TITLE: $C[0,1]$ is not Hilbert space
QUESTION [19 upvotes]: Prove that the space $C[0,1]$ of continuous functions from $[0,1]$ to $\mathbb{R}$ with the inner product $ \langle f,g \rangle  =\int_{0}^{1} f(t)g(t)dt \quad $ is not Hilbert space.
I know that I have to find a Cauchy sequence  $(f_n)_n$ which converges to a function $f$ which is not continuous, but I can't construct such a sequence $(f_n)_n$.
Any help?

REPLY [5 votes]: An alternative way (sledgehammer): if $C[0,1]$ where an Hilbert space then the linear continuous map $L\colon f\mapsto \int_0^{\frac 12}f(t)dt-\int_{\frac 12}^1f(t)dt$ would be represented by $g_0$. 
Let $x\in (0,1/2)$ be fixed. Then consider a function $f_n$ such that $f_n(t)=1$ if $t\lt x$, $0\leqslant f_n\leqslant 1$ and $f_n(t)=0$ if $t\gt x+1/n$. We have $$\lim_{n\to +\infty}L(f_n)=x=\lim_{n\to +\infty}
\int_0^{x+1/n}g_0(t)f(t)\mathrm dt,$$
and the last limit is $\int_0^xg_0(t)\mathrm dt$. This proves that $g_0(t)=1$ for each $t\in (0,1/2)$. Similarly, we can prove that $g_0(t)=-1$ for each $t\in(1/2,1)$, hence $g_0$ cannot be continuous.