Weak convergence for subsequence + pointwise a.e. convergence for the full sequence

Question

Suppose a sequence of functions on a finite measure space is bounded in $H^1$. That is, $\lVert f_n \rVert_{H^1} \leq C$, so that $f_{n_j} \to f$ in $L^2$ for asubsequence $n_j$ and it also converge weakly in $H^1$.

If I know that $f_n \to f$ pointwise a.e., and monotonically, is it enough to conclude that $f_n \to f$ in $L^2$? That is, the whole sequence converges?

Answer 1

I would like to put this as a comment but my reputation is too low to do it.

To show the whole sequence convergence, we take arbitrary sub-sequence of $f_n$, say $f_{n_k}$, and use the argument provided by @fourierwho you can show that there exists a subsequence of $f_{n_k}$, say $f_{n_{k_m}}$ such that it converges in $L^2$ to a function $\bar f$.

Now, since we know that $f_n\to f$ a.e., we must have $\bar f=f$. Thus, we showed that for arbitrary subsequence of $f_n$, we always have a further sequence which convergences to $f$, the same limiting function.

Hence, you may apply the subsequence principle to conclude that the whole sequence does convergence in $L^2$ to $f$.

If you does not know subsequence principle, here it's how it works. Suppose the conclusion is not right, that is, there exits a subsequence of $f_n$ such that it does not convergence to $f$ in $L^2$. Then apply @fourierwho's argument, we have the contradiction.

Answer 2

See the Rellich-Kondrachov Theorem. https://en.wikipedia.org/wiki/Rellich%E2%80%93Kondrachov_theorem

I'm assuming by "finite measure space" you mean something like "bounded open Lipschitz domain."

First, note that since $f_{n} \to f$ a.e. and $(f_{n})_{n \in \mathbb{N}}$ is bounded in $L^{2}(U)$, it follows that $f_{n} \rightharpoonup f$ (weakly) in $L^{2}(U)$.

In fact, $f \in H^{1}(U)$ and $f_{n} \rightharpoonup f$ in $H^{1}(U)$. To see this, pick $\varphi \in C_{c}^{\infty}(U)$ and $j \in \{1,2,\dots,n\}$ and observe that $$\lim_{n \to \infty} \int_{U} \frac{\partial f_{n}}{\partial x_{j}}(x) \varphi(x) \, dx = \lim_{n \to \infty} \int_{U} f_{n}(x) \frac{\partial \varphi}{\partial x_{j}}(x) \, dx = \int_{U} f(x) \frac{\partial \varphi}{\partial x_{j}}(x)\, dx.$$ Therefore, by density of $C_{c}^{\infty}(U)$ in $L^{2}(U)$ and boundedness of $(\frac{\partial f_{n}}{\partial x_{j}})_{n \in \mathbb{N}}$ in $L^{2}(U)$, $\varphi \mapsto \lim_{n \to \infty} \int_{U} \frac{\partial f_{n}}{\partial x_{j}}(x) \varphi(x) \, dx$ extends to a bounded linear functional on $L^{2}(U)$. The Riesz Representation Theorem implies there is a $g_{j}$ such that $$\forall \varphi \in C_{c}^{\infty}(U) \quad \int_{U} g_{j}(x) \varphi(x) \, dx = \lim_{n \to \infty} \int_{U} \frac{\partial f_{n}}{\partial x_{j}}(x) \varphi(x) \, dx = \int_{U} f(x) \frac{\partial \varphi}{\partial x_{j}}(x) \, dx.$$ Therefore, $\frac{\partial f}{\partial x_{j}} = g_{j}$ and $\frac{\partial f_{n}}{\partial x_{j}} \rightharpoonup g_{j}$. Since this works for each $j$, we deduce $f \in H^{1}(U)$ and $f_{n} \rightharpoonup f$ in $H^{1}(U)$.

We now conclude $f \in H^{1}(U)$ and, moreover, $f_{n} \rightharpoonup f$ (weakly) in $H^{1}(U)$. Since the embedding $H^{1}(U) \hookrightarrow L^{2}(U)$ is compact, it follows that $f_{n} \to f$ in $L^{2}(U)$.