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I'm working on Reproducing Kernel Hilbert Spaces and I had a problem proving the the continuity of the evaluation functional $e_{t}$ ($e_{t}(\phi) = \phi(t)$).

Theorem A Hilbert Space of complex valued functions on E has a reproducing kernel then all the evaluation functionals $e_{t}$, $t \in E$, are continuous on $\mathcal{H}$.

Proof: If $\mathcal{H}$ has a reproducing kernel $K$ then for any $t \in E$, we have

$$\forall \phi \in \mathcal{H} \ \ \ \ \ \ \ \ e_{t}(\phi) = \langle \phi, K(\cdot, t) \rangle_{\mathcal{H}}$$

Thus the evaluation functional $e_{t}$ is linear and by the Cauchy-Schwarz inequality.

$$|e_{t}(\phi)| = |\langle\phi, K(\cdot, t) \rangle_{\mathcal{H}}| \ \ \leq \ \ ||\phi|| \ \ ||K(\cdot, t)|| = ||\phi|| \ \ [K(t, t)]^{1/2}$$

$$ |e_{t}(\phi)| \leq ||\phi|| \ \ [K(t, t)]^{1/2}$$

Then $e_{t}$ is bounded then $e_{t}$ is continuous.

How can I conclude that $e_{t}$ is bounded from the last inequality? As far as I know $[K(t, t)]^{1/2}$ is not bounded.

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  • $\begingroup$ This shows that $e_t$ is a bounded linear functional for each $t$. $t$ is fixed. $\endgroup$
    – user38355
    Dec 11, 2013 at 18:59
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    $\begingroup$ That what I was thinking. How can I ensure that $K(t, t)$ is not infinity? $\endgroup$
    – matiskay
    Dec 11, 2013 at 21:34

1 Answer 1

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You have the assumption that
$$\forall \phi \in \mathcal{H} \quad e_{t}(\phi) = \langle \phi, K(\cdot, t)\rangle _{\mathcal{H}}$$ Consider the meaning of the above statement. The thing on the right implicitly says that $K(\cdot,t)$ is an element of $\mathcal H$, because otherwise the inner product is not defined. Any functional of this form is bounded. Specifically, $$|e_{t}(\phi)| \le \|\phi\|_{\mathcal H}\, \| K(\cdot, t)\|_{\mathcal{H}}$$

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  • $\begingroup$ It's the reproducing property of the Reproducing Kernel Hilbert Space. $$\forall t \in E, \forall \phi \in \mathcal{H} \langle \phi, K(\cdot, t) \rangle = \phi(t) = e_{t}(\phi)$$ $\endgroup$
    – matiskay
    Dec 15, 2013 at 16:51
  • $\begingroup$ @matiskay That was a rhetorical question. I reworded the post to clarify this point. $\endgroup$ Dec 15, 2013 at 17:00

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