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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 $2<p<\infty$.

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

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  • $\begingroup$ 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. $\endgroup$ – sharpe Apr 11 at 9:43
  • $\begingroup$ How to apply Rellich Theorem directly to the subsequence? $\endgroup$ – Evan William Chandra Apr 11 at 11:28
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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 $y=Tx$.

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

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