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Exercise :

Let $H$ be a Hilbert space and $A \in \mathcal{L}_c(H)$. Show that $\exists x \in H : \|A(x)\| = \|A\|_\mathcal{L}$.

Attempt - Thoughts :

Note : The space $\mathcal{L}_c(H)$ is the space of all the Linear Compact Operators $L : H \to H$.

Let $A \in \mathcal{L}_c(H)$ such that :

$$\langle A(y),x\rangle = \langle y, A(x) \rangle, \quad \text{for} \quad x,y \in H$$

Then : \begin{align} \|A\|_\mathcal{L} &= \sup\left\{ \|A(y)\| : \|y\| \leq 1 \right\} \\ &= \sup\left\{\sup_{\|x\| \leq 1}|\langle A(y),x\rangle| : \|y\| \leq 1 \right\} \\ &= \sup\left\{\sup_{\|x\| \leq 1} |\langle y, A(x)\rangle| : \|y\| \leq 1 \right\} \\ &= \sup_x \|A(x)\| \end{align}

So for that $x$ that the supremum is achieved, it indeed is $\|A(x)\| = \|A\|_\mathcal{L}$.

But isn't that more like showing that $\exists A \in \mathcal{L}_c(X,Y)$ such that $\|A(x)\| = \|A\|_\mathcal{L}$ ? I got the intuition above from the way we defined the Dual-Operator $A^* \in \mathcal{L}(X,Y)$ over two Banach spaces.

If my approach is not legit, how would one show that $\exists x \in H : \|A(x)\| = \|A\|_\mathcal{L}$ ?

Any elaboration or hint will be greatly appreciated.

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  • $\begingroup$ I'm guessing you want $\|x\| \le 1$, not just $x \in H$? Otherwise the result is trivial. $\endgroup$ – Theo Bendit Feb 24 at 10:52
  • $\begingroup$ @TheoBendit Hi, thanks for input. Well, the exercise says $x \in H$. $\endgroup$ – Rebellos Feb 24 at 10:52
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    $\begingroup$ In that case, two options: $\|Ax\| = 0$ for all $x$, in which case $\|A\| = 0$, so pick any $x$, or $\|Ax\| > 0$, for some $x$, in which case choose $y = \frac{\|A\|}{\|Ax\|} x$. Then $\|Ay\| = \|A\|$ as required. $\endgroup$ – Theo Bendit Feb 24 at 10:54
  • $\begingroup$ @TheoBendit Way simpler than what I tried. By the way, is my approach even legitimate? $\endgroup$ – Rebellos Feb 24 at 10:57
  • $\begingroup$ @user123 Well, the final part is $\sup_y \sup_x |\langle y, A(x)\rangle| = \sup_x \|A(x)\|_{y \in H}$ but $x \in H$ too so it is just $\sup_x \|A(x)\|$ and we are talking about a compact operator. This means that it transfers bounded sets from $H$ to sets with compact closure on $H$. Yes, I am sure that it is not $\| x\| \leq 1$. $\endgroup$ – Rebellos Feb 24 at 11:13
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It seems that you assume that $A$ is self-adjoint, which is not necessarily the case.

As mentioned in the comments, it seems that we want $x$ to have a norm smaller or equal to one. Otherwise, we play with $\lambda y$, $\lambda \gt 0$ where $y$ is such that $Ay\neq 0$ (if it exists) and we are not, without needing the fact that we work on a Hilbert space or with a compact operator.

In order to see that the operator norm is reached by an element of the closed unit ball, we consider a sequence $\left(x_n\right)_{n\geqslant 1}$ of elements of the unit ball such that $\left\lVert Ax_n\right\rVert\geqslant \left\lVert A\right\rVert-n^{-1}$. We extract a weakly convergence subsequence to some $x$; by compactness, we extract from this subsequence a further subsequence, denotes $\left(x_{n_k}\right)_{k\geqslant 1}$, such that $\left(Ax_{n_k}\right)_{k\geqslant 1}$ is convergent to some $y$. Using adjoint, we can see that $\langle Ax_{n_k},z\rangle \to \langle Ax,z\rangle$ for all $z\in H$ hence $y=Ax$. From $$ \left\lVert Ax_{n_k}\right\rVert\geqslant \left\lVert A\right\rVert-n_k^{-1} $$ and the convergence of $\left(Ax_{n_k}\right)_{k\geqslant 1}$, we derive that $\left\lVert Ax \right\rVert=\left\lVert A\right\rVert$.

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