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Let $M$ be a self-adjoint operator on a complex Hilbert space $H$.

Let $s\in \mathbb{R}$. Is $$\|M^s\|\leq \|M\|^s\;?$$

Since $M$ is selfadjoint, then by the Spectral Theorem we have $$ M^s=\int_{\sigma(M)}\lambda^s\,dE(\lambda). $$ Then $$ \|M^s\|\leq\int_{\sigma(M)}|\lambda|^s\,\|dE(\lambda)\|. $$

Schüler
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1 Answers1

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Consider $$M = \pmatrix{2 & 0 \\ 0 & 3}, \quad M^{-1} = \pmatrix{\frac12 & 0 \\ 0 & \frac13}$$ so $\|M\| = 3$ and $\|M^{-1}\| = \frac12$ so $$\|M^{-1}\| \le \|M\|^{-1}$$ yields $\frac12 \le \frac13$ which is not true.

For positive $s > 0$ the claim is true since the function $x \mapsto x^s$ is continuous and increasing so $$\|M^s\| = \max_{\lambda \in \sigma(M^s)}|\lambda| = \max_{\lambda\in\sigma(M)}|\lambda^s| = \max_{\lambda\in\sigma(M)}|\lambda|^s = \left(\max_{\lambda\in\sigma(M)}|\lambda|\right)^s = \|M\|^s$$ so even equality holds.

mechanodroid
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  • Thanks for the answer. It is trivial that when $M$ is selfadjoint then so is $M^s$ with $s$ real? – Schüler Jan 17 '21 at 16:12
  • @Schüler Yes, a normal operator $A$ is self-adjoint if and only if $\sigma(A) \subseteq \Bbb{R}$. Since $M$ is self-adjoint, we have $\sigma(M)\subseteq \Bbb{R}$ so $$\sigma(M^s) = \sigma(M)^s \subseteq \Bbb{R}^s \subseteq \Bbb{R}$$ and $M^s$ is normal by definition so $M^s$ is self-adjoint. – mechanodroid Jan 17 '21 at 16:31