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This was the proof from a textbook "Hilbert Space Operators in Quantum Physics" by Jiří Blank, Pavel Exner, Miloslav Havlíček.

3.5.1 Theorem: An operator is compact iff it maps any weakly convergent sequence to a convergent one.

Proof: Suppose that $C\in K(H)$ and $ x_n \to^{w} x$ does not imply $Cx_n → Cx.$ Then there is a positive and a growing sequence $\{n_k\}$ of natural numbers such that $||Cx - y_k||\ge ε$ holds for $y_k = Cx_{n_k}, k = 1, 2,....$ $\{x_{n_k}\}$ is bounded, so one can select from $\{y_k\}$ a subsequence $\{y_{k_j}\}$ which converges to some $y.$ This means, in particular, $y_{k_j} \to^{w} y$ . On the other hand $Cx_n\to^{w} Cx$ since $C$ is bounded, so together we get $y = Cx$ in contradiction with the assumption. The opposite implication follows from the reflexivity of H one can select a weakly convergent subsequence $\{x_{n_k}\}$ from any bounded $\{x_n\}\subset H$, and $\{Cx_{n_k}\}$ converges by assumption.

My attempt of understanding the proof:- We are trying to prove using the method of contradiction. Suppose that $C\in K(H)$(This means $C$ is a compact operator.) and $ x_n \to^{w} x$ does not imply $Cx_n → Cx.$

We know that $Cx_n$does not converge to $Cx.$ Hence there exists a subsequence of $x_n$ such that there is a positive and a growing sequence $\{n_k\}$ of natural numbers such that $||Cx - y_k||\ge ε$ holds for $y_k = Cx_{n_k}, k = 1, 2,....$ (Am I correct?) Using the logic given below:-

We know that $x_n\to^w x \implies \{x_n\}$ is a bounded sequence. Hence $\overline{\{Cx_n\}}$ is a compact set.(Because $C$ is a compact set.) Hence, $\{Cx_n\}$ is bounded. $\{Cx_n\}$ is a sequence in Hilbert space. Hence, it is a sequence in complete metric space. Hence, Bounded sequence has convergent subsequence.

$\{x_{n_k}\}$ is bounded, so one can select from $\{y_k\}$ a subsequence $\{y_{k_j}\}$ which converges to some $y.$

Does convergence implies weakly convergence? why do we use weakly convergence in the range space? We kneed to prove $Cx_n$ converges to $Cx$ strongly. Right?

I am not able to understand the proof. Could you help me to understand the proof better?

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  • $\begingroup$ The argument is: the compactness of $\overline{\{Cx_n\}}$ implies that there is a converging subsequence $y_{n_k}$. In general, a bounded subsequence does not have a stronlgy converging subsequence. $\endgroup$
    – daw
    Commented Dec 21, 2022 at 7:08

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Convergence always implies weak convergence . If $x_{n}\to x$ then $f(x_{n})\to f(x)$ for any continuous linear functional just by continuity.

"Hence, it is a sequence in complete metric space. Hence, Bounded sequence has convergent subsequence. "

This is wrong. This is true for finite dimensional normed spaces and is called the Bolzano Weiresstrass theorem. You can just check this yourself as it would imply that any sequence in the unit ball has a convergent subsequence and thus the unit ball would be compact which is not possible. The unit ball is compact if and only if the normed space is finite dimensional. Many proofs are available of this fact . You can try it yourself using Riesz Lemma.

Here is the complete proof of the Theorem

Let $x_{n}$ be an arbitrary sequence such that $||x_{n}||\leq 1$ . So $C(x_{n})$ is an arbitrary sequence in $C(B_{H})$ (the image of the unit ball) .

Using your very own question here you have that a bounded sequence in a Hilbert Space has a weakly convergent subsequence. Let $x_{n_{k}}$ be a weakly convergent subsequence of $x_{n}$ .

Then as $C$ maps weakly convergent sequences to convergent ones, we have $C(x_{n_{k}})$ is convergent. Thus any sequence in $C(B_{H})$ has a convergent subsequence and thus $C(B_{H})$ is sequentially compact . Thus $C$ is a compact operator.

Conversely let $C$ be a compact operator. Let $x_{n}\rightharpoonup x$ . Then $x_{n}$ is bounded by Uniform Boundedness Principle and as $C$ is bounded, $C(x_{n})$ is bounded. Let $C(x_{n_{k}})$ be an arbitrary subsequence of $C(x_{n})$ . Then by compactness of $C$, there exists a further subsequence $C(x_{n_{k_{l}}})$ that converges to some $z$ say.

But for any continuous linear functional $g\in Y^{*}$ you have $g\circ C\in X^{*}$ . Thus $g\circ C(x_{n})\to g\circ C(x)$ . Thus $C(x_{n})\rightharpoonup C(x)$. Thus as convergence implies weak convergence, you have $C(x_{n_{k_{l}}})\rightharpoonup z$ . But by uniqueness of weak limit, you have $z=C(x)$ and hence $C(x_{n_{k_{l}}})\to z= C(x)$

Here we can use a basic result from real analysis that a sequence $x_{n}$ converges to $x$ in a metric space if and only if given any subsequence $x_{n_{k}}$ of $x_{n}$ there exists a further subseuqnece $x_{n_{k_{l}}}$ such that $x_{n_{k_{l}}}\to x$. Thus using this result we get $C(x_{n})\to C(x)$ .

The proof of the above is exactly the fact that is being used in your proof. Suppose $x_{n}$ does not converge to $x$. Then there exists a subsequence $x_{n_{k}}$ and a fixed $\epsilon >0$ such that $d(x_{n_{k}},x)\geq \epsilon\,,\forall k\in\Bbb{N}$ . But $x_{n_{k}}$ has a subsequence $x_{n_{k_{l}}}\to x$ . Thus there exists $L$ such that $l\geq L\implies d(x_{n_{k_{l}}},x)<\frac{\epsilon}{2}$ which is a contradiction. Thus $x_{n}\to x$. Conversely if $x_{n}\to x$ then for any subseuquence $x_{n_{k}}\to x$.

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  • $\begingroup$ Hi, you have proved in the direct implication that any sequence in $C(B_H)$ has a convergent subsequence, but is it correct that you have not proved here that the limit of this subsequence needs to be in $C(B_H)$? I suppose the way to show this is to let $x$ be the limit of the weakly convergent $x_{n_k}$ and then $\lVert x \rVert \leq \lim \! \inf_{k \to \infty} \lVert x_{n_k} \rVert \leq 1$, but this is not so trivial and would need the uniform boundedness principle applied to the dual I believe. Or is there a shorter argument? $\endgroup$ Commented Jan 30 at 19:11
  • $\begingroup$ @NeckverseHerdman The limit of the subsequence need not be in $C(B_{H})$. Definition of compact operator $T$ is that $\overline{T(B)}$ should be compact. And obviously if $C(x_{n})$ converges then the limit lies in $\overline{T(B)}$. In the context of Banach spaces (or complete metric spaces) this amounts to only showing $C(B)$ is totally bounded by showing every sequence has a Cauchy subsequence. $\endgroup$ Commented Jan 30 at 20:28
  • $\begingroup$ Okay, that clears it up a bit, but shouldn't we show $\overline{C(B_H)}$ is compact by taking the sequence in $\overline{C(B_H)}$? In the argument you take the sequence in $C(B_H).$ I understand that when $C(x_{n_k})$ converges then the limit lies in $\overline{C(B_H)}$. $\endgroup$ Commented Jan 30 at 20:47
  • $\begingroup$ Why don't you brush up on the theory of compactness for metric spaces? These are basic facts that a set is totally bounded if and only if it's closure it totally bounded. So to show total boundedness (i.e. sequential precompactness), you just take sequence from the set itself and show that it has a cauchy subsequence. It does not matter whether you take it from the set itself or it's closure. These are assumed facts while studying functional analysis and one just does the shorter steps instead of justifying all details from a first analysis course once again. @NeckverseHerdman $\endgroup$ Commented Jan 30 at 21:01
  • $\begingroup$ @NeckverseHerdman Additionally, look up the wikipedia defintion of compact operators. To be precise, I am using the fifth bullet point/characterization there. $\endgroup$ Commented Jan 30 at 21:14

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