Identification of $L^2$ limits with distributional convergence

Question

I just read the thread on "too much effort" and I would like to be more specific.

Is the following reasoning correct: Let $g,g_\delta\in H^1(D)$, $D$ some domain in $\mathbb{R}^n$ with the following convergences $g_\delta\to g\in L^2(D)$, $$\nabla g_\delta \to \chi_{E}\nabla g $$ in $L^2(D)$, where $E\subset D$ and $E\neq D$. $D\setminus E$ might have positive measure. Furthermore, one has distributional convergence $$\int \nabla g_\delta \xi= - \int g_\delta \nabla\cdot \xi \to -\int g \nabla\cdot \xi = \int \nabla g \xi.$$

With the fundamental theorem of calculus of variations one obtains even $$\chi_E\nabla g =\nabla g$$ a.e. in $D$ and hence even $g_\delta \to g$ strongly in $H^1(D)$. Is this reasoning correct? So far, I haven't used distibutional convergence to identify limits. Of course I know that strong convergence implies weak convergence, but what puzzles me is that the strong convergence to the desired limit only occurse on a relatively small part of the domain.

Answer 1

The presence of $\chi_E$ is confusing. Here's a simpler version:

If $f_n$ converge to something in $L^2$ and $f_n\to f$ in the sense of distributions, then $f_n\to f$ in $L^2$.

This sort of argument (with different kinds of convergence) is pretty standard. The proof is what you wrote: let $\tilde f$ be the $L^2$ limit; then it's also a distributional limit; but the distributional limit is unique; so $\tilde f = f$.