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?

  • 10
    $\begingroup$ Please, oh please, write the inner product using \langle f,g\rangle, resulting in $\langle f,g\rangle$. < and > are for inequalities. $\endgroup$ Feb 22, 2012 at 21:29
  • 2
    $\begingroup$ Try to find a sequence converging to a step function with just two values. $\endgroup$ Feb 22, 2012 at 21:31
  • 1
    $\begingroup$ @HaraldHanche-Olsen That's one of my pet peeves. I've even had professors that wrote $(\cdot,\cdot)$ for the inner product. How hard is it to use the correct, unambiguous notation? $\endgroup$
    – Math1000
    May 6, 2016 at 10:18

2 Answers 2


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] 

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

Can you show it does not converge?

  • $\begingroup$ 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? $\endgroup$
    – passenger
    Feb 22, 2012 at 21:53
  • $\begingroup$ For $ t > 1/n$ when $n \to \infty$ it is $f_n \to 0$ and for $t=0$ it is $f_n(0) \to 1$ $\endgroup$
    – passenger
    Feb 22, 2012 at 21:59
  • $\begingroup$ (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? $\endgroup$ Feb 22, 2012 at 22:01
  • $\begingroup$ The sequence you constructed is a sequence of continuous functions right? $\endgroup$
    – passenger
    Feb 22, 2012 at 22:04
  • $\begingroup$ Well... what do you think? $\endgroup$ Feb 22, 2012 at 22:05

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.

  • 1
    $\begingroup$ I think your $h_n$ should be defined differently, i.e. "$1$ on $[0,1/2 - 1/(2n)]$" or something similar. Could you elaborate how this implies that $g_0 = 1$ on [0,1/2)? $\endgroup$
    – el_tenedor
    May 1, 2016 at 8:53
  • $\begingroup$ @el_tenedor You are right. I have edited. Thanks. $\endgroup$ May 6, 2016 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.