I have the following statement I need to prove.
Let $f:D \to \mathbb{R}$. Then the function $f$ is lower semicontinuous at some point $x_0 \in D$ if
$$\lim_{x \to x_0 } \inf f(x) \geq f(x_0)$$.
I have the studied the proof on this link, but have trouble understanding most of it.
It starts off by defining the following set, which is a neighbourhood of $x_0$.
$$B(x_0;\delta) = (x_0-\delta, x_0) \cup (x_0, x_0 + \delta)$$.
If $f$ is lower semicontinuous, for every $\epsilon > 0$ there exists a $\delta$ such that
$$\forall x \in B(x_0; \delta) \left( f(x) \geq f(x_0) -\epsilon \right)$$ holds true.
This implies that the infimum of $f(x)$ where $x \in B(x_0; \delta)$ is also greater.
$$\inf_{x \in B(x_0; \delta)} f(x) \geq f(x_0) - \epsilon$$.
Define $h(\delta) = \inf_{x \in B(x_0; \delta)} f(x)$.
Now the proof states that $\sup h(\delta) = \lim_{x \to x_0 } \inf f(x) $ which then makes use the fact that $\sup h(\delta) \geq h(\delta)$.
This is one of the things I don't understand. $\sup h(\delta)$ is a decreasing function. If you consider $B(x_0, d\delta)$ for some infinitesimal $d\delta$, $\inf_{x \in B(x_0; d\delta)} f(x) = f(x_0)$. And every $B(x_0; \delta)$ contains $B(x_0; d\delta)$. This just tells me that $\sup h(\delta) = f(x_0)$.
But $\lim_{x \to x_0} \inf f(x)$ is the infimum of $f$ in $[x_0, \infty)$. That is $\lim_{x \to x_0} \inf f(x) = \lim_{y \to x_0 } \inf_{x \geq y} f(x)$ and this value does not have to be equal to $f(x_0)$.
The second thing I don't understand is that even if you show that $\lim_{x \to x_0} \inf f(x) \geq f(x_0) - \epsilon$ how does that prove that $\lim_{x \to x_0} \inf f(x) \geq f(x_0)$.
In summary ...
- Why is $\lim_{x \to x_0 } \inf f(x) = \sup h(\delta)$?
- How do you show that $\lim_{x \to x_0} \inf f(x) \geq f(x_0) - \epsilon \implies \lim_{x \to x_0} \inf f(x) \geq f(x_0)$