Strongly convergent subsequence $+$ point-wise convergence $\Rightarrow$ strong convergence?

Question

Let $X$ be a Hilbert space containing functions defined over a bounded region $\Omega\subset \mathbb{R}^N$. Assume $f_n\in X$ has a strongly convergent subsequence, say $f_{n_k}$. Also, assume $f_n\to f$, $f\in X$, point-wise over $\Omega$. Can we conclude that $f_n\to f$ strongly?

Answer 1

No. To see this, take any sequence $g_n$ such that $g_n\to f$ pointwise, but not strongly. Then take the sequence $f_n=\{f,g_1,f,g_2,...\}$. The odd indices form a strongly convergent subsequence, but $f_n$ clearly converges pointwise, but not strongly.

Answer 2

I will answer a slightly different question, because this question might be what you are wanting to ask.

1. Let us additionally assume that the convergence in $X$ has some relation to pointwise convergence, otherwise it is impossible to conclude anything. In particular, we assume that every convergent sequence in $X$ possesses a pointwise convergent subsequence. (For $L^p$-spaces, you can relax this to pointwise a.e.).

2. We even require that every subsequence $(f_{n_k})$ of $(f)$ has a strongly convergent subsequence $(f_{n_{k_\ell}})$. This follows, e.g., if the range of the sequence $(f)$ is precompact.

Under these requirements, the entire sequence $(f_n)$ converges towards $f$ strongly in $X$. We argue by contradiction. Assume that $f_n \not\to f$ in $X$. Then, there is $\varepsilon >0$ and a subsequence $(f_{n_k})$ such that $\| f_{n_k} - f\| \ge \varepsilon$. By assumption (2.), there is a strongly convergent subsequence $(f_{n_{k_\ell}})$, i.e., $f_{n_{k_\ell}}\to g$ for some $g \in X$. By assumption (1.), there is a further subsequence $(f_{n_{k_{\ell_m}}})$ with $f_{n_{k_{\ell_m}}} \to g$ pointwise in $\Omega$. But this also implies $f_{n_{k_{\ell_m}}} \to f$ pointwise in $\Omega$ since the original sequence $(f_n)$ converges pointwise towards $f$. Hence, $f = g$. This is a contradiction to $\| f_{n_k} - f\| \ge \varepsilon$ and $f_{n_{k_\ell}}\to g$ in $X$.