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I'm reading the following proof, but I can't seem to follow one line (the last line listed). The proof is as follows:

Assume by contradiction that $f$ is not lower semicontinuous, meaning that there exists $x^{*} \in E$ and $\{x_n\} \subseteq E$ such that $x_n \rightarrow x^{*}$ and $\liminf_{n \rightarrow \infty} f(x_n)<f(x^{*})$. Take $\alpha$ that satisfies:$$\liminf_{n \rightarrow \infty} f(x_n)<\alpha<f(x^{*})$$

Then, there exists a subsequence $\{x_{n_k}\}_{k \geq 1}\rightarrow x^{*}$ such that $f(x_{n_k}) \leq \alpha$ for all $k \geq 1$.

I'm having trouble seeing why does there exist a subsequence $\{x_{n_k}\}$ such that $f(x_{n_k}) \leq \alpha$. I revised my undergraduate real analysis notes but I couldn't figure it out. I'm guessing it is related to $\liminf_{n \rightarrow \infty}$ but I can't seem to figure it out. Where does it come from? Thanks.

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Suppose there is no such subsequence $\{x_{n_k}\}$; then we know that $f(x_n) \gt \alpha$ for every $n \in {\mathbb N}$. But then $\liminf_{n\rightarrow \infty} f(x_n) $ cannot be less than $\alpha$, which contradicts how we chose $\alpha$.

The key idea here is that the strict inequality $\liminf f(x_n) \lt f(x^*)$ guarantees us some number lying between our two values (a bit like finding a separating hyperplane), and then the limiting process guarantees us that we must cross the hyperplane.

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  • $\begingroup$ This is a really nice answer thanks. If I understand correctly if $f(x_n)>\alpha$ for every $n\in \mathbb{N}$ then this implies $\liminf_{n \rightarrow \infty} f(x_n)>\alpha$ which is a contradiction. Is this correct? $\endgroup$
    – yessssir
    Apr 28 '20 at 14:02
  • $\begingroup$ @yessssir It implies $\liminf_{n\rightarrow \infty} f(x_n) \geq \alpha$ but yes (think about $1/n$ for $n\in {\mathbb N}$ which is $\gt 0$ but the $\liminf$ is $0 $) $\endgroup$
    – postmortes
    Apr 28 '20 at 14:04
  • $\begingroup$ Got it thanks makes sense. $\endgroup$
    – yessssir
    Apr 28 '20 at 14:57

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