Let $l^2(V)$ be the space of real square-summable functions defined on a discrete set $V$ (e.g. the node set on a graph), with inner product: \begin{align} \langle f, g \rangle = \sum_{v\in V} f(v)g(v) \end{align} Moreover, let $L$ be a linear self-adjoint, positive semidefinite operator acting on $l^2(V)$. I have checked that this operator fulfills the following property for any $f\in l^2(V)$: \begin{align} \langle Lf,f\rangle \leq C ||f||^2 \end{align} where $C$ is a constant. Based on this, I have been told that it follows trivially that $L$ is bounded. However, based on my (very limited) knowledge of operator theory, to prove an operator is bounded I need to show that $\langle Lf, Lf\rangle \leq C ||f||^2$. I don't see how the second claim follows from the first. Is it necessary to use some of the other properties of $L$, like the fact that it is self-adjoint?
3 Answers
Well $|\langle Lf, g \rangle|^2 \leq \langle Lf, f \rangle \langle Lg, g\rangle$. (By the usual proof of CS.) So $L$ is bounded because $\|Lf\|=\sup_{\|g\|=1}|\langle Lf,g \rangle|.$
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Since CS has many proof, the one I had in mind was: Let $a$ be a scalar such that $|a|=1$ and $a\langle Lf,g\rangle = | \langle Lf,g\rangle |.$ For real $t$, let $p(t):= \langle L(taf+g),taf+g\rangle.$ Since $0\leq p(t) = \langle Lf,f\rangle t^2 +2 | \langle Lf,g\rangle | t + \langle Lg,g\rangle $, we have $| \langle Lf,g\rangle | \leq \langle Lf,f\rangle \langle Lg,g\rangle $. – Steen82 Aug 09 '24 at 11:56
This is basically the polarization identity. Since $L$ is self-adjoint, we have,
$$\langle Lf, g \rangle = \frac{1}{2}(\langle Lf, g \rangle + \langle Lg, f \rangle) = \frac{1}{2}(\langle L(f+g), f+g \rangle -\langle Lf, f \rangle - \langle Lg, g \rangle)$$
Thus,
$$\begin{split} |\langle Lf, g \rangle| &\leq \frac{C}{2}(\|f+g\|^2 + \|f\|^2 + \|g\|^2)\\ &\leq \frac{C}{2}(2\|f\|^2 + 2\|g\|^2 + 2\|f\|\|g\|)\\ &\leq \frac{3C}{2}(\|f\|^2 + \|g\|^2) \end{split}$$
Letting $f$ and $g$ range over unit vectors yields,
$$\|L\| = \sup_{\|f\|=1} \|Lf\| = \sup_{\|f\|=1, \|g\| = 1} |\langle Lf, g \rangle| \leq \frac{3C}{2} \cdot 2 < \infty$$
So $L$ is bounded.
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On a real Hilbert space, polarization identity only works for self-adjoint operators, so using the fact that $L$ is self-adjoint is more or less necessary. On a complex Hilbert space, polarization identity holds for all operators, so you don’t even need $L$ being self-adjoint in that case. – David Gao Aug 08 '24 at 17:26
Yes, it's a fact of self-adjoint operators that $L$ is bounded iff $M := \sup_{\lVert x \rVert = 1} |\langle L x, x\rangle| < \infty$, and further if it is bounded then $\lVert L \rVert = \sup_{\lVert x \rVert = 1} |\langle L x, x\rangle|$. You can even initially take $L$ to be just "densely-defined".
You can extract a proof of this from Norm of self-adjoint operator for example. Here it's shown, with the parallelogram identity, that $|\langle L x, y\rangle| \le \sup_{\lVert x \rVert = 1} |\langle L x, x\rangle| = M$ for each $x, y$ with $\lVert x\rVert = \lVert y\rVert = 1$. Then (if $L x \ne 0$), set $y = (L x)/\lVert L x\rVert$ to get that $\lVert L x \rVert \le M$ for each $x$ with $\lVert x \rVert = 1$. You can then swap $x$ with $x/\lVert x\rVert$ to finish.
Let me know if I can clarify any details.
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