Let $H$ be a reproducing-kernel Hilbert space (RKHS) of bounded continuous functions $g:X \to \mathbb{R}$, where $X$ is some topological space.

The norm of $g \in H$ is given by

$$ \lvert\lvert g \rvert\rvert_H = \sqrt{\langle g, g \rangle_H}. $$

This paper (in the final equality of the second proof on p. 727) effectively states the following property

$$ \sup_{\lvert\lvert f \rvert\rvert \le 1} \langle g, f \rangle_H = \lvert\lvert g \rvert\rvert_{H}. \quad\quad\quad (\star) $$

How does $(\star)$ follow?

One thought is that the sup-norm of $g$ is given by

$$ \lvert\lvert g \rvert\rvert_{\infty} = \sup_{x \in X}|g(x)| = \sup_{x\in X}\langle g, \phi_x \rangle_H,$$

where $\phi_x \in H$ is the projection of $x$ by RKHS. If $\phi: X \to H$ is onto (i.e.,, $\{\phi_x \mid x \in X\} \equiv H$) then $$\sup_{x \in X} \langle g, \phi_x \rangle_H = \sup_{f \in H}\langle g, f \rangle_H.$$

However, it is not clear to me that the sup-norm is relevant, that $\phi$ is onto, or how the unit ball in $(\star)$ relates to the norm.


By Cauchy-Schwarz, we have that $|\langle g,f\rangle_H|\leq\|g\|_H\|f\|_H\leq \|g\|_H$ if $\|f\|_H\leq 1$. On the other hand, every Hilbert spaces are reflexive. By Kakutani's Theorem, the closed unit ball is weakly compact, hence ($\star$) says that the function $f\mapsto\langle g,f\rangle_H$ attains its maximum over all $f$ with norm $\leq 1$. There is no square as in the question.

  • 1
    $\begingroup$ I see; and it is the unit vector $\hat{g} := g/||g||_H$ attains the maximum, since for $g \ne 0$ we have $$ \begin{align} \langle g, \hat{g} \rangle_H &= \langle g, g/||g||_H \rangle_H \\ &= 1/||g||_H \langle g, g \rangle_H \\ &= 1/||g||_H (||g||_H)^2 \\ &= ||g||_H \end{align} $$ $\endgroup$ – jII Jan 20 at 14:26
  • $\begingroup$ Yes indeed, this unit vector realizes the maximum. $\endgroup$ – Nicolas Jan 20 at 16:25

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