Let $H$ denote the set of functions $f:[0, 1]\to\mathbb R$ that are absolutely continuous, with $f' ∈ L_2 [0, 1]$ and $f(0) = f(1) = 0$. Define an inner product on H by $\langle f, g\rangle = \int_0^1 (f(t)g(t) + f'(t)g'(t))dt$. Of course this implies
$$\lVert f\rVert^2=\lVert f\rVert_2^2+\lVert f'\rVert_2^2$$
I want to show that the evaluation functional, i.e. the map $f\to f(x)$ is bounded for any $x$. The way I see it this basically comes down to saying that if $f$ and $f'$ have small $L_2$ norm, and $f$ is absolutely continuous, then $f$ must have small $L_\infty$ norm. Under normal circumstances information about $\lVert f\rVert_2$ tells you nothing about $\lVert f\rVert_\infty$, but here since $f$ is absolutely continuous it does: if $f(x)>y$, then we must have $f(x)>\frac y2$ on an neighborhood of $x$. Having a bound on $\lVert f'\rVert_2$ means that that neighborhood must be fairly wide since $f'$ can't be "too big". However, when I try to convert these intuitions into a proof, I feel like things get very convoluted very quickly. For example, I'm not even sure how to convert an upper bound on $\lVert f'\rVert_2$ into a lower bound for how wide that neighborhood must be.
This is Exercise 1.3 in An introduction to the theory of reproducing kernel Hilbert spaces by Paulsen and Raghupathi.
If $\|f_n-f\|\to 0$ then $f_n(x)=\int_0^{x} f_n' (t)dt \to \int_0^{x}f'(t)dt=f(x)$. This is because $$| \int_0^{x} f'_n (t)dt -\int_0^{x}f'(t)dt|$$ $$=|\int_0^{x} [f'_n (t)-f'(t)]dt| \leq \int_0^{x} |f'_n (t)-f'(t)|dt$$ $$ \leq \int_0^{1} |f'_n (t)-f'(t)|dt\leq \|f'_n-f'\|_2$$ by Cauchy-Schwarz inequality and $\|f'_n-f'\|_2 \leq \|f_n-f\|$.