Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Functional Analysis of Peter Lax there are the following exercise

Show that if $\bf C$ is compact and $\{{\bf M}_n \}$ tends strongly to $\bf M$, then $\bf CM_n$ tends uniformly to $\bf CM$.

Assumptions are that ${\bf C}: {\bf X} \rightarrow {\bf X},{\bf M_n}: {\bf X} \rightarrow {\bf X}$ where $\bf X$ is a Banach space

I was thinking that one could use that if let $x_i$ be such that $|({\bf M_i-M})x_i|\ge||{\bf M_i-M}|| - \epsilon $. Then from compactness of $\bf C$ we have that there is a finite sequence $j=1\ldots,n$ of ${\bf C}({\bf M_j-M})x_j$ s.t $\min_j ||{\bf C}({\bf M_j-M})x_j - {\bf C}({\bf M_i-M})x_i||<\epsilon$ for all i. But I dont get anywhere.

share|improve this question
    
Maybe try to approximate $C$ by suitable finite rank operators? (Assuming you work in Hilbert space) It might also be useful to note also that a strongly convergent sequence must be norm-bounded by uniform boundedness. –  Martin Jan 2 '13 at 13:43
    
@Davide from X->X, where X is Banach. –  Jonas Wallin Jan 2 '13 at 13:58
    
@Martin not in a Hilbert space. Yes, norm-bounded might be needed. –  Jonas Wallin Jan 2 '13 at 14:01
add comment

2 Answers 2

As Jonas Wallin points out in his answer, it's not necessarily true that $CM_n-MC\to 0$ uniformly even in the context of Hilbert spaces. It's however true in finite dimensional vector spaces, as strong convergence is the same thing as uniform convergence in this particular case.

However, maybe Lax meant to show that $M_nC-MC\to 0$ uniformly. In this case, we can follow the following steps.

  1. We assume that $M=0$, otherwise consider $M_n-M$.
  2. Using the principle of uniform boundedness, we obtain that $R:=\sup_{n\in\Bbb N}M_n$ is finite.
  3. Let $\delta$ such that $\delta\leqslant\limsup_{n\to +\infty}\lVert M_nC\rVert$. Let $\{n_k\}$ a strictly increasing sequence of integers such that $\delta\leqslant \lVert M_{n_k}C\rVert$, and $x_k$ of norm $1$ such that $\lVert M_{n_k}C\rVert\leqslant k^{-1}+\lVert M_{n_k}Cx_k\rVert$.
  4. The sequence $\{Cx_k\}$ lies in a compact set, hence we extract a converging subsequence $\{x_{k'}\}$ to some $y$.
  5. We have for all $k'$ that $$\delta\leqslant k'^{-1}+\lVert M_{n_{k'}}(Cx_{k'}-y)+M_{n_{k'}}y\rVert\leqslant k'^{-1}+R\lVert Cx_{k'}-y\rVert+\lVert M_{n_{k'}}y\rVert.$$
  6. Taking the limit $k'\to +\infty$ in the later inequality, we get that $\delta=0$.
share|improve this answer
1  
Actually Lax's question was two show it both ways that is both $M_nC-MC \rightarrow 0 $ and $CM_n-CM \rightarrow 0$. –  Jonas Wallin Jan 17 '13 at 12:42
    
So compact operators correct strong convergence only "by the right side". –  Davide Giraudo Jan 17 '13 at 14:46
add comment

I believe that the problem is wrong. Here is an counter example:

Assume that ${\bf X}=l^2$, ${\bf C}x=(x,e_1)e_1$ and ${\bf M}_nx=(x,e_n)e_1$.

${\bf M_n} x\rightarrow 0$ for all $x$, so ${\bf M_n} \rightarrow {\bf 0}$.

But $||{\bf C M}_n - {\bf C 0}|| = ||{\bf C M}_n|| \geq ||{\bf C M}_ne_n||=||(e_n,e_n)(e_1,e_1)e_1||=1.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.