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Let $E$ be a normed vector space and $H = \{x \in E; f(x) = \alpha\}$. If $f$ is continuous then because $\{\alpha\}$ is closed $f^{-1}(\alpha) = H$ is also closed.

Now assume that $H$ is closed which means it contains all its limits points. Let $x_n \rightarrow x$. I claim that $f(x) = \alpha$. Since $f(x_n) = \alpha$ for all $n$, then for all $n$ and all $\epsilon > 0$ $$|f(x_n) - \alpha| < \epsilon$$ which shows that $f(x_n) \rightarrow \alpha$.

Is my proof correct? In Brezis' book this is proved without using sequences but it's a little more complicated.

Mathematics
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    https://math.stackexchange.com/questions/tagged/solution-verification: For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is too broad or missing context. Instead, the question must identify precisely which step in the proof is in doubt, and why so. – Anne Bauval Apr 18 '25 at 04:08

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No this is not correct. First, this theorem is only true if $f$ is linear, and you do not even specify it explicitly. Moreover, your proof cannot be correct. You are right that you need to prove $$x_n\to x \Rightarrow f(x_n)\to f(x)$$ but you need to do so for any sequence $x_n$ that converges to a generic $x\in E$. Your proof only works for $x_n\in H$, hence also $x\in H$.

Giuseppe Negro
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  • So my proof only shows that $f$ is continuous on $H$? – Mathematics Apr 18 '25 at 04:03
  • It does, but it is a trivial conclusion, since $\left. f \right|_H$ is just a constant function. – Giuseppe Negro Apr 18 '25 at 04:05
  • Also, it is unclear whether you can assume that $H$ is closed for any $\alpha$ or just for one. – Giuseppe Negro Apr 18 '25 at 04:06
  • In Brezis' book he says that $f$ is a linear functional that does not vanish identically and $\alpha \in \mathbb{R}$ is a given constant. Is it possible to prove this theorem using a sequence argument? – Mathematics Apr 18 '25 at 04:07
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    So the theorem goes like: there exists $\alpha$ such that $H_\alpha$ is closed $\Rightarrow$ $f$ is continuous. Of course you can prove this via sequences. There are several proofs here, I found this for example. – Giuseppe Negro Apr 18 '25 at 04:15