1
$\begingroup$

Let $T$ be bounded operators between Hilbert Spaces and define the Schatten p-norm $(p \geq 1)$ \begin{equation*} \sigma_p(T) = \left( \sum_{n=1}^\infty a_n(T)^p \right)^{1/p}, \end{equation*} where $a_n(T)$ are the singular values of $T$.

Suppose that $T \in \mathcal{K}(H_1,H_2)$, $S \in \mathcal{K}(H_1,H_3)$, and \begin{equation*} ||Tx|| \leq ||Sx|| \quad \text{for all} \ x \in H_1, \end{equation*} Does it follow that $ \sigma_p(T) \leq \sigma_p(S)$?

The case $p = 2$ it's clear, because the 2-Schatten class are the Hilbert-Schmidt operators and \begin{equation*} \sigma_2(T)^2 = \sum_{n=1}^\infty ||Te_n||^2 \leq \sum_{n=1}^\infty ||Se_n||^2 = \sigma_2(S)^2, \end{equation*} where $(e_n)$ is an orthonormal basis of $H_1$.

What can we say if $p \neq 2$?

$\endgroup$
1
$\begingroup$

Yes the $p$-Schatten norm is monotone. This can be seen from the following fact: A mapping $T:H_1\rightarrow H_2$ is of $p$-Schatten class if and only if$$\{||T\psi_j||_{H_2}\}_{j=1}^{\infty}\in\ell^p$$ for all($2\leq p<\infty$) orthonormal bases/for some ($0<p<2$) orthonormal basis $\{\psi_j\}_j$ of $H_1$ and the $p$-Schatten norm is obtained by maximizing($2\leq p<\infty$) or minimizing($0<p<2$)the expression $$\left(\sum_{j=1}^{\infty}||T\psi_j||^p_{H_2}\right)^{\frac{1}{p}}$$ where the maximum or minimum is taken over all orthonormal bases $\{\psi_j\}_j$ of $H_1$. I proved this fact in my master thesis:

https://www.fernuni-hagen.de/analysis/download/diplomarbeit_melech.pdf

You find the proof in chapter 6 (p.36). Since: $$\left(\sum_{j=1}^{\infty}||T\psi_j||^p_{H_2}\right)^{\frac{1}{p}}\leq \left(\sum_{j=1}^{\infty}||S\psi_j||^p_{H_3}\right)^{\frac{1}{p}}$$ this proves the monotonicity of the $p$-Schatten norm for $0<p<\infty $.

$\endgroup$
  • $\begingroup$ Ok, I found the proof of maximizing for $2 \leq p < \infty$. The case $ 0 < p < 2$ is equivalent? $\endgroup$ – Javier González Feb 28 '18 at 17:28
  • $\begingroup$ Yes the minimum is obtained by the orthonormal basis consisting of eigenvectors of $T^*T$ in this case $\endgroup$ – Peter Melech Feb 28 '18 at 17:33
  • $\begingroup$ $\sum_{j=1}^{\infty}||T\psi_j||^p=\sum_{j=1}^{\infty}(T\psi_j,T\psi_j)^{\frac{p}{2}}=\sum_{j=1}^{\infty}(T^*T\psi_j,\psi_j)^{\frac{p}{2}}=\sigma_p(T)^p$ for this basis. This is actually sufficient for what You want to show. In both cases!! $\endgroup$ – Peter Melech Feb 28 '18 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.