Is the gradient of a function in $H^2_0$ in $H^1_0$? Suppose we have $f\in H^2_0(U)$, so $f$ is the limit of some sequence $(g_n)$ of smooth compactly supported functions on $U\in\mathbb{R}^n$ (assume bounded & smooth boundary) and $f$ is in the Sobolev space $H^2(U)$. Does this imply that $\frac{\partial f}{\partial x_i}\in H^1_0(U)$ for all $i \in \{1,...,n\}$? 
Clearly $f\in H^1(U)$ but I'm not sure if it's the limit of a sequence in $C^{\infty}_c(U)$. Can we just take the sequence $(\frac{\partial g_n}{\partial x_i})$, or does this derivative not commute with the limit $n\rightarrow\infty$?

If $g_n$ converges towards $f$ in $H^2(U)$, you have
$$ \sum_{|\alpha|\le 2}\left\| \partial^\alpha f - \partial^\alpha g_n \right\|^2_{L^2(U)} \to 0,$$
where the sum ranges over all multiindices $\alpha$, with $|\alpha| \le 2$.
In order to prove $\partial g_n / \partial x_i \to \partial f / \partial x_i$ in $H^1(U)$, you have to prove$$ \sum_{|\alpha|\le 1}\left\| \partial^\alpha \partial_{x_i} f - \partial^\alpha \partial_{x_i} g_n \right\|^2_{L^2(U)} \to 0.$$
Now you use
$$\sum_{|\alpha|\le 1}\left\| \partial^\alpha \partial_{x_i} f - \partial^\alpha \partial_{x_i} g_n \right\|^2_{L^2(U)} \le \sum_{|\alpha|\le 2}\left\| \partial^\alpha f - \partial^\alpha g_n \right\|^2_{L^2(U)}.$$
Hence, the (components  of the) gradient belongs to $H_0^1(U)$.
