Let $B$ be the unit ball in the Euclidean space $\mathbb{R}^n$. Consider the set of functions $$X=\{u\in C^2(\bar B) \mid u|_{\partial B}=0 \text{ and } \|\Delta u\|_{L^2(B)}\leq 1\},$$ where $\Delta$ is the Laplacian.
Is it true that the set of first derivatives $\{\frac{\partial u}{\partial x_1}|\, u\in X\}$ is pre-compact in $L^2(B)$?
Yes, this is true and the "proof" is as follows (I wouldn't really dare to call that a proof, though). For any $u\in X$ write $f=-\Delta u$, and observe that trivially $u$ solves the elliptic PDE $$ \left\{ \begin{array}{ll} -\Delta u=f & \mbox{in }B\\ u=0 & \mbox{on }\partial B \end{array} \right.. $$ (This is a tautology, if you will.) By standard elliptic regularity (see e.g. Evans' textbook on PDEs, section 6.3.2 and in particular remark (i) just after theorem 4) we get that $$ \|u\|_{H^2(B)}\leq C \|f\|_{L^2(B)}=C \|\Delta u\|_{L^2(B)}\leq C $$ for some purely dimensional $C\equiv C(n)$. In other words, your set $X$ is bounded in $H^2$. The precompactness then follows from the Rellich-Kondrachov theorem applied to $u_i=\partial_{x_i}u$, with of course $\|\nabla u_i\|_{L^2}\leq \|u\|_{H^2}$ (the full Hessian of $u$ controls any first derivative of $u_i$).
For a more "serious" reference (by that I mean not a textbook) I can recommend Gilbarg-Trudinger.
