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Show that f is lower semi-continuous at x if, and only if, $$ \liminf f(x_n) \geq f(x). $$ for any sequence $(x_n)$ converging to $x$.


Proof (only if)*:

Assume that $f$ is lower semi-continuous at $x$. Let $(x_n)$ be an arbitrary sequence converging to $x$. Fix an arbitrary $\varepsilon > 0$. Since $(x_n)$ converges to $x$, $\exists N \in \mathbb{N}$ s.t. $\forall n\geq\mathbb{N}, \ \ |x_n - x| < \delta$

We have that by lower semi-continuity, this implies that (for every $n$ bigger than $N$) $f(x_n) - f(x) > -\varepsilon$. Then it is the case that $$ \liminf f(x_n) - f(x) > -\varepsilon. $$ Since we can take $\varepsilon$ to be as small as we want, we have that $$ \liminf f(x_n) \geq f(x). $$

Since if we assume by contradiction that $f(x) >$ lim inf $f(x_n)$, we get the existence of an $\varepsilon$ and an $\delta$ such that lim inf $f(x_n) - f(x) < -\varepsilon$.


Does this proof hold?

If not, which are the mistakes and the alternatives?

Thank you

*Proof of if comes by contradiction without much worries.

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1 Answer 1

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Your argument looks fine to me.

Now, after you established that $\liminf_nf(x_n)\geq f(x)$, you keep going with "since if we assume by contradiction..." I don't see how that contributes anything to the proof.

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