RKHS rough definition

Question

Wikipedia provides an intuitive explanation of RKHS as a Hilbert space of functions where point evaluation is a continuous linear functional:

Roughly speaking...if two functions $f$ and $g$ in the RKHS are close in norm, i.e., $\|f-g\|$ is small, then $f$ and $g$ are also pointwise close, i.e., $|f(x)-g(x)|$ is small for all $x$. The reverse need not be true.

I see the first statement holds because of the continuity/boundedness of pointwise evaluation. Why does the reverse not necessarily hold -- what's a simple example I should be thinking of?

Answer 1

In $\ell^2(\mathbb N)$, take $g=0$, and $$f_m(n)=\begin{cases} 1/\sqrt{m},&\ n\leq m\\ \ \\ 0,&\ n>m\end{cases}$$ Then, for each $n$, we have $\|f_m(n)-g(n)\|\leq1/\sqrt m$, while $$ \|f_m-g\|_2=1 $$ for all $m$. That is, the sequence $f_m$ converges uniformly to zero, but in the two-norm it remains a fixed distance from the origin.

Answer 2

A continuous example to complement Martin's one. Take the Brownian motion RKHS over $[0,1]$, i.e., the set of functions $f$ over $[0,1]$ with $||f||=\sqrt{\int f'(x)^2dx}$. Let $g=0$ and $$ f_n(x)=\frac{\sqrt{2} }{n\pi}\sin (n\pi x). $$ Then, for each $x$, we have $|f_n(x)-g(x)|\leq 1/\frac{\sqrt{2} }{n\pi}$, while $$ \|f_n-g\|_2=1 $$ for all $n$.