紧的连续自映射T必有不动点] 1.5.41.5.5证明：设C = {x = (x1, x2, ..., x n)∈n | ∑ 1 ≤i ≤n x i = 1，x i ≥ 0 ( i = 1, 2, ..., n) }．则C是有界闭集，且是凸集，因此C是紧凸集．因为∀x∈C，x i 不全为0，而a ij> 0，故Ax的各分量也非负但不全为零．∀x∈C，设f (x) = (Ax)/( ∑ 1 ≤i ≤n (Ax)i )，则f (x)∈C．容易验证f : C→C还是连续的．由Brouwer不动点定理，存在f的不动点x0∈C．即f (x0) = x0，也就是(Ax0)/( ∑ 1 ≤i ≤n (Ax0)i ) = x0．令λ= ∑ 1 ≤i ≤n (Ax0)i，则有Ax0 = λ x0．1.5.6证明：设B = { u∈C[0, 1] | ⎰[0, 1]u(x) dx = 1，u(x) ≥ 0 }，则B是C[0, 1]中闭凸集．设max (x, y)∈[0, 1]⨯[0, 1]K(x, y) = M，min (x, y)∈[0, 1]⨯[0, 1]K(x, y) = m，⎰[0, 1] (⎰[0, 1]K(x, y) dy) dx = N，max x∈[0, 1] | ⎰[0, 1]K(x, y) dy |= P．令(S u)(x) = (⎰[0, 1]K(x, y) u(y) dy)/(⎰[0, 1] (⎰[0, 1]K(x, y) u(y) dy) dx )则⎰[0, 1] (S u)(x) dx = 1，u(x) ≥ 0；即S u∈B．因此S是从B到B的映射．∀u, v∈B，|| ⎰[0, 1]K(x, y) u(y) dy -⎰[0, 1]K(x, y) v(y) dy ||= || ⎰[0, 1]K(x, y) (u(y)-v(y)) dy ||= max x∈[0, 1] | ⎰[0, 1]K(x, y) (u(y)-v(y)) dy |≤M·|| u -v ||；因此映射u #⎰[0, 1]K(x, y) u(y) dy在B上连续．类似地，映射u #⎰[0, 1] (⎰[0, 1]K(x, y) u(y) dy) dx也在B上连续．所以，S在B上连续．下面证明S(B)列紧．首先，证明S(B)是一致有界集．∀u∈B，|| S u || = || (⎰[0, 1]K(x, y) u(y) dy )/(⎰[0, 1] (⎰[0, 1]K(x, y) u(y) dy) dx )||= max x∈[0, 1] | ⎰[0, 1]K(x, y) u(y) dy |/(⎰[0, 1] (⎰[0, 1]K(x, y) u(y) dy) dx )≤ (M ·⎰[0, 1]u(y) dy |/(m ⎰[0, 1] (⎰[0, 1]u(y) dy) dx ) = M/m，故S(B)是一致有界集．其次，证明S(B)等度连续．∀u∈B，∀t1, t2∈[0, 1]，| (S u)(t1) - (S u)(t2)|= | ⎰[0, 1]K(t1, y) u(y) dy-⎰[0, 1]K(t2, y) u(y) dy |/(⎰[0, 1] (⎰[0, 1]K(x, y) u(y) dy) dx )≤⎰[0, 1] | K(t1, y) -K(t2, y) | u(y) dy /(m⎰[0, 1] (⎰[0, 1]u(y) dy) dx )≤ (1/m) ·max y∈[0, 1] | K(t1, y) -K(t2, y) |由K(x, y)在[0, 1]⨯[0, 1]上的一致连续性，∀ε > 0，存在δ> 0，使得∀(x1, y1), (x2, y2)∈[0, 1]，只要|| (x1, y1) - (x2, y2) || < δ，就有| K(x1, y1) -K(x2, y2) | < m ε．故只要| t1-t2 | < δ时，y∈[0, 1]，都有| K(t1, y) -K(t2, y) | < m ε．此时，| (S u)(t1) - (S u)(t2)| ≤ (1/m) ·max y∈[0, 1] | K(t1, y) -K(t2, y) |≤ (1/m) ·m ε = ε．故S(B)是等度连续的．所以，S(B)是列紧集．根据Schauder不动点定理，S在C上有不动点u0．令λ= (⎰[0, 1] (⎰[0, 1]K(x, y) u0(y) dy) dx．则(S u0)(x) = (⎰[0, 1]K(x, y) u0(y) dy)/λ= (T u0)(x)/λ．因此(T u0)(x)/λ= u0(x)，T u0 = λ u0．显然上述的λ和u0满足题目的要求．1.6.1 (极化恒等式)证明：∀x, y∈X，q(x + y) -q(x-y) = a(x + y, x + y) -a(x-y, x-y) = (a(x, x) + a(x, y) + a(y, x) + a(y, y)) - (a(x, x) -a(x, y) -a(y, x) + a(y, y))= 2 (a(x, y) + a(y, x))，将i y代替上式中的y，有q(x + i y) -q(x-i y) = 2 (a(x, i y) + a(i y, x))= 2 (-i a(x, y) + i a( y, x))，将上式两边乘以i，得到i q(x + i y) -i q(x-i y) = 2 ( a(x, y) -a( y, x))，将它与第一式相加即可得到极化恒等式．1.6.2证明：若C[a, b]中数|| ·||是可由某积( ·, ·)诱导出的，则数|| ·||应满足平行四边形等式．而事实上，C[a, b]中数|| ·||是不满足平行四边形等式的，因此，不能引进积( ·, ·)使其适合上述关系．数|| ·||是不满足平行四边形等式的具体例子如下：设f(x) = (x–a)/(b–a)，g(x) = (b–x)/(b–a)，则|| f || = || g || = || f + g || = || f –g || = 1，显然不满足平行四边形等式．1.6.3证明：∀x∈L2[0, T]，若|| x || = 1，由Cauchy-Schwarz不等式，有| ⎰[0, T]e- ( T-τ)x(τ) dτ |2≤ (⎰[0, T] (e- ( T-τ))2dτ ) (⎰[0, T] ( x(τ))2dτ )= ⎰[0, T] (e- ( T-τ))2dτ = e- 2T ⎰[0, T]e 2τdτ= (1-e- 2T )/2．因此，该函数的函数值不超过M = ((1-e- 2T )/2)1/2．前面的不等号成为等号的充要条件是存在λ∈，使得x(τ) = λ e- ( T-τ)．再注意|| x || = 1，就有⎰[0, T] (λ e- ( T-τ))2dτ= 1．解出λ= ±((1-e- 2T )/2)- 1/2．故当单位球面上的点x(τ) = ±((1-e- 2T )/2)- 1/2 ·e- ( T-τ)时，该函数达到其在单位球面上的最大值((1-e- 2T )/2)1/2．1.6.4证明：若x∈N⊥，则∀y∈N，(x, y) = 0．而M⊆N，故∀y∈M，也有(x, y) = 0．因此x∈M⊥．所以，N⊥⊆M⊥．1.6.51.6.6解：设偶函数集为E，奇函数集为O．显然，每个奇函数都与正交E．故奇函数集O ⊆E⊥．∀f∈E⊥，注意到f总可分解为f = g + h，其中g是奇函数，h是偶函数．因此有0 = ( f, h) = ( g + h, h) = ( g, h) + ( h, h) = ( h, h)．故h几乎处处为0．即f = g是奇函数．所以有E⊥⊆O．这样就证明了偶函数集E的正交补E⊥是奇函数集O．1.6.7证明：首先直接验证，∀c∈，S = {e2π i n x| n∈ }是L2[c, c + 1]中的一个正交集．再将其标准化，得到一个规正交集S1 = {ϕn(x) = d n e2π i n x| n∈ }．其中的d n= || e2π i n x|| (n∈)，并且只与n有关，与c的选择无关．(1) 当b–a =1时，根据实分析结论有S⊥ = {θ}．当b–a <1时，若u∈L2[a, b]，且u∈S⊥，我们将u延拓成[a, a + 1]上的函数v，使得v(x) = 0 (∀x∈(b, a + 1])．则v∈L2[a, a + 1]．同时把S = {e2π i n x| n∈ }也看成L2[a, a + 1]上的函数集．那么，在L2[a, a + 1]中，有v∈S⊥．根据前面的结论，v = θ．因此，在L2[a, b]中就有u = θ．故也有S⊥ = {θ}；(2) 分成两个区间[a, b–1)和[b–1, b]来看．在[a, b–1)上取定非零函数u(x) = 1 ( ∀x∈[a, b–1) )．记p n = ⎰[a, b–1)u(x)ϕn(x) dx．我们再把u看成是[b–2, b–1]上的函数(u在[b–2, a)上去值为0)．那么p n就是u在L2[b–2, b–1]上关于正交集S1 = {ϕn(x)| n∈ }的Fourier系数．由Bessel不等式，∑n∈ | p n |2 < +∞．再用Riesz-Fischer定理，在L2[b–1, b]中，∑n∈p n ϕn收敛．并且，若令v = -∑n∈p n ϕn，则(v, ϕn)= -p n (∀n∈)．设f : [a, b] →为：f(x) = u(x) (当x∈[a, b–1))，f(x) = v(x) (当x∈[b–1, b])．则f∈L2[a, b]，f≠θ，但( f, ϕn) = ⎰[a, b–1)f(x)ϕn(x) dx + ⎰[b–1, b]f(x)ϕn(x) dx= ⎰[a, b–1)u(x)ϕn(x) dx + ⎰[b–1, b]v(x)ϕn(x) dx= p n -p n = 0，因此，f∈S1⊥= S⊥，故S⊥≠ {θ}．1.6.8证明：( z n/(2π)1/2, z n/(2π)1/2 ) = (1/i)⎰| z | = 1 ( z n/(2π)1/2·(z*)n/(2π)1/2 )/z dz= (1/(2πi))⎰| z | = 1z n·(z*)n/z dz = (1/(2πi))⎰| z | = 1 1/z dz = 1．若n > m，则n- m - 1 ≥ 0，从z n -m - 1而解析．( z n/(2π)1/2, z m/(2π)1/2 ) = (1/i)⎰| z | = 1 ( z n/(2π)1/2·(z*)m/(2π)1/2 )/z dz= (1/(2πi))⎰| z | = 1z n·(z*)m/z dz = (1/(2πi))⎰| z | = 1z n -m - 1dz = 0．因此，{ z n/(2π)1/2 }n ≥ 0是正交规集．1.6.91.6.10证明：容易验证{e n}⋂{ f n}是正交规集，下面只证明{e n}⋂{ f n}是X的基．∀x∈X，由正交分解定理，存在x关于X0的正交分解x = y + z，其中y∈X0，z∈X0⊥．因{e n}, { f n}分别是X0和X0⊥的正交规基，故y = ∑ n∈( y, e n ) e n，z = ∑ n∈( z, f n ) f n．因z∈X0⊥，故(x, e n) = ( y + z, e n) = ( y, e n) + ( z, e n) = ( y, e n)．因y∈X0，故(x, f n) = ( y + z, f n) = ( y, f n) + ( z, f n) = ( z, f n)．故x = y + z = ∑ n∈( y, e n ) e n + ∑ n∈( z, f n ) f n= ∑ n∈( x, e n ) e n + ∑ n∈( x, f n ) f n．因此{e n}⋂{ f n}是X的正交规基．1.6.11证明：首先，令ϕk (z) = (( k +1 )/π)1/2 z k ( k≥ 0 )，则{ ϕk }k≥ 0是H2(D)中的正交规基．那么，∀u(z)∈H2(D)，设u(z) = ∑k≥ 0 a k z k，则∀k∈，有(u, ϕk) = ⎰D u(z) ·ϕk(z)*dxdy= ⎰D (∑j≥ 0 a j z j) ·ϕk(z)*dxdy= ∑j≥ 0 a j(π/( j +1 ))1/2⎰D (( j +1 )/π)1/2 z j·ϕk(z)*dxdy= ∑j≥ 0 a j(π/( j +1 ))1/2⎰Dϕj(z) ·ϕk(z)*dxdy= ∑j≥ 0 a j(π/( j +1 ))1/2 (ϕj, ϕk)= a k(π/( k +1 ))1/2．即u(z)的关于正交规基{ ϕk }k≥ 0的Fourier系数为a k(π/( k +1 ))1/2( k≥ 0 )．(1) 如果u(z)的Taylor展开式是u(z) = ∑k≥ 0 b k z k，则u(z)的Fourier系数为b k(π/( k +1 ))1/2( k≥ 0 )．由Bessel不等式，∑k≥ 0| b k(π/( k +1 ))1/2|2≤ || u || < +∞，于是有∑k≥ 0| b k|2/( k +1 ) < +∞．(2) 设u(z), v(z)∈H2(D)，并且u(z) = ∑k≥ 0 a k z k，v(z) = ∑k≥ 0 b k z k．则u(z) = ∑k≥ 0 a k(π/( k +1 ))1/2ϕk (z)，v(z) = ∑j≥ 0 b j(π/( j +1 ))1/2ϕj (z)，(u, v) = ( ∑k≥ 0 a k(π/( k +1 ))1/2ϕk (z), ∑j≥ 0 b j(π/( j +1 ))1/2ϕj (z) )= ∑k≥ 0∑j≥ 0 (a k(π/( k +1 ))1/2ϕk (z), b j(π/( j +1 ))1/2ϕj (z))= ∑k≥ 0∑j≥ 0 (a k(π/( k +1 ))1/2 ·b j*(π/( j +1 ))1/2) (ϕk (z), ϕj (z))= ∑k≥ 0 (a k(π/( k +1 ))1/2 ·b k* (π/( k +1 ))1/2) = π∑k≥ 0 (a k·b k* )/( k +1 )．(3) 设u(z)∈H2(D)，且u(z) = ∑k≥ 0 a k z k．因1/(1 -z) = ∑k≥ 0z k，1/(1 -z)2 = ∑k≥ 0 (k +1) z k，其中| z | < 1．故当| z | < 1时，有1/(1 - | z | )2 = ∑k≥ 0 (k +1) | z | k．根据(2)，|| u(z) ||2 = π∑k≥ 0 (a k·a k* )/( k +1 ) = π∑k≥ 0 | a k|2/( k +1 )．|| u ||2/(1 - | z |)2 = (π∑k≥ 0 | a k|2/( k +1 )) ·( ∑k≥ 0 (k +1) | z | k )≥ (π∑k≥ 0 | a k|2/( k +1 ) | z | k) ·( ∑k≥ 0 (k +1) | z | k )≥π ( ∑k≥ 0 ( | a k|/( k +1 )1/2 | z | k/2) ·((k +1)1/2 | z | k/2))2 (Cauchy-Schwarz不等式) = π ( ∑k≥ 0 | a k| ·| z | k )2≥π | ∑k≥ 0a k z k |2 = π | u(z)|2，故| u(z) | ≤ || u ||/(π1/2 ( 1 - | z | ))．(4) 先介绍复分析中的Weierstrass定理：若{ f n }是区域U ⊆上的解析函数列，且{ f n }在U上闭一致收敛到f，则f在U上解析．（见龚升《简明复分析》）回到本题．设{ u n }是H2(D)中的基本列．则∀z∈D，由(3)知{ u n(z) }是中的基本列，因此是收敛列．设u n(z) →u(z)．对中任意闭集F⊆D，存在0 < r < 1使得F⊆B(0, r) ⊆D．∀ε > 0，存在N∈+，使得∀m, n > N，都有|| u n-u m|| < επ1/2 ( 1 -r )．再由(3)，∀z∈F，| u n(z) -u m(z) | ≤ || u n-u m||/(π1/2 ( 1 - | z | )) ≤ || u n-u m||/(π1/2 ( 1 -r )) < ε．令m→∞，则| u n(z) -u(z) | ≤ε．这说明{ u n }在D上闭一致收敛到u．由前面所说的Weierstrass定理，u在D上解析．把{ u n }看成是L2(D)中的基本列，因L2(D)，故{ u n }是L2(D)中的收敛列．设{ u n }在L2(D)中的收敛于v．则v必然与u几乎处处相等．即{ u n }在L2(D)中的收敛于u．因此{ u n }在H2(D)中也是收敛的，且收敛于u．所以，H2(D)完备．1.6.12证明：由Cauchy-Schwarz不等式以及Bessel不等式，∀x, y∈X，有| ∑n≥ 1 (x, e n) ·(y, e n)* |2≤ (∑n≥ 1 | (x, e n) |·| (y, e n)* | )2= (∑n≥ 1 | (x, e n) |·| (y, e n) | )2≤ (∑n≥ 1 |(x, e n) |2) ·(∑n≥ 1 | (y, e n)|2)≤ || x ||2·|| y ||2．因此，| ∑n≥ 1 (x, e n) ·(y, e n)* | ≤ || x || ·|| y ||．1.6.13证明：(1) 因数是连续函数，故C = { x ∈X | || x - x0 || ≤r }是闭集．∀x, y∈C，因|| x - x0 || ≤r，|| x - x0 || ≤r }，故∀λ∈[0, 1]，|| (λ x + (1-λ) y ) - x0 || = || λ( x-x0 ) + (1-λ) (y - x0)||≤ || λ( x-x0 ) + (1-λ) (y - x0)|| ≤λ|| x-x0 || + (1-λ) || y - x0 || ≤λ r + (1-λ) r = r．所以，C是X中的闭凸集．(2) 当x ∈C时，y = x．显然y是x在C中的最佳逼近元．当x ∈C时，y = x0 + r (x - x0)/|| x - x0 ||．∀z∈C，|| x-y || = || ( x-x0 -r (x - x0)/|| x - x0 ||) ||= || (1 -r/|| x - x0 ||) (x - x0) || = || x - x0 || -r．≤ || x - x0 || - || z - x0 || ≤ || x - z||．因此，y是x在C中的最佳逼近元．1.6.14解：即是求e t在span{1, t, t2}中的最佳逼近元(按L2[0, 1]数)．将{1, t, t2}正交化为{1, t- 1/2, (t- 1/2)2 - 1/12 } (按L2[0, 1]积)再标准化为{ϕ0(t), ϕ1(t), ϕ2(t)}，则所求的a k= (e t, ϕ k(t)) = ⎰[0, 1]e tϕ k(t) dt，k = 0, 1, 2．1.6.15证明：设g(x) = (x-a) (x-b)2，则g(a) = g (b) = 0，g’(a) = (b-a)2，g’(b) = 0．由Cauchy- Schwarz不等式，我们有(⎰[a, b] | f’’(x) |2 dx)·(⎰[a, b] | g’’(x) |2 dx)≥ (⎰[a, b]f’’(x) ·g’’(x) dx )2．因g’’(x) = 3x- (a + 2b)，故⎰[a, b] | g’’(x) |2 dx = ⎰[a, b] (3x- (a + 2b))2 dx = (b-a)3；又⎰[a, b]f’’(x) ·g’’(x) dx = ⎰[a, b] (3x- (a + 2b)) ·f’’(x) dx = ⎰[a, b] (3x- (a + 2b))d f’(x)= (3x- (a + 2b)) ·f’(x)| [a, b] - 3⎰[a, b]f’(x) dx = 2(b-a)；故(b-a)3 ·⎰[a, b] | f’’(x) |2 dx ≥ (2(b-a))2 = 4(b-a)2．所以⎰[a, b] | f’’(x)|2 dx≥ 4/(b-a)．1.6.16 (变分不等式)证明：设f(x)= a(x, x) - Re(u0, x)．则f(x) = a(x, x) - Re(u0, x) ≥δ || x ||2 - | (u0, x) |≥δ || x ||2 - || u0 || ·|| x || ≥- || u0 ||2/(4δ) > -∞．即f在X上有下界，因而f在C有下确界μ = inf x∈C f(x)．注意到a(x, y)实际上是X上的一个积，记它所诱导的数为|| x ||a = a(x, x)1/2，则|| ·||a与|| ·||是等价数．因此f(x) = a(x, x) - Re(u0, x) = || x ||a2- Re(u0, x)．设C中的点列{ x n }是一个极小化序列，满足μ≤f(x n ) < μ + 1/n ( ∀n∈+ )．则由平行四边形等式，|| x n-x m ||a2 = 2(|| x n ||a2 + || x m ||a2 ) - 4|| (x n + x m)/2||a2= 2( f(x n) + Re(u0, x n) + f(x m) + Re(u0, x m) ) - 4( f((x n + x m)/2) + Re(u0, (x n + x m)/2))= 2( f(x n) + f(x m)) - 4 f((x n + x m)/2) + 2 Re( (u0, x n) + (u0, x m) - (u0, x n + x m) )= 2( f(x n) + f(x m)) - 4 f((x n + x m)/2)≤ 2( μ + 1/n + μ + 1/m ) - 4 μ= 2(1/n + 1/m) → 0 ( m, n→∞ )．因此|| x n-x m ||2≤ (1/δ) || x n-x m ||a2→ 0 ( m, n→∞ )．即{ x n }为X中的基本列．由于X完备，故{ x n }收敛．设x n→x0 ( n→∞ )．则|| x n-x0 ||a2≤M || x n-x0 ||2→ 0 ( m, n→∞ )．而由积a( ·, ·)，( ·, ·)的连续性，有a( x n , x n) →a( x0 , x0 )，且(u0, x n) → (u0, x0)，( n→∞ )．因此f(x n) = a(x n, x n) - Re(u0, x n) →a(x0, x0) - Re(u0, x0) = f(x0)，( n→∞ )．由极限的唯一性，f(x0) = μ = inf x∈C f(x)．至此，我们证明了f在C上有最小值．下面说明最小值点是唯一的．若x0, y0都是最小值点，则交错的点列{ x0, y0, x0, y0, x0, ... }是极小化序列．根据前面的证明，这个极小化序列必须是基本列，因此，必然有x0 = y0．所以最小值点是唯一的．最后我们要证明最小点x0∈C满足给出的不等式．∀x∈C，∀t∈[0, 1]，有x0 + t ( x - x0)∈C，因此有f(x0 + t ( x - x0)) ≥f(x0)．即|| x0 + t ( x - x0) ||a2- Re(u0, x0 + t ( x - x0)) ≥ || x0 ||a2- Re(u0, x0)．展开并整理得到t Re ( 2a(x0, x - x0) - (u0, x - x0) ) ≥-t2 || x - x0 ||a2．故当∀t∈(0, 1]，有Re ( 2a(x0, x - x0) - (u0, x - x0) ) ≥-t|| x - x0 ||a2．令t→ 0就得到Re ( 2a(x0, x - x0) - (u0, x - x0) ) ≥ 0．2.1.22.1.32.1.42.1.52.1.62.1.72.1.82.1.92.2.22.2.52.3.12.3.3-22.3.42.3.52.3.72.3.82.3.92.3.112.3.122.3.132.3.142.4.42.4.52.4.62.4.72.4.82.4.92.4.102.4.112.4.122.4.132.4.142.5.42.5.52.5.72.5.82.5.10。