Does weak convergance in $H_{0}^{1}(0,1)$ implies strong convergance in $C(0,1)$?

Question

I was wondering whether one can prove the following statement

$(x_{n})\subset H_{0}^{1}(0,1)$ and $x_{n}\rightarrow x$ weakly $\Rightarrow x_{n}\rightarrow x$ in $C(0,1)$?

As far as I know, if $x_{n}\rightarrow x$ in $H_{0}^{1}(0,1)$ then follows the convergance in $C(0,1)$ because $H_{0}^{1}(0,1)\subset C(0,1)$. But now, only weak convergance is given.