$f_n$ defined on $[0;1]$ by $f_n(x)=x^n$, $f_n(x)$ weakly-converges if $\forall x \in I$ $(f_n(x_0))_{n\in \mathbb{N}}$ have a finite limit.
And it goes without saying (tell me if that's too weak to prove) that $\forall x\in [0,1],f_n(x)\rightarrow 0$
$f_n(x)$ converges uniformly to $f(x)$ if $\forall\epsilon\ge 0,\exists N\in \mathbb{N}|f_n(x)-f(x)|\le\epsilon$
An it exists such a N because $\lim\limits_{n\rightarrow +\infty}|x^n-f(x)|=0$
but I'm not quite sure about the uniform-convergence as far as I don't know what's hidding behind $f(x)$ alone without $n$.