Concrete identification of a functional

Question

On, say, $H_0^1$, consider a map of the type

$$ f \mapsto \phi(x_0)f'(x_0) $$

for a fixed $\phi$ and $x_0$. Can this functional be identified as

$$ f \mapsto \int f\psi $$

for some $\psi$---either just formally or otherwise?

Just formally, can anything be said? The issue seems to be that integration by parts doesn't work with general measures.

More than just formally, is pointwise evaluation of the derivative a bounded functional on $H_0^1$? Does Riesz representation hold here?

Added: As pointed out below, this map does not lie in $(H_0^1)^*$ for essentially the same reason that pointwise evaluation is not bounded on $H_0^1$. However, can anything formally be said? What if one replace Sobolev spaces with a rigged Hilbert space---i.e. is more willing to treat $f \mapsto \int f \delta_x$ as a bounded functional?

Answer 1

Pointwise evaluation of derivative is not a bounded functional on $H^1_0$, for the same reason that pointwise evaluation of function itself is not a bounded functional on $L^2$. It cannot even be defined on the entire space in a natural way (short of arbitrary choices based on a Hamel basis).

For a concrete example (in one dimension) let $f_n(x) = n^{-1/2}\sin nx$ for $|x|\le \pi/n$, and $f_n(x)=0$ otherwise. Then the $H^1$ norm of $f_n$ is bounded independently of $n$, since $$ \int_{|x|\le \pi/n} (f_n'(x))^2\,dx = \int_{|x|\le \pi/n} n \cos^2 nx \,dx = \pi $$ but $f_n'(0)= n^{1/2} \to \infty$ as $n\to\infty$.