# Can we show that $\sup_{\left\|x\right\|_H=1}\langle Ax,x\rangle_H$ is attained at the supremum of $\sigma(A+A^\ast)$?

Let $$H$$ be a $$\mathbb R$$-Hilbert space and $$A\in\mathfrak L(H)$$. Consider the following optimization problem: $$\sup_{\left\|x\right\|_H=1}\langle Ax,x\rangle_H.\tag1$$ We may note that $$A+A^\ast$$ is self-adjoint and hence there is an unique compactly supported spectral measure $$E$$ on $$\mathcal B(\mathbb R)$$ associated with $$A+A^\ast$$. Now, $$\langle Ax,x\rangle_H=\frac12\langle(A+A)^\ast x,x\rangle_H=\frac12\int_{\sigma(A+A^\ast)}\lambda\:\langle E({\rm d}\lambda)x,x\rangle_H\;\;\;\text{for all }x\in H.\tag2$$

In the case $$H=\mathbb R^d$$, as it was shown in this answer, the supremum in $$(1)$$ is attained at the unit eigenvector $$z_{\text{max}}$$ associated with the largest eigenvalue $$\lambda_{\text{max}}$$ of $$A+A^*$$ and the optimal objective value is the logarithmic norm of $$A$$. This is easily seen from $$(2)$$.

Question: Can we infer a similar result in the general case?

We should be able to argue in the same manner by noting that $$V:=\left\{\langle(A+A^\ast)x,x\rangle_H:\left\|x\right\|_H=1\right\}$$ is bounded and convex and $$\sigma(A+A^\ast)\subseteq\left[\inf V,\sup V\right]\tag3,$$ but I'd need some help to figure out the details.

Your equation $$(2)$$ reduces the problem to $$A$$ selfadjoint. You also see from $$(2)$$ that $$\sup\langle Ax,x\rangle\leq\sup\sigma(A)$$. For the reverse inequality, fix $$\varepsilon>0$$. Then the Spectral Theorem (or the definition of integral, if $$(2)$$ is a given) gives you projections $$P_1,\ldots,P_n$$ with $$\sum_jP_j=I$$, and scalars $$\lambda_1\geq\ldots\geq\lambda_n\subset\sigma(A)$$ with $$\|A-\sum_j\lambda_jP_j\|<\varepsilon.$$ We may also choose $$\lambda_1$$ such that $$|\lambda_1-\sup\sigma(A)|<\varepsilon$$. Take $$x\in P_1H$$ a unit vector. Then $$\langle Ax,x\rangle=\lambda_1>\sup\sigma(A)-\varepsilon.$$ As $$\varepsilon$$ is arbitrary, it follows that $$\sup\langle Ax,x\rangle\geq\sup\sigma(A)$$.
• Everything is clear to me except why we can assume that $\lambda_1,\ldots,\lambda_n\in\sigma(A)$ and that $|\lambda_1-\sup\sigma(A)|<\varepsilon$. (I'm using the definition of the integral to obtain pairwise disjoint $P_1,\ldots,P_n\in\mathcal B(\mathbb R)$ and $\lambda_1,\ldots,\lambda_n\in\mathbb R$ such that $\int|\operatorname{id}_{\sigma(A)}-f|\:|E|({\rm d}\lambda)<\varepsilon$, where $f=\sum_{i=1}^n\lambda_i1_{B_i}$. In your description, $P_i=E(B_i)$. It's clear to me that $E(\mathbb R\setminus\sigma(A))=0$.) Oct 9, 2019 at 17:23
• Take $n>\tfrac1\varepsilon$. If $a=\inf\sigma(A)$, $b=\sup\sigma(A)$, define $$B_j=[\tfrac {j-1}n,\tfrac jn]\cap\sigma(A),\ \ j=1,\ldots,n$$ and take $\lambda_j=\tfrac jn$. Oct 9, 2019 at 17:48
• (a) Thank you for your comment, but I still don't see why your choice of $\lambda_j$ should belong to $\sigma(A)$; unless $\sigma(A)$ is connected. (b) You've proved that the maximum objective value is $\sigma(A)$, but what can we say about the maximizer $x$? (It's obvious that if $\sigma(A)$ is finite, then the maximizer is any eigenvector corresponding to the eigenvalue $\lambda=\max\sigma(A)$.) Oct 9, 2019 at 18:05
• Right, replace my last line with taking any $\lambda_j\in B_j$. Oct 9, 2019 at 18:44
• Well, the conditions have to be either very explicit ("$\sup\sigma(A)$ is an eigenvalue") or very restrictive ("$A$ is finite-rank"). Oct 10, 2019 at 5:12