Help proving the equivalence of lower semicontinuity and sequential lower semicontinuity involving the negation of a stament with multiple quantifiers

Question

Let $X$ be a metric space, $G:X \to \mathbb{R}$ a functional

Def 1) $G$ is slsc(sequentially lower semicontinuos) at $x$ $\Longleftrightarrow$ $G(x) \leq \liminf_{x_n\rightarrow x} G(x_n)$ $\forall x_n \rightarrow x$ in $X$.

Def 2) $G$ is lsc(lower semicontinuos) if for every $x \in X $and every $t \in \mathbb{R}$ with $G(x ) > t$, there exists $r > 0$ such that $G(y) > t$ for all $y ∈ B(x , r )$;

I wanto prove that they are equivalent

To prove $1) \implies 2)$ I argue by contradiction.

Suppose 1) holds and 2) is false. 2) being false means that its negation is true. First I rewrite 2) in a more clear way, specifying the quantificators and moving them to the begining of the sentence I think the statement has an if-then structure:

Def 2') $G$ is lsc(lower semicontinuos) if $\forall x \in X $, $\forall t \in \mathbb{R}$, $\exists r > 0$: $\forall y ∈ B(x , r )$, if $G(x ) > t \implies G(y) > t $

So the negation should be

$G$ is NOT lsc(lower semicontinuos) if $\exists x \in X $, $\exists t \in \mathbb{R}$, $\forall r > 0$: $\exists y ∈ B(x , r )$, $G(x ) > t$ and $G(y) \le t $

and I chain up these last two inequalities together with that from the hypothesis of slsc:

$\forall x_n \rightarrow x$ in $X$, $G(y) \le t<G(x) \leq \liminf_{x_n\rightarrow x} G(x_n)$

1) From here I don't know how to continue. In particular, am I correctly negating the statement ?I am a bit unsure if it was ok to move the quantifiers around as I did

2) How do I prove $2) \implies 1)$?

Answer 1

For $ t \in \mathbb{R}$, let $ {S}_{t} = {G}^{{-1}} \left(\left(t , \infty \right)\right)$. We can rephrase your question's properties 1) and 2) as $\ $

1. For every $ t < G \left(x\right)$ and every sequence $ {x}_{n} \rightarrow x$, one has $ t \leqslant {\liminf }_{n} G \left({x}_{n}\right)$
2. For every $ t < G \left(x\right)$, there is a $ r > 0$ such that $B \left(x , r\right) \subset {S}_{t}$

Proof that 2) $\Longrightarrow $ 1)

Suppose 2), let $ t < G \left(x\right)$ and $ {x}_{n} \rightarrow x$. Choose $ r$ as in 2). For $ n$ large enough, one has $ {x}_{n} \in B \left(x , r\right)$, hence $ t < G \left({x}_{n}\right)$. By taking the limit inf, it follows that $ t \leqslant {\liminf }_{n} G \left({x}_{n}\right)$.

Proof that 1) $\Longrightarrow $ 2)

We use the following property: for every $ x \in X$ and every set $ S \subset X$, exactly one of the two properties holds $\ $

• a) There is a $ r > 0$ such that $B \left(x , r\right) \subset S$
• b) There is a sequence $ {x}_{n} \rightarrow x$ such that $ \forall n , {x}_{n} \notin S$

Indeed case a) means that $x \in \text{int} \left(S\right)$ and case b) means that $ x \in \text{cl} \left(X \setminus S\right)$, but it is well known that the complement of the interior is the closure of the complement.

Now, suppose that 2) is false, let $t < t' < G \left(x\right)$ such that a) doesn't hold for $ {S}_{t}$. By b) one can find a sequence $ {x}_{n} \rightarrow x$ such that $ {x}_{n} \notin {S}_{t}$, that is to say $ G \left({x}_{n}\right) \leqslant t$. It follows that $ {\liminf }_{n} G({x}_{n}) \leqslant t < t'$ and 1) is false for $t'$.