# Characterization of lower semicontinuity [duplicate]

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 $$y, there exists an open neighbourhood $$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 $$y such that for all open neighbourhood $$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

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 $$\mathbb{R}$$.

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 $$n\to \infty$$:

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

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

• They use another definition Commented Apr 7, 2022 at 10:55
• 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. Commented Apr 7, 2022 at 11:03
• Wow, thanks a lot. Really helpful. Commented Apr 7, 2022 at 11:05
• 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. Commented Apr 7, 2022 at 11:11
• @SarveshRavichandranIyer Why not an official answer? Commented Apr 7, 2022 at 22:46

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 $$x\in F_\alpha$$.