I know that function evaluation in $H^1(0,1)$ is continuous (see, e.g., Is the Delta distribution a continuous functional on $H^1(\mathbb R)$).
So, $\delta_x : H^1(0,1)\to\mathbb C$ is a continuous operator. Since it has finite-dimensional range, it is compact. In other words, $\delta_x$ is $D$-compact in $L^2(0,1)$, where $Df = f'$, $f\in H^1(0,1)$.
As $D$-compact operators have $D$-bound zero, for any $\varepsilon>0$, I should find an $a > 0$ such that $$ |f(x)|\,\le\,a\|f\|_2 + \varepsilon\|f'\|_2,\quad f\in H^1(0,1). $$ But I can't find such $a$ for given $\varepsilon$. Can anyone help me out?
A somehow clumsy argument: By approximation assume $f$ is $C^1$. First assume $x\leq \frac 12$. If $|f(y)|>\frac 12 |f(x)$| for all $y\in [x, x+\frac 14 \epsilon^2]$, then $$\int_0^1f^2(y)dy\geq \int_x^{x+\frac 14 \epsilon^2}f(y)^2 dy\geq \frac 14\epsilon^2\frac 14 f(x)^2; $$ thus in this case $|f(x)|\leq 4\epsilon^{-1}\|f\|_2$. If for some $y\in [x, x+\frac 14 \epsilon^2]$ we have $|f(y)|\leq \frac 12 |f(x)|$, then $$ f(x)^2\leq 4[f(x)-f(y)]^2=4\Big(\int_x^y -f'(t) dt\Big)^2 \leq 4\int_x^y|f'(t)|^2dt \cdot |y-x| \leq \epsilon^2\int_0^1|f'(t)|^2dt, $$ i.e. $|f(x)|\leq \epsilon \|f'\|_2$.
If $x>\frac 12$ we can argue similarly by considering $y\in [x-\frac 14 \epsilon^2, x]$.
Overall we see that one can take $a=4 \epsilon^{-1}$.