$C[0,1]$ 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?

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.
