# Norm of a reproducing kernel Hilbert space in terms of supremum over unit ball

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.

## 1 Answer

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.

• 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} – jII Jan 20 at 14:26
• Yes indeed, this unit vector realizes the maximum. – Nicolas Jan 20 at 16:25