# Is Schatten p-norm a monotone ideal norm?

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 $p \neq 2$?

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