Prove the following claim:$f$ is lower semi-continuous at $x_0$ if and only if $\liminf_{k\rightarrow \infty} f(x_k) \geq f(x_0)$ for all $x_k \rightarrow x_0$ as $k\rightarrow \infty$.
First, assume that $f$ is lower semi-continuous at $x_0$. By definition:
$$\liminf_{k\rightarrow \infty} f(x_k) = \sup_n \inf_{m \geq n} f(x_m)$$
Then, by definition of lower semicontinuity, since $x_m$ is arbitrarily close to $x_0$ for $m$ large enough, the supremum is strictly greater than $f(x_0) – \epsilon$ where $\epsilon > 0$ is arbitrary*. Taking $\epsilon \rightarrow 0$ we have that $\liminf_{k\rightarrow \infty} f(x_k) \geq f(x_0)$ as claimed.
Next, assume that $\liminf_{k\rightarrow \infty} f(x_k) \geq f(x_0)$ for every sequence $x_k$ such that $x_k \rightarrow x_0$ as $k\rightarrow \infty$. By definition:
$$\liminf_{k\rightarrow \infty} f(x_k) < f(x_k) + \epsilon$$
for $k$ large enough. Since this is true for all sequences $x_k\rightarrow x_0$, it is also true for every $x$ in a $\delta$ neighborhood of $x_0$ for some $\delta > 0$**. So, there exists a $\delta > 0$ such that:
$$f(x_0) < f(x) + \epsilon$$
for all $x$ in a $\delta$ neighborhood of $x_0$. But this is the definition of lower semi-continuity.
My question
I am not sure that the claim marked by * is correct. $x_k$ is eventually in any $\delta$ neighborhood of $x_0$, so given any fixed $\epsilon$ I should be able to find a $k$ large enough such that the supremum is greater than $f(x_0) -\epsilon$ as claimed. But I'm not sure if I've missed something.
I am also not sure about the claim at **. Intuitively it makes sense to me since given a point in the neighborhood, I can construct a sequence going through that point and converging to $x_0$.
Are these two claims correct?
Best Answer
You need to be explicit when dealing with *.
Suppose $f$ is lsc. at $x_0$. Suppose $x_k \to x_0$. Pick $\epsilon>0$, then for some $\delta>0$ if $x \in B(x_0,\delta)$ then $f(x) > f(x_0)-\epsilon$. For $k$ large enough we have $f(x_k) > f(x_0)-\epsilon$ and hence $\liminf_k f(x_k) \ge f(x_0)-\epsilon$. Since $\epsilon$ was arbitrary we are finished.
For the other direction **, you can't just leap from sequences to the $x$ in a neighbourhood conclusion, you need to prove it.
One way is the contrapositive. Suppose $f$ is not lsc. at $x_0$. Then there exists $\epsilon>0$ such that for all $\delta>0$ there is some $x \in B(x_0,\delta)$ such that $f(x) \le f(x_0)-\epsilon$. By choosing $\delta={1 \over k}$ we obtain a sequence $x_k \to x_0$ and $f(x_k) \le f(x_0)-\epsilon$. Hence $\liminf_k f(x_k) \le f(x_0)-\epsilon$.