$F$ is lower semicontinuous $\iff F(x)=\sup_{r>0}\inf_{y\in B(x,r)}F(y)$ for all $x\in X$

Question

Let (X,d) be a metric space. $F: X\to \overline{\mathbb{R}}$. The definitions that I have to use are:

(1) $F$ is sequentially lower semicontinuous if for all sequences $(x_n)_n \subseteq X$ s.t $ x_n\to x$ , $\liminf_n F(x_n)\ge F(x)$

(2) $F$ is lower semicontinuous if $\forall x \in X$ and $t \in \mathbb{R} $ with $F(x)>t$, there exists $r>0$ s.t $F(y)>t$ for all $y\in B(x,r)$

(3) It holds $F(x)=\sup_{r>0}\inf_{y\in B(x,r)}F(y)$ for all $x\in X$

I have already proven $(1)\implies (3)$ and that $ (1)\iff (2)$ I have to prove $(3)\implies (1)$:

My tries:

I want to proof that $F(x) \le \liminf_n F(x_n)=\sup_k\inf_{n>k}F(x_k)$. Take $x_n \to x.$

try 1: I have tried to use the definition of $\sup$ in (3):

Calling $\inf_{y\in B(x,r)}F(y):=\alpha(r)$

$\forall \varepsilon >0 \exists \alpha_{\varepsilon}(r)$ s.t $ F(x)-\varepsilon < \alpha_{\varepsilon}(r) \le F(x) $

... no clue if this is helpful

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try 2

Since $x_n \to x$, for large $n, x_n \in B(x,r)$, so $\inf_{y\in B(x,r)}F(y)\le F(x_n)$, for large n. Then taking the sup over $r>0$: $F(x)=\sup_{r>0}\inf_{y\in B(x,r)}F(y)\le F(x_n)$ and taking the $\liminf $ over $ n$:

$F(x)\le \liminf_n F(x_n)$

This looks correct to me, but I am unsure,specially when taking the sup and then the liminf, maybe it's not allowed? It feels too easy, what do you think?

Answer 1

Your second attempt is not quite correct, for sloopiness on quantifiers. Let us make it rigorous.

Since $x_n\to x$ we have $$\forall r>0\quad\exists N\quad\forall n\ge N\quad x_n\in B(x,r).$$ Therefore, $$\forall r>0\quad\exists N\quad\forall n\ge N\quad F(x_n)\ge\inf_{y\in B(x,r)}F(y)$$ or equivalently $$\forall r>0\quad\exists N\quad\inf_{n\ge N}F(x_n)\ge\inf_{y\in B(x,r)}F(y).$$ This implies $$\forall r>0\quad\sup_N\inf_{n\ge N}F(x_n)\ge\inf_{y\in B(x,r)}F(y),$$ which is equivalent to $$\liminf F(x_n)\ge\sup_{r>0}\inf_{y\in B(x,r)}F(y).$$ This proves $(3)\implies(1).$ However, it is much easier to prove $(3)\implies(2)$: assume $$t<F(x)=\sup_{r>0}\alpha(r),\quad\text{where}\quad\alpha(r):=\inf_{y\in B(x,r)}F(y).$$ By definition of $\sup,$ $$\exists r>0\quad t<\alpha(r)$$ and by definition of $\alpha,$ the conclusion follows.

Answer 2

This should be easier when argued via contradiction, i.e. assume that (a) fails, then you can extract a subsequence which converges to a value strictly lower than $F(x)$. Now the subsequence is eventually in every neighborhood of $x$. Use this and the claim should follow.