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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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    $\begingroup$ Please, oh please, write the inner product using \langle f,g\rangle, resulting in $\langle f,g\rangle$. < and > are for inequalities. $\endgroup$ – Harald Hanche-Olsen Feb 22 '12 at 21:29
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    $\begingroup$ Try to find a sequence converging to a step function with just two values. $\endgroup$ – Harald Hanche-Olsen Feb 22 '12 at 21:31
  • $\begingroup$ @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 '16 at 10:18
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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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  • $\begingroup$ 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 '12 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 '12 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? $\endgroup$ – Mariano Suárez-Álvarez Feb 22 '12 at 22:01
  • $\begingroup$ The sequence you constructed is a sequence of continuous functions right? $\endgroup$ – passenger Feb 22 '12 at 22:04
  • $\begingroup$ Well... what do you think? $\endgroup$ – Mariano Suárez-Álvarez Feb 22 '12 at 22:05
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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.

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  • $\begingroup$ 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 '16 at 8:53
  • $\begingroup$ @el_tenedor You are right. I have edited. Thanks. $\endgroup$ – Davide Giraudo May 6 '16 at 9:33

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