# Relation between strong convergence in $L^{p}$ and weak convergence in $H_{0}^{1}(\Omega)$

Let $$\{u_{n}\}_{n\in\mathbb{N}}$$ be a bounded sequence in $$H_{0}^{1}(\Omega)$$ for a bounded interval $$\Omega \subset \mathbb{R}$$. By weak compactness of Hilbert Space, we can extract a subsequence of $$\{u^{1}_{n_{k}}\}_{k\in\mathbb{N}}$$ such that $$u^{1}_{n_{k}}\to u^{1}$$ weakly in $$H_{0}^{1}(\Omega)$$.

On the other hand, by Rellich Theorem, we can obtain another subsequence $$\{u^{2}_{n_{k}} \}_{k\in\mathbb{N}}$$ such that $$u^{2}_{n_{k}}\to u^{2}$$ strongly in $$L^{p}(\Omega)$$ for $$2.

My question is how can we show that $$u^{1} = u^{2}$$? At the very least, I want to obtain a subsequence which satisfies both properties.

My attempt so far is first to extract subsequence $$\{u^{1}_{n_{k}} \}_{k\in\mathbb{N}}$$ from $$\{u_{n}\}_{n\in\mathbb{N}}$$. Then, I extract subsequence $$\{u^{2}_{n_{k}} \}_{k\in\mathbb{N}}$$ as another subsequence from $$\{u^{1}_{n_{k}} \}_{k\in\mathbb{N}}$$. Is this enough to ensure that $$u^{1} = u^{2}$$? If it is enough, in what sense can I say that $$u^{1} = u^{2}$$?

Any help is much appreciated! Thank you very much

• You should write $u_{n_k}$ instead of $u_{n_k}^1$. We assume that $u_n$ is a bounded sequence in $H^{1}_0 (\Omega)$. Then, there exists a subsequence $u_{n_k}$ of $u_n$ and $u^1 \in H^{1}_{0}(\Omega)$ such that $u_{n_k} \to u^1$ weakly in $H_{0}^{1}(\Omega)$. Then, by the Rellich theorem, $u_{n_k} \to u^1$ in $L^{2}(\Omega)$, where you do not have to take a subsequnce. – sharpe Apr 11 at 9:43
• How to apply Rellich Theorem directly to the subsequence? – Evan William Chandra Apr 11 at 11:28

Recall the following result.

Let $$X,Y$$ be Hilbert spaces and $$T:X \to Y$$ a compact operator.

Let $$\{x_n\}$$ be a sequence in $$X$$ such that $$x_n \to x$$ weakly in $$X$$.

Then, it holds that $$Tx_n \to Tx$$ in $$Y$$.

Proof.

Let $$y_n=Tx_n$$. It is enough to show that every subsequence of $$y_n$$ has a convergent subsequence which converges to $$Tx$$.

Let $$y_{n_k}$$ be a subsequence of $$y_n$$. Since $$T$$ is a compact operator, $$y_{n_k}$$ is a sequence in a compact metric space. Therefore, there exists a convergent subsequence $$y_{n_{k_l}}$$. We denote by $$y \in Y$$ the limit.

It remains to show $$y=Tx$$. Since $$x_{n_{k_l}} \to x$$ weakly in $$X$$, $$y_{n_{k_l}}=Tx_{n_{k_l}} \to Tx$$ weakly in $$Y$$. Hence $$y=Tx$$.

• Thank you, now I can work it on myself! – Evan William Chandra Apr 12 at 3:02
• @EvanWilliamChandra You're welcome. – sharpe Apr 12 at 7:31