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Let $A$ be a linear operator on $H$ Hilbert space (with a scalar product $(\cdot,\cdot)$ ), and let's define $r_{A} =\sup_{\|x\|=1}(Ax, x)$. We want to show that $r_A=\|A\|$, where we endow the space $L(H)$ of the linear continuous operator on H with the uniform norm. From Cauchy-Schwartz we easily obtain $r_A \leq \|A\|$. I'm stuck with the other inequality; on the (rather dubious) note it should follow somehow from the polarization equality $(x, y)= \frac 1 4 [\|x+y\|^2-\|x-y\|^2]$, but I can't manage to get it. Any help is appreciated.

mechanodroid
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Bargabbiati
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2 Answers2

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This isn't true in general. Consider $A = \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$. We have

$$\left\langle\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}, \begin{bmatrix} x \\ y\end{bmatrix} \right\rangle = \left\langle \begin{bmatrix} y \\ -x\end{bmatrix}, \begin{bmatrix} x \\ y\end{bmatrix} \right\rangle = 0$$

so $r_A = 0$ but $\|A\| \ne 0$.

However, this is true if $A$ is assumed to be self-adjoint.

Moreover, it is even true for normal operators if the Hilbert space is complex.

mechanodroid
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  • If $A$ is self-adjoint, does it follow from the equality $||A||=||A^{2^n}||^{\frac 1 {2^n}}$ and from the fact that $r_A= lim _{n \to \infty}||A^n||^{\frac 1 n}$? – Bargabbiati Jul 17 '18 at 12:28
  • (And that I know that the last sequence admits limit). – Bargabbiati Jul 17 '18 at 12:33
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    @GiuseppeBargagnati $\lim_{n\to\infty} |A^n|^{\frac1n}$ is the spectral radius of $A$, not the numerical radius $r_A$. But yes, for normal operators we have $|A^n| = |A|^n$ so the spectral radius of $A$ is equal to $|A|$. – mechanodroid Jul 17 '18 at 14:19
  • Ok, I've some problems with the notation with the notes I'm studying, thank you very much. – Bargabbiati Jul 17 '18 at 14:22
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I assume from your version of the polarization identity that we're working over a real Hilbert space.

Hint: Note that (if $A$ is self-adjoint) $$ (Ax,y) = \frac 14 \left[ (x+y,A(x+y)) - (x-y,A(x-y))\right] $$ you can regard this as a generalization of the usual polarization identity. Thus, we have $$ 4\sup_{\|x\| = \|y\| = 1} (Ax,y) \leq \sup_{\|x\| = \|y\| = 1}(A(x+y),(x+y)) - \inf_{\|x\| = \|y\| = 1} A(x-y,x-y) $$

Ben Grossmann
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