# maximization of trace between two operators with respect to different norm constraints

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.

• 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. Feb 24, 2022 at 15:21
• @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 Feb 24, 2022 at 17:26
• @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. Feb 24, 2022 at 17:45
• 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? Feb 25, 2022 at 10:24

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).

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.