Prove the expression $\|A\|_p=\max_{\|B\|_q\le 1} {\rm tr}(A^{\dagger}B)$ for the $p$-norm in terms of its Hölder dual $q$-norm

Question

I'm trying to prove

\begin{align} \|A\|_p = \max_{\|B\|_q\leq 1} {\rm tr}\{A^{\dagger}B\}, \end{align} where $\|\cdot\|_p$ is the Schatten p-norm, and p and q are Hölder conjugates.

What I have now is \begin{align} {\rm tr}\{A^{\dagger}B\}\leq {\rm tr}\{|A^{\dagger}B|\} = \|A^{\dagger}B\|_1\leq \|A\|_p\|B\|_q\leq \|A\|_p \end{align} for all $\|B\|_q\leq 1$. By the proof of the vector case, it seems like I need to construct such a $B$ to achieve equality. However, it seems not so straightforward to apply the argument for the vector case to the Schatten p-norm case.

In Quantum Information Theory, Mark Wilde, it claims the result can be obtained by considering the sufficient equality condition for the Hölder inequality. That is the case when we choose a $B$ such that $A^{\dagger}=a|B|^{q/p}U^{\dagger}$, where $U^{\dagger}$ is a unitary such that $B=U|B|$ is the left polar decomposition of B. In this setting, I have ${\rm tr}\{A^{\dagger}B\}=a \|B\|_q^q$, and it is not clear to me how to proceed.

Answer 1

Here is how you can choose the optimal $B$. As John suggested in his comment, start from a singular value decomposition $A=\sum_js_j|\phi_j\rangle\langle\psi_j|$ of a given $A\in\mathbb C^{n\times n}\setminus\{0\}$ (i.e., $s_j\geq 0$ are the singular values of $A$ and $\{\phi_j\}_j,\{\psi_j\}_j$ are orthonormal bases of $\mathbb C^n$).

First let us assume $p\in(1,\infty)$. Choose $$ B=\sum_j \frac{ s_j^{p-1} }{ \|A\|_p^{p-1} } |\phi_j\rangle\langle\psi_j|\,. $$ This operator $B$ has $q$-norm $1$: because $\frac1p+\frac1q=1$ is equivalent to $q(p-1)=p$ one computes $$ \|B\|_q=\Big( \sum_j \frac{ s_j^{q(p-1)} }{ \|A\|_p^{q(p-1)} } \Big)^{1/q}=\Big( \frac{ \sum_js_j^{p} }{ \|A\|_p^{p} } \Big)^{1/q}=\Big( \frac{\|A\|_p^{p} }{ \|A\|_p^{p} } \Big)^{1/q}=1\,. $$ More importantly, this $B$ is the "correct" choice because \begin{align*} |{\rm tr}(A^\dagger B)|=\sum_js_j\cdot \frac{ s_j^{p-1} }{ \|A\|_p^{p-1} }=\frac{\sum_js_j^p}{\|A\|_p^{p-1}}=\frac{\|A\|_p^p}{\|A\|_p^{p-1}}=\|A\|_p\,. \end{align*}

The edge cases can be obtained by taking suitable limits; if $p=\infty$, then choose $B=|\phi_k\rangle\langle\psi_k|$ where $k$ is chosen such that $s_k=\max_j s_j$ is the largest singular value of $A$. On the other hand, if $p=1$ you just choose $B=\sum_j|\phi_j\rangle\langle\psi_j|$.