# Showing that $\exists x \in H : \|A(x)\| = \|A\|_\mathcal{L}$ if $H$ is Hilbert and $A \in \mathcal{L}_c(X,Y)$.

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.

• I'm guessing you want $\|x\| \le 1$, not just $x \in H$? Otherwise the result is trivial. – Theo Bendit Feb 24 at 10:52
• @TheoBendit Hi, thanks for input. Well, the exercise says $x \in H$. – Rebellos Feb 24 at 10:52
• 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. – Theo Bendit Feb 24 at 10:54
• @TheoBendit Way simpler than what I tried. By the way, is my approach even legitimate? – Rebellos Feb 24 at 10:57
• @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 $\| x\| \leq 1$. – Rebellos Feb 24 at 11:13

## 1 Answer

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 $$\left\lVert Ax \right\rVert=\left\lVert A\right\rVert$$.