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I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.

The Schatten-$p$ norm of a matrix $X$ is defined as $$\lVert X\rVert_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\lVert X\rVert_p = \lVert\sigma(X)\rVert_p$, where $\sigma(X)$ is the vector of singular values.

From von Neumann's trace inequality we know that $$|{\operatorname{trace}(XY)}| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have $$\langle \sigma(X), \sigma(Y)\rangle \le \lVert\sigma(X)\rVert_p\lVert\sigma(Y)\rVert_q,$$ where $1/p + 1/q = 1$.

On combining the above two, we immediately get the alleged Schatten-$p$ norm Hölder inequality.

I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.

The Schatten-$p$ norm of a matrix $X$ is defined as $$\lVert X\rVert_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\lVert X\rVert_p = \lVert\sigma(X)\rVert_p$, where $\sigma(X)$ is the vector of singular values.

From von Neumann's trace inequality we know that $$|{\operatorname{trace}(XY)}| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have $$\langle \sigma(X), \sigma(Y)\rangle \le \lVert\sigma(X)\rVert_p\lVert\sigma(Y)\rVert_q,$$ where $1/p + 1/q = 1$.

On combining the above two, we immediately get the alleged Schatten-$p$ norm Hölder inequality.

I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.

The Schatten-$p$ norm of a matrix $X$ is defined as $$\|X\|_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\|X\|_p = \|\sigma(X)\|_p$, where $\sigma(X)$ is the vector of singular values.

From von Neumann's trace inequality we know that $$|\mbox{trace}(XY)| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have $$\langle \sigma(X), \sigma(Y)\rangle \le \|\sigma(X)\|_p\|\sigma(Y)\|_q,$$ where $1/p + 1/q = 1$.

On combining the above two, we immediately get alleged Schatten-p norm Hölder inequality.

I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.

The Schatten-$p$ norm of a matrix $X$ is defined as $$\lVert X\rVert_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\lVert X\rVert_p = \lVert\sigma(X)\rVert_p$, where $\sigma(X)$ is the vector of singular values.

From von Neumann's trace inequality we know that $$|{\operatorname{trace}(XY)}| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have $$\langle \sigma(X), \sigma(Y)\rangle \le \lVert\sigma(X)\rVert_p\lVert\sigma(Y)\rVert_q,$$ where $1/p + 1/q = 1$.

On combining the above two, we immediately get the alleged Schatten-$p$ norm Hölder inequality.

I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.

The Schatten-$p$ norm of a matrix $X$ is defined as $$\|X\|_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\|X\|_p = \|\sigma(X)\|_p$, where $\sigma(X)$ is the vector of singular values.

From von Neumann's trace inequality we know that $$|\mbox{trace}(XY)| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have $$\langle \sigma(X), \sigma(Y)\rangle \le \|\sigma(X)\|_p\|\sigma(Y)\|_q,$$ where $1/p + 1/q = 1$.

On combining the above two, we immediately get alleged Schatten-p norm Hölder inequality.

I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.

The Schatten-$p$ norm of a matrix $X$ is defined as $$\|X\|_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\|X\|_p = \|\sigma(X)\|_p$, where $\sigma(X)$ is the vector of singular values.

From von Neumann's trace inequality we know that $$|\mbox{trace}(XY)| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have $$\langle \sigma(X), \sigma(Y)\rangle \le \|\sigma(X)\|_p\|\sigma(Y)\|_q,$$ where $1/p + 1/q = 1$.

On combining the above two, we immediately get alleged Schatten-p norm Hölder inequality.