泛函分析英文习题2

Prof.Dr.Wolfgang Arendt Wintersemester2005/2006 Dr.Delio Mugnolo Ulm,den25.10.2005
Functional Analysis
Exercise5.Consider the space of thefinite sequences
c00:={x∈ ∞:x n=0for almost every n}.
Show that c00is dense in c0with respect to · ∞.
Solution:First observe that c00is a subspace of the Banach space c0,since every vector of c00is a sequence that only hasfinitely many non-zero coordinates.The Banach space c0is a subspace of the Banach space ∞with respect to the induced norm · ∞,hence it is closed.Thus,c00⊂c0=c0.
In order to check the converse inclusion,i.e.,that every vector of c0is limit(with respect to · ∞)of a suitable sequence in c00,take x=(x n)n∈N∈c0.Consider the sequence(x k)k∈N=((x k n)n∈N)k∈N defined as follows:
x1:=(x1,0,0,...),
x2:=(x1,x2,0,0,...),
x3:=(x1,x2,x3,0,0,...),
..
.
Since for all k∈N the sequence x k has at most k non-zero coordinates,x k∈c00. Moreover,
x k−x ∞=sup
n∈N |x k n−x n|=sup
n≥k
|x n|for all k∈N.
Since by assumption x∈c0,letting k→∞we conclude that lim k→∞ x k−x ∞=0. Exercise6.Consider the space C1[0,1]of all continuously differentiable functions on[0,1]equipped with the norm| · |introduced in the Exercise3.Show that
C1[0,1]is a Banach space.
(Hint:Recall that f(t)=f(0)+ t
f (s)ds(t∈[0,1])and that if lim n→∞f n=f
in C[0,1],then also lim n→∞ t
f n(s)ds=
1
f(s)ds.)
Solution:Let(f n)n∈N be a Cauchy sequence in C1[0,1]with respect to| · |. Then,it is clear that(f n)n∈N is a Cauchy sequence in C[0,1]with respect to · ∞. Moreover,
|f n(x)−f m(x)|≤|f n(0)−f m(0)|+|
x
f n(s)−f m(s)ds|
≤|f n(0)−f m(0)|+|
1
f n(s)−f m(s)ds|=| f n−f m |
for all x∈[0,1]and all n,m∈N.Taking the supremum over x∈[0,1]one concludes that
f n−f m ∞≤| f n−f m |
for all n,m∈N,hence also(f n)n∈N is a Cauchy sequence in C[0,1]with respect to · ∞.Since(C[0,1], · ∞)is a Banach space,we conclude that(f n)n∈N is a sequence of continuous functions such that both(f n)n∈N and(f n)n∈N converge uniformly to f,g∈C[0,1],respectively.It is then well-known that f∈C1[0,1] and f =g:this shows that the Cauchy sequence(f n)n∈N actually converges with respect to| · |.
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Exercise7.Let y∈ 1and x k∈ 1such that|x k n|≤y n for all k∈N.Show that if lim k→∞x k n=x n,then there exists x∈ 1such that lim n→∞x k=x with respect to the norm of 1.
Solution:Since|x k n|≤y n for all n,k∈N,for k→∞it also holds|x n|≤y n:thus,
∞
n=1x n≤
∞
n=1
y n= y 1<∞.
We can therefore say that x:=(x n)n∈N∈ 1.Accordingly,
lim k→∞ x k−x =lim
k→∞
∞
n=1
|x k n−x n|=
∞
n=1
lim
n→∞
|x k n−x k|=0,
where we have used the fact that x k−x is absolutely summable for all k∈N. Exercise8.Consider the sequence(e k)k∈N defined in Exercise1.Show that it does not have any convergent subsequence.Deduce that the unit ball of 1is not compact in ∞.
Solution:For all k∈N one has e k 1=1,thus the sequence(e k)k∈N is contained in the unit ball of 1.Moreover,it is known from Exercise1that(any subsequence of)the sequence(e k)could only converge to its coordinatewise limit,i.e.,to the zero sequence.However, e k ∞=1for all k∈N,so that in fact no subsequence of (e k)k∈N converges with respect to · ∞.This yields the claim.。