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I want to maximize $\text{Tr}(XY)$ over $X$ for fixed $Y$, where $X$ and $Y$ are both hermitian (but doesn't necessarily positive) operators, and $X$ is constrained by its p-norm bounded by $1$, i.e. $|X|_p \leq 1$. I want to prove that optimized operators, the one when the norm constraint is $p=1$ and the other one when $p=2$, are related to each other.

Update: My question turns out to be very misleading, so please ignore the q! Apologize for the inconvenience.

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  • $\begingroup$ Isn't $\vert X_{opt,2}\vert_2 = 1$ without the factor of $\sqrt{d}$? That is, if $X_{opt,p}$ is a rank-1 projection onto $Y$'s maximum eigenvalue (I assume finite-dimensional), then $\vert X_{opt,p}\vert_p=1$ for all $p$, right? And it will saturate the bound you found, which seems to apply for all $p$ as well. $\endgroup$
    – Sam Jaques
    Feb 24, 2022 at 15:21
  • $\begingroup$ @SamJaques $|X_{opt,2}|_2$ is 1. Since $max \text{Tr}(XY) = |X|_1 |Y|_\infty \leq \sqrt{d} |X|_2 |Y|_\infty$ (now, just for general $X$, and using $|X|_2 \leq \sqrt{d}|X|_1$ inequaility), if we optimize with $p=2$-norm constraint, we end up getting $\sqrt{d} |X_{opt,2}|_2 |Y|_\infty = \sqrt{d} |Y|_\infty$. Does this make senes? I could be entirely wrong though! that's why I want discussions in the forum $\endgroup$
    – Jon Megan
    Feb 24, 2022 at 17:26
  • $\begingroup$ @SamJaques Now I think about it again, and I think $X_{opt,p}$ should not be always rank-1 projection onto $Y$'s maximum eigenvalue. Rather, $X_{opt,p}$ can be a sum of projections to + eigenspace of $Y$ and - eigenspace of $Y$ with -1 factor, then divide by 2, i.e. $X_{opt,p} = \frac{|Y+><Y+| - |Y-><Y-|}{2}$ where $|Y+>, |Y->$ are eigenspace for largest and smallest eigenvalues respectively. $\endgroup$
    – Jon Megan
    Feb 24, 2022 at 17:45
  • $\begingroup$ When you say $\vert A\vert_p$, do you mean $\max_{x:\Vert x\vert_p=1}\{\Vert Ax\Vert_p\}$ or do you mean $\left(\text{tr}(A^p)\right)^{1/p}$? For the second (the Schatten $p$-norm) there is a general Holder inequality (mathoverflow.net/questions/158881/…) and I think this could be used directly in your case (take $X=Y/\vert Y\vert_2$). For the first I'm not sure. But if you're thinking of the first, $\vert Y\vert_\infty$ isn't the same as it's maximum eigenvalue anyway, is it? $\endgroup$
    – Sam Jaques
    Feb 25, 2022 at 10:24

2 Answers 2

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If I understand your question correctly, you're trying to prove something that is false. Consider the operator $$ Y = \begin{pmatrix} 2 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{pmatrix}. $$ In the case $p = 2$, the optimizer is $X_{\mathrm{opt},2} = Y/\sqrt{6}$, but for $p = 1$ we have $$ X_{\mathrm{opt},1} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}. $$ These optimizers are not related in the way you have expressed.

Let me point out a mistake in the second paragraph of the question, in case it is causing confusion: the maximum of $\operatorname{Tr}(XY)$ over all $X$ with $\|X\|_2 \leq 1$ is not equal to $\|X\|_1 \|Y\|_{\infty}$ in general. It is equal to $\|Y\|_2$, and the optimizer is $X = Y/\|Y\|_2$. You can prove this using Cauchy-Schwarz (for the Hilbert-Schmidt inner product).

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They are related when $Y$ is a rank-2 matrix with eigenvalues. (say $\pm \lambda$)

Also, you need to prove the equality condition for the Holder's inequality in each case. You are using the equality condition without proving it.

Suppose $Y$ is a rank-2 matrix. Then equality case for Holder's inequality $Tr(XY) \leq |X|_1|Y|_\infty$ holds when $|Y|$ has an eigenvalue of $|Y|_\infty$, which is easy to show. So $max Tr(XY) = |Y|_\infty =\lambda$, and you're right that $|X|_{opt,1}$ is one of the suggested solutions.

The equality case for Holder's inequality $Tr(XY) \leq |X|_2|Y|_2 = \sqrt{2}\lambda|X|_2$ holds when $|X|^2 = a|Y|^2$ for some scaler $a>0$. When you think about maximizing trace, a natural choice would be aligning $X$ to $Y$, so that $X = a' Y$ for some scalar $a'$. This covers the equality condition trivially. Therefore, since $max Tr(XY) = |X_{opt,2}|_2|Y|_2 = \sqrt{2}\lambda $, $X_{opt,2}$ is just $\sqrt{2}$ multiplied by one of the $|X|_{opt,1}$ solutions suggested earlier. So they are related.

But in general case of $Y$ being a random Hermitian operator, this is not true at all.

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