# $C[0,1]$ 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?

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Please, oh please, write the inner product using \langle f,g\rangle, resulting in $\langle f,g\rangle$. < and > are for inequalities. – Harald Hanche-Olsen Feb 22 '12 at 21:29
Try to find a sequence converging to a step function with just two values. – Harald Hanche-Olsen Feb 22 '12 at 21:31

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?

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 So I have to prove that $f_n(t)=\begin{cases}1-nt, & \text{if$t\in[0,\tfrac1n]$;} \\ 0, & \text{otherwise.}\end{cases}$ does not converge. How can I do this? – passenger Feb 22 '12 at 21:53 What have you tried? – Mariano Suárez-Alvarez♦ Feb 22 '12 at 21:54 For $t > 1/n$ when $n \to \infty$ it is $f_n \to 0$ and for $t=0$ it is $f_n(0) \to 1$ – passenger Feb 22 '12 at 21:59 (Notice I have changed the functions in my answer...) And why does that imply that the sequence I constructed does not converge with respect to the norm induced by your inner product? – Mariano Suárez-Alvarez♦ Feb 22 '12 at 22:01 The sequence you constructed is a sequence of continuous functions right? – passenger Feb 22 '12 at 22:04
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 $g_0$. Taking a continuous function $h_n$ which is $1$ on $[0,1/2-1/2]$ and $0$ for $x\geq 1/2$ we get that $g_0=1$ on $[0,1/2)$ and $g_0=-1$ on $(1/2,1]$, therefore $g$ cannot be continuous.