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I have a question related to lower semicontinuity of real valued function.

Let $X$ be a metric space, $x_0\in X$ and $f:X\rightarrow\overline{\mathbb{R}}$ a function. We say that $f$ is lower semicontinuous in $x_0$ if for all $y<f(x_0)$, there exists an open neighbourhood $U$ of $x_0$ such that $f(x)>y$ for all $x\in U$.

If $E$ is a subspace of $X$, $x_0\in E'$ and $f:E\rightarrow\overline{\mathbb{R}}$ is a function, we define the limit inferior of $f(x)$ when $x$ goes to $x_0$ as follows:

$$\liminf_{x\to x_0}{f(x)}=\left\{L\in\mathbb{R}\middle|\exists\{x_n\}_{n\in\mathbb{N}}\subset E,\lim_{n\to\infty}{x_n}=x_0\text{ and }\lim_{n\to\infty}{f(x_n)}=L\right\}.$$

I know there are other definitions for inferior limit (using infimum of balls), but this is the only I could understand and work with.

What I want to prove is the following:

Let $X$ be a metric space, $f:X\rightarrow\overline{\mathbb{R}}$ a function and $x_0\in X$. Then $f$ is lower semicontinuous in $x_0$ if and only if $\liminf_{x\to x_0}{f(x)}\geq f(x_0)$.

What I've tried is this:

Suppose that $f$ is semicontinuous in $x_0$ and let $L\in\mathbb{R}$ such that there exists $\{x_n\}_{n\in\mathbb{N}}\subset X$ which converges to $x_0$ and $\lim_{n\to\infty}{f(x_n)}=L$. I have to prove that $L\geq f(x_0)$.

By reduction to the absurd, let suppose that $f(x_0)>L$. Then there exists an open neighbourhood $U$ of $x_0$ such that $f(x)>L$ for all $x\in U$. As $\lim_{n\to\infty}{x_n}=x_0$, there exists $n_0\in\mathbb{N}$ such that $x_n\in U$ for all $n\geq n_0$. Then:

$$f(x_n)>L,\forall n\geq n_0.$$

From this I don't have anything, because when I do $n\to\infty$, the strict inequality is lost and we have $L\geq L$.

How could you prove it? (First part solved!)

For the other direction, I don't have anything (I need to use the definitions shown above). I was thinking something like this:

By reduction to the absurd, suppose that $f$ is not lower semicontinuous in $x_0$. Then, there exists $y<f(x_0)$ such that for all open neighbourhood $U$ of $x_0$, there exists $x\in U$ satisfying $f(x)\leq y$. For each $n\in\mathbb{N}$, let be $x_n\in B(x_0,1/n)$ such that $f(x_n)\leq y$. We know that $\lim_{n\to\infty}{x_n}=x_0$.

If there's $L\in\mathbb{R}$ such that $\lim_{n\to\infty}{f(x_n)}=L$, we've got a contradiction because:

$$y<f(x_0)\leq\liminf_{x\to x_0}{f(x)}\leq L\leq y.$$

What if there's no $L\in\mathbb{R}$? I need to construct a sequence $\{x_n\}_{n\in\mathbb{N}}$ so that the sequence $\{f(x_n)\}_{n\in\mathbb{N}}$ is convergent in $\mathbb{R}$.

Thanks in advance.

Proof.

$\boxed{\Rightarrow}$ Let $L\in\mathbb{R}$ such that there exists $\{x_n\}_{n\in\mathbb{N}}\subset X$ verifying $\lim_{n\to\infty}{x_n}=x_0$ and $\lim_{n\to\infty}{f(x_n)}=L$. We must prove that $L\geq f(x_0)$.

By reduction to the absurd, suppose that $f(x_0)>L$. Then:

$$f(x_0)=\dfrac{f(x_0)}{2}+\dfrac{f(x_0)}{2}>\dfrac{f(x_0)}{2}+\dfrac{L}{2}.$$

As $f$ is lower semicontinuous, there is a open neighbourhood $U$ of $x_0$ such that:

$$f(x)>\dfrac{f(x_0)}{2}+\dfrac{L}{2}.$$

As a consequence of $\lim_{n\to\infty}{x_n}=x_0$, there exists $n_0\in\mathbb{N}$ so that for all $n\geq n_0$, $x_n\in U$. Therefore:

$$f(x_n)>\dfrac{f(x_0)}{2}+\dfrac{L}{2}.$$

Taking limits when $n\to \infty$:

$$L\geq\dfrac{f(x_0)}{2}+\dfrac{L}{2}>\dfrac{L}{2}+\dfrac{L}{2}=L.$$

We've got a contradiction.

$\boxed{\Leftarrow}$ [...]

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  • $\begingroup$ They use another definition $\endgroup$ Commented Apr 7, 2022 at 10:55
  • $\begingroup$ Thanks for the clarification. Your argument above can be instantly improved, because it's very weak at one point. If $f(x_0)>L$ , then $\lim_{n \to \infty} f(x_n) = f(x_0)$ will tell you something stronger : the point is that $f(x_0) > \frac{f(x_0)}2+ \frac{L}{2}>L$ as well, so use this to see that there is an open set $U$ on which $f(x_n) > \frac{f(x_0)}2+ \frac{L}{2}$ whenever $x_n \in U$. That argument will then work out. $\endgroup$ Commented Apr 7, 2022 at 11:03
  • $\begingroup$ Wow, thanks a lot. Really helpful. $\endgroup$ Commented Apr 7, 2022 at 11:05
  • $\begingroup$ Welcome. Try the other way, if you have already got it then write an answer, otherwise edit it as an attempt and we can fix it. $\endgroup$ Commented Apr 7, 2022 at 11:11
  • $\begingroup$ @SarveshRavichandranIyer Why not an official answer? $\endgroup$
    – Paul Frost
    Commented Apr 7, 2022 at 22:46

1 Answer 1

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Suppose $f$ is lower semicontinuous and let $\{x_n:n\in D\}$ be a sequence (more generally net if the space is just a Hausdorff top. space) that converges to $x$. For any $\alpha>f(x)$ the set $V=\{f>\alpha\}$ is an open neighborhood of $x$. Hence there is $n_0\in D$ such that $n\geq n_0$ implies that $f(x_n)>\alpha$; this implies that $\alpha\leq\liminf_nf(x_n)$. Necessity follows by letting $\alpha\rightarrow f(x)$.

Conversely, suppose $f(x)\leq \liminf_nx_n$ for any $x\in X$ and any sequence (net) $x_n\rightarrow x$. We will show that for each $\alpha\in\mathbb{R}$, the set $F_\alpha:=\{f\leq \alpha\}$ is closed. Let $\{x_n:n\in D\}$ be a sequence (net) in $F_\alpha$ that converges to a point $x\in X$. Then $f(x_n)\leq \alpha$ for all $n\in D$, and so $$ f(x)\leq \liminf_nf(x_n)= \sup_{n\in D}\inf_{m\in D: m\geq n}f(x_m)\leq \alpha. $$ Therefore $x\in F_\alpha$.

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