Is $H^2\cap H^1_0$ dense in $H_0^1$?

Question

Let $-\infty<a<b<+\infty$ and $I=(a,b)$. Consider $H^1_0(I)$ equipped with the norm $\|\cdot\|_{H^1}$ given by

$$\|f\|_{H^1}=\|f\|_{L^2}+\|f'\|_{L^2}.$$

Is $H^2(I)\cap H_0^1(I)$ dense in $(H_0^1(I),\|\cdot\|_{H^1})$? If so, how to prove it?

Thanks.

Answer 1

YES it is dense.

Even better, $H^1_0(I)$ is defined as the closure of $C_0^\infty(I)$ is $H^1(I)$.

Hence, not only $H^1_0(I)\cap H^2(I)$ is dense in $H^1_0(I)$, but the set of $C^\infty$ functions in $H^1_0(I)\cap H^2(I)$ is dense in $H^1_0(I)$.