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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.

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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 $\sup\langle Ax,x\rangle\geq\sup\sigma(A)$.

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  • $\begingroup$ 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 $E(\mathbb R\setminus\sigma(A))=0$.) $\endgroup$
    – 0xbadf00d
    Oct 9, 2019 at 17:23
  • $\begingroup$ 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 $\lambda_j=\tfrac jn$. $\endgroup$
    – Martin Argerami
    Oct 9, 2019 at 17:48
  • $\begingroup$ (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 $\lambda=\max\sigma(A)$.) $\endgroup$
    – 0xbadf00d
    Oct 9, 2019 at 18:05
  • $\begingroup$ Right, replace my last line with taking any $\lambda_j\in B_j$. $\endgroup$
    – Martin Argerami
    Oct 9, 2019 at 18:44
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    $\begingroup$ Well, the conditions have to be either very explicit ("$\sup\sigma(A)$ is an eigenvalue") or very restrictive ("$A$ is finite-rank"). $\endgroup$
    – Martin Argerami
    Oct 10, 2019 at 5:12

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