$C[0,1]$ is not Hilbert space 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?
 A: 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.
A: Let $f_n:[-1,1]\to\mathbb R$ be such that $$f_n(t)=\begin{cases}1, & \text{if $t\in[-1,0];$} \\1-nt, & \text{if $t\in[0,\tfrac1n]$;} \\ 0, & \text{otherwise.}\end{cases}$$
According to Mathematica, we have $\lVert f_n-f_m\rVert=\frac{(m-n)^2}{3 m^2 n}$ if $1<n<m$ so this is indeed a Cauchy sequence. 
In[1]:= f[n_] := Piecewise[{{1, t < 0}, {1 - n t, 0 <= t <= 1/n}}];

In[2]:= Integrate[(f[n]-f[m])^2, {t, -1, 1},  Assumptions-> 1<n<m] 

               2
        (m - n)
Out[2]= --------
            2
         3 m  n

Can you show it does not converge?
A: You are right to claim that in order to prove that the subspace $C[0,1]$ of $L^2[0,1]$ is not complete, it is sufficient to "find [in $C[0,1]$] a Cauchy sequence $(f_n)_n$ [i.e. Cauchy for the $L^2$-norm] which converges [in $L^2[0,1]$] to a function $f$ which is not continuous". It will even be useless to check that $(f_n)_n$ is $L^2$-Cauchy: this will result from the $L^2$-convergence.
The sequence of functions $f_n\in C[0,1]$ defined by
$$f(x)=\begin{cases}n\left(x-\frac12\right)&\text{if }\left|x-\frac12\right|\le\frac1n\\+1&\text{if }x-\frac12\ge\frac1n\\-1&\text{if }x-\frac12\le-\frac1n\end{cases}$$
satisfies $|f_n|\le1$ and converges pointwise to the function $f$ defined by
$$f(x)=\begin{cases}0&\text{if }x=\frac12\\+1&\text{if }x>\frac12\\-1&\text{if }x<\frac12.\end{cases}$$
By the dominated convergence theorem, we deduce $\|f_n-f\|_2\to0.$
Because of its jump, the function $f$ is discontinuous, and more precisely: not equal almost everywhere to any continuous function.
