I don't understand the implication in the following demonstration which shows that if $f_n(x)$ uniformly converges to $f(x)$ then $(f_n(x))_{n\in \mathbb{N}}$ weakly-converges to f(x)
$f_n$ converges uniformly to $f(x) \Leftrightarrow \lim\limits_{n\rightarrow +\infty}\sup\limits_{x\in I}|f_n(x)-f(x)|=0$
$\Rightarrow\forall x\in I \lim\limits_{n\rightarrow +\infty}f_n(x)=f(x)\forall x\in I$
Can you help me understand it? I don't understand why the $\sup$ disapears for a simple $\forall$...
Let $x\in I$, one has: $$\forall n\in\mathbb{N},|f_n(x)-f(x)|\leqslant\sup_{x\in I}|f_n(x)-f(x)|.$$ Letting $n$ goes to $+\infty$, one gets: $$\lim_{n\to+\infty}|f_n(x)-f(x)|\leqslant 0.$$ Therefore, one has $\lim\limits_{n\to+\infty}|f_n(x)-f(x)|=0$ and $\lim\limits_{n\to+\infty}f_n(x)=f(x)$. Hence the result since $x$ is any element of $I$.
The point is that for $n\in\mathbb{N}$ and for every $x\in I$, $\sup\limits_{x\in I}|f_n(x)-f(x)|$ is greater than $|f_n(x)-f(x)|$. The behaviour near infinity of $|f_n(x)-f(x)|$ is controlled by the behaviour of $\sup\limits_{x\in I}|f_n(x)-f(x)|$.