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Let $A \colon X \to X$ be a compact, self-adjoint operator that is linear as well, on the real Hilbert space $X$. Let $R \colon X \setminus\{0\} \to \mathbb{R}$ be defined as $R(x) = ( Ax, x)_X$ for $\| x \| = 1$. (better known as the Rayleigh-quotient).

We can assume that $A$ is non-negative, that is $R(x) \geq 0$ and $\sup_{x} R(x) > 0$ and that $\| A \| = \sup_{x \in X \setminus\{0\}} R(x)$. Previously, I showed that $R$ is bounded (Cauchy-Schwarz, quite easy).

I want to show that there exists $x_0$ so that $R(x_0) = \sup_{y} R(y)$.

The difficulty seems to be that a compact operator maps bounded sets to relatively compact sets. I tried to show that the image is closed using that, unsuccessfully. Working with limits wasn't conclusive either, due to the image not being necessarily closed. Another option I considered was using that for any $f\colon X \to Y$ continuous, $X$ compact, $Y$ Hausdorff implies $f$ closed (mapping closed sets to closed sets), but I could not show that the unit circle is compact in $X$. Can someone post any hint how to get to this conclusion?

This problem occurs in a proof about the spectral theorem, so any hint relying on results of this fact are unhelpful.

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mdot
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1 Answers1

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Here's a sketch.

Let's suppose, by rescaling $A$ if necessary, that $\sup_y R(y) = 1$, so that $\|A\| = 1$ as well as you have shown. Choose a sequence $x_n$ with $\|x_n\|=1$ such that $R(x_n) \to 1$. By the compactness of $A$, by passing to a subsequence, we can assume that $A x_n$ converges to some $z$. You can now verify that $\langle z, x_n \rangle \to 1$. By writing $\|x_n - z\|^2 = \|x_n\|^2 - 2 \operatorname{Re} \langle z, x_n \rangle + \|z\|_2$ and noting $\|x_n\|=1$, $\|z\| \le 1$, conclude that $x_n \to z$ as well, and therefore $\langle Az, z\rangle = 1$. (In fact, you conclude for free that $Az=z$ so that $z$ is an eigenvector with eigenvalue 1.)

Nate Eldredge
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