The following lemma is from Brezis' text. Throughout we take $\Omega=\mathbb{R}^{N}_{+}$.
Let $u\in H^{2}(\Omega)\cap H^{1}_{0}(\Omega)$ satisfying,
\begin{align}
\int\nabla u\cdot\nabla\varphi+\int u\varphi=\int f\varphi\quad\forall \varphi\in H^{1}_{0}(\Omega).
\end{align}
Then $Du\in H^{1}_{0}(\Omega)$ and, moreover,
\begin{align}
\int\nabla(Du)\cdot\nabla\varphi+\int(Du)\varphi=\int f\varphi\quad\forall\varphi\in H^{1}_{0}(\Omega).
\end{align}
The proof is as follows:
Let $h=|h|e_{j}$, $1\leq j\leq N-1$, so that $D_{h}u\in H^{1}_{0}$. By Lemma 9.6 we have,
\begin{align}
\|D_{h}u\|_{H^{1}}\leq\|u\|_{H^{2}}.
\end{align}
Thus there exists a sequence $h_{n}\rightarrow 0$ such that $D_{h_{n}}u$ converges weakly to some $g$ in $H^{1}_{0}$. In particular, $D_{h_{n}}u\rightharpoonup g$ weakly in $L^{2}$. For $\varphi\in C_{c}^{\infty}(\Omega)$ we have,
\begin{align}
\int(D_{h}u)\varphi=-\int u(D_{-h}\varphi),
\end{align}
and in the limit, as $h_{n}\rightarrow 0$, we obtain,
\begin{align}
\int g\varphi=-\int u\frac{\partial\varphi}{\partial x_{j}}\quad\forall\varphi\in C_{c}^{\infty}(\Omega).
\end{align}
Therefore, $\frac{\partial u}{\partial x_{j}}=g\in H^{1}_{0}(\Omega)$.
My Question: Why does a bound on the $H^{1}$ norm of $D_{h}u$ guarantee a weakly convergent sequence $D_{h_{n}}u$? Also in the last part where he is taking the limit $h_{n}\rightarrow 0$ in the integrals, there are no $h_{n}$ terms appearing, so what are we really doing here?
Best Answer
Since this is also a boundedness of $D_{h_n} u$. Take a sequence $h_n=|h_n|e_j$ and now let $|h_n| \to 0$. Hence also $h_n \to 0$ and $h_n$ corresponds to a tangential translation operator i.e. on the same lines from the proof we get $\|D_{h_n} u\|_{H^1}$ bounded.
He writes down the formula for a general $h$, yes, but he applies the limit to the equation where we replace $h$ by $h_n$, i.e.
$$\int g \varphi \leftarrow \int (D_{h_n} u) \varphi = \int u (D_{-h_n} \varphi) \rightarrow -\int u \partial_{x_j}\varphi. $$