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Consider the inner product by $\langle f,g \rangle_{H^1} = \langle f, g \rangle_{L^2} + \sum_{|\alpha|=1} \langle D^\alpha f, D^\alpha g \rangle_{L^2}$ where $\alpha$ is a multi-index and $D$ denotes the weak derivative. Define $H^1(\Omega)$ as the space of functions that are finite under the norm induced from this inner product. It can be shown that $H^1(\Omega)$ is a Hilbert space.

Now, suppose there exists a sequence of functions $\{ f_n \} \subset H^1(\Omega)$ that converges weakly in $H^1$ to some limit $f \in H^1(\Omega)$. Can I then say that this sequence converges weakly to the same limit under the $L^2$ inner product?

By an application of the Banach-Alaoglu Theorem, I know that weak convergence of this sequence in $H^1$ will imply strong convergence of a subsequence in $H^1$. And then I think strong convergence of this subsequence in $H^1$ will imply strong convergence in $L^2$ as well.

However, I'm not sure if anything can be said about the entire sequence under the $L^2$ inner product and its weak/strong convergence properties.

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up vote 8 down vote accepted

$L^2(\Omega)$ is a subspace of $H^{-1}(\Omega)$, the dual of $H^1(\Omega)$. Thus, a sequence converging weakly in $H^1$ converges weakly in $L^2$ to the same limit.

In general, nothing can be said about strong convergence in $L^2$.

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I don't see how a simple statement '$L^2(\Omega)$ is a subspace of $H^{-1}(\Omega)$' proves that the weakly converging sequence in $H^1$ converges to the same limit in $L^2$ because the inner products on $L^2$ and $H^1$ are different. Here is my attempt:

Let $(u_n)_n$ be a sequence converging weakly to $u$ in $H^1$, hence $(u_n)_n$ is bounded in $L^2$, and we can extract a subsequence $(u_{\varphi(n)})_n$ that converges weakly to $v \in L^2$. $\forall \phi \in C^1_c(\Omega)$, $\left|\int_\Omega u_{\varphi(n)} \phi'\right| = \left|\int_\Omega u_{\varphi(n)}' \phi\right|\leq C ||\phi||_{L^2}\Rightarrow \left|\int_\Omega v\phi'\right|\leq C ||\phi||_{L^2}$, hence $v$ is weakly differentiable. Furthermore, $\left|\int_\Omega (u_{\varphi(n)}' - v')\phi\right| = \left|\int_\Omega (u_{\varphi(n)} - v)\phi'\right| \xrightarrow{n\rightarrow \infty} 0$ by the weak convergence of $(u_{\varphi(n)})_n$. Thus we have that $(u_{\varphi(n)}')_n$ converges weakly to $v'$ in $L^2$ by the density of $C^1_c$ in $L^2$. To conclude, $\forall f\in H^1, \int_\Omega u_{\varphi(n)} f + \int_\Omega u_{\varphi(n)}' f' \xrightarrow{n\rightarrow \infty} \int_\Omega v f + \int_\Omega v' f'$ by the respective weak convergence of $u_{\varphi(n)}$ and $u_{\varphi(n)}'$. The uniqueness of weak limit then implies $v=u$ in $H^1\subset L^2$.

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Weak convergence in either function space implies convergence in the sense of distributions: that is, $\int u_n\phi \to \int u \phi$ for any $\phi\in C^\infty_c(\Omega)$. The distributional limit is unique. Therefore, the limits in $H^1$ and in $L^2$ represent the same distribution, i.e., they are the same. // Distributions are a convenient common denominator of function spaces. – user53153 Feb 28 '13 at 1:52
The statement "$L^2(\Omega)$ is a subspace of $H^{-1}(\Omega)$, the dual of $H^1(\Omega)$" should IMHO be interpreted as follows: for each $g \in L^2(\Omega)$, the map $f \mapsto \int_{\Omega} fg$ is a continuous linear functional on $H^1(\Omega)$. Indeed, $\lvert \int_\Omega fg \rvert \le \|f\|_{H^1} \|g\|_{L^2}$. This is easy to verify and immediately implies the desired statement about weak convergence. – Nate Eldredge Mar 9 '13 at 15:28

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