# Compact operator in Hilbert spaces reach the maximum in the sphere.

I found the following question in my textbook:

(QUESTION) Let $\mathcal{H}$ a Hilbert space and $T: \mathcal{H} \rightarrow \mathcal{H}$ a compact operator. Show that exists $x \neq 0$ in $\mathcal{H}$, such that

$\|Tx\| = \|T\|\|x\|$

A few days ago, I found a similar question in another textbook, but I had $f: \mathcal{H} \rightarrow \mathbb{R}$ bounded. This one, I can use Hanh-Banach theorem to show that exists $x \in \mathcal{H}$, such that $f(x) = \|f\| \|x\|$.

Thanks.

• What is your question? Commented Feb 3, 2016 at 19:36
• by definition a compact operator is bounded and $||T|| = \max_{||x||=1} ||T x||$ (which is finite because the operator is bounded), so I guess you have a different definition for a compact operator and you want to prove that it implies it is bounded ? Commented Feb 3, 2016 at 20:34

## 4 Answers

We will prove that $\exists x_0\in S(0,1)=\{x\in H:\|x\|=1\}:\|Tx\|=\|T\|$.

You know that $\|Tx\|\leq \|T\|\,\forall x\in B(0,1)\Rightarrow \exists$ a sequence $\{x_n\}\subset B(0,1): \|Tx_n\|\to \sup\limits_{x\in B(0,1)} \|Tx\|=:\|T\|$. From this sequence you can chose a weakly convergent subsequence, still denoted by $\{x_n\}$, in $B(0,1)$ , say convergent to $x_0\in B(0,1)$ (because $B(0,1)$ is convex and closed and $H$ is reflexive $\Rightarrow B(0,1)$ is weakly compact). But $T$ maps weakly convergent sequences to strongly convergent ones, so $\|Tx_n-Tx_0\|\to 0$. Therefore we have $\|Tx_n\|\to \|Tx_0\|$ and also $\|Tx_n\|\to \sup\limits_{x\in B(0,1)} \|Tx\|\Rightarrow \|Tx_0\|=\sup\limits_{x\in B(0,1)}{\|Tx\|}=\|T\|$, which follows by the uniqueness of limits. It is only left to show that $\|x_0\|=1$. This is done by observing $\|T\|=\|Tx_0\|\leq \|T\|\|x_0\|\Rightarrow \|x_0\|\ge 1$.

If you prove that the image of the closed unit ball of $\mathcal H$ by $T$ is compact in $\mathcal H$, then there exists $x\in \mathcal H$ with $\|x\|=1$ such that $\|Tx\|=\|T\|.$

Indeed, since $T(B_\mathcal H)$ is compact, the norm $\|\cdot\|$ atains its maximum on $T(B_\mathcal H)$. Therefore, the set $\{\|Tx\|:\; x\in B_\mathcal H\}$ is closed. This means that the supremum when calculating operator norm is attained.

To prove that $T(B_\mathcal H)$ is compact note first that $T(B_\mathcal H)$ is a convex set. Therefore, its norm closure coincides with its weak closure. Since $B_\mathcal H$ is weakly compact due to the reflexivity of the space $\mathcal H$, and since $T$ is also weakly continuous, then $T(B_\mathcal H)$ is weakly compact as well. Therefore it is also weakly closed. By the argument above it is closed. Since $T$ is compact, $T(B_\mathcal H)$ is relatively compact. And finally, since it is closed, it needs to be compact proving the claim.

• $T(B_{H})$ need not be norm closed in $H$. It is only a precompact set, which means that after norm closing it, it becomes compact. Commented Feb 5, 2016 at 10:44
• In order to make a clarification, I denoted the closed unit ball of $\mathcal H$ by $B_\mathcal H$. Again the argument goes as follows: Because $\mathcal H$ is reflexive, $B_\mathcal H$ is weakly compact. Since $T$ is also continuous with respect to the weak topology, the image $T(B_\mathcal H)$ is weakly compact as well. Therefore,it is weakly closed. Since for convex sets closure agrees with weak closure, $T(B_\mathcal H)$ is closed. Since it is also relatively compact due to compactness of $T$, $T(B_\mathcal H)$ is a compact set. Commented Feb 6, 2016 at 7:45
• Yes, I agree that $T(B_H)$ is compact and so $\|Tx\|$ attains its supremum over $T(B_H)$, which is $\sup\limits_{x\in B_H}{\|Tx\|=\|Tx_0\|}$ for some $x_0\in B_H$. But it is not clear why $\|x_0\|=1$. Commented Feb 6, 2016 at 12:40
• Suppose that $\|x_0\|<1$ and denote $y=\frac{x_0}{\|x_0\|}.$ Then $\|y\|=1$ and $\|Ty\|=\frac{\|Tx_0\|}{\|x_0\|}=\frac{\|T\|}{\|x_0\|}>\|T\|.$ Since $y\in B_{\mathcal H}$, this contradicts the fact that $\|T\|$ is the operator norm of $T$. Commented Feb 6, 2016 at 12:56
• @Svetoslav: I admit the fact that the vector is of norm one needed to be clarified. Thanks for the comment. Commented Feb 6, 2016 at 13:09

Regarding your comment about bounded functionals. Your answer is not correct. Hahn-Banach theorem works in other direction.

Given $x\in X$, then there exists a bounded functional $f$ with $\|f\|=1$ and $f(x)=\|x\|.$ Your argument proves what I claimed. What you need to do on Hilbert spaces, just apply Riesz representation theorem to obtain a vector $y\in \mathcal H$ such that $$f(x)=\langle x,y\rangle.$$ Then $\|f\|=\|y\|$, and if you take $x=\frac{y}{\|y\|}$, then $\|x\|=1$ and $f(x)=\|y\|=\|f\|.$

• you argument are right. But, I use Hahn-Banach theorem in $\hat{x}(f) = f(x)$
– BBVM
Commented Feb 3, 2016 at 20:18

Let $S=\{x\in\mathcal{H}:\|x\|=1\}$. Since $S$ is closed and bounded, then $S$ is weakly compact.

But $T$ is compact. Then $T$ is weak-norm continous. Thus $f:H\to\mathbb{R}$ given by $f(x)=\|Tx\|$ is weak-continous. Since $H$ equipped with the weak topology is Hausdorff and $S$ is weakly compact, then $f$ attains a globlal maxima on $S$. That is, there is $x_0\in S$ such that $\|Tx_0\|\ge \|Tx\|$ for all $x\in S$.

Hence $\|Tx_0\|=\sup\{Tx:x\in S\}=\|T\|$ and done.

EDIT a simpler way

Moreover, note that if $T$ is compact and selfadjoint, it is simpler:

If $T=0$, done. Suppose $T\neq0$. Hence $\|T\|\neq0$

Let $\sigma(T)$ the spectrum of $T$. Since $T$ is continous, then $\|T\|\in\sigma(T)$.

Suppose $T$ is compact and selfadjoint. Then $\sigma(T)$ is discrete. Thus, only $0$ can be a limit point. Hence elements of $\sigma(T)$ are eigenvectors. In particular $\|T\|$ is an eigenvector. That is, $Tx=\|T\|x$ for some $x\in\mathcal{H}$.

Now, if $T$ is not selfadjoint, then $T=UA$ for some partial isometry and nonnegative $A$. Applying the above argument to $A$, done.

• If a set in a Hilbert space is closed and bounded, it does not follow that it is weakly compact. It follows, if in addition we assume that it is convex. The unit sphere $S$ in $H$ is not convex. Also $S$ is not weakly compact. Indeed, let $\{e_i\}$ be an orthonormal sequence in $H$. Then from Bessel's inequality $\sum\limits_{i=1}^{\infty}{|\langle x,e_i\rangle |^2}\leq \|x\|^2,\,\forall x\in H$. Therefore $\langle x,e_i\rangle \to 0,\,\forall x\in H\Rightarrow \{e_i\}$ is weakly convergent to $0$, which is not an element of $S$. Commented Feb 5, 2016 at 10:34
• @sinbadh: If $T$ is negative-definite, then $\sigma(T)\subseteq (-\infty,0]$. Therefore $\|T\|$ cannot be in the spectrum. In general, $\|T\|$ or $-\|T\|$ is contained in $\sigma(T)$. Regarding your last comment, I do not know if I understand it correctly. By the polar decomposition you proved that $\|T\|$ is an eigenvalue of $T$? Commented Feb 6, 2016 at 7:52