Prove that a given subspace of $C[-1,1]$ with $L^2$ norm is closed Let $H= C[-1,1]$ with $L^2$ norm and consider $G=\{f \in H \mid f(1) = 0\}$. Show that $G$ is a closed subspace of $H$.
I've been trying to prove this for a while but i can't establish that given $\|f_n-f\|$ < $\epsilon$ (where the norm is the $L^2$ one)  we have uniform convergence for the sequence $(f_{n})$. If I could prove this the result would follow easily given that $G \subset \text{closure}(G)$ and if $(f_{n})$ converged uniformly we would have  $\displaystyle f(1)=\lim_{x \to 1} \lim_{n \to \infty} f_n(x)=\lim_{n \to \infty} \lim_{x \to 1} f_n=0$ and thus that $f \in G.$

$G$ is the kernel of the discontinuous linear map $L(f):=f(1)$. Indeed, $\ker f$ is is dense in $C[-1,1]$.
To see that, fix $f\in H$. For a fixed integer $n$, consider $f_n(x):=f(x)$ if $x\leqslant 1-n^{-1}$ and the graph of $f$ is a straight line joining $(1-n^{-1},f(1-n^{-1}))$ and $(1,0)$. Using the fact that $f$ is bounded, we can show that $f_n\to f$ in $H$, and $f_n\in G$ for all $n$. 
