Why isn't it a Hilbert space

Let $X$ be the vector space of all the continuous complex-valued functions on $[0,1]$. Then $X$ has an inner product $$(f,g) = \int_0^1 f(t)\overline{g(t)} dt$$ to make it an inner product space. But this is not a Hilbert space.

Why isn't is complete? Which Cauchy sequence in it is not convergent?

Thanks.

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In this norm, you can have a sequence of continuous functions converging to a $L^2$ function. –  Tomás Nov 21 '12 at 11:35
The functions $\exp(-ax)$ as $a\to\infty$ converge pointwise on $[0,1]$ to the function $$\begin{cases}1&x=0\\0&x\ne0,\end{cases}$$ which is not continuous at $x=0$. –  anon Nov 21 '12 at 11:38
anon, we do not care of pointwise convergence here. In the given norm, your sequence converges to $0$, which is continuous. However, you can easily fix your example by adding a chunk of meat on $[-1,0]$ ;). –  fedja Nov 21 '12 at 11:44
So you mean, "why is this space not complete in the norm induced by this inner product?" –  JohnD Dec 16 '12 at 21:38

Consider the sequence of continuous functions $f_n$ defined by $$f_n(x) = \left\{ \begin{array}{rl} 0 &\mbox{ if x\in \Big[0,\frac{1}{2}-\frac{1}{n}\Big]} \\ \frac{nx}{2}-\frac{n}{4}+\frac{1}{2} &\mbox{ if x\in\Big[\frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac{1}{n}\Big]}\\ 1 &\mbox{ if x\in\Big[\frac{1}{2}+\frac{1}{n},1\Big]} \end{array} \right.$$

Let $f(x)=0$ if $x\in[0,\frac{1}{2}]$ and $f(x)=1$ if $x\in (\frac{1}{2},1]$. Im gonna prove that $f_n\rightarrow f$ in your norm. In fact, we have \begin{eqnarray} \int_0^1|f_n-f|^2 &=& \int_{\frac{1}{2}-\frac{1}{n}}^{\frac{1}{2}+\frac{1}{n}}( \frac{nx}{2}-\frac{n}{4}+\frac{1}{2}-f)^2 \nonumber \\ &=& \int_{\frac{1}{2}-\frac{1}{n}}^{\frac{1}{2}} (\frac{nx}{2}-\frac{n}{4}+\frac{1}{2})^2+\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}(\frac{nx}{2}-\frac{n}{4}-\frac{1}{2})^2 \nonumber \\ &=& \frac{1}{12n}+\frac{7}{12n}\\ &\rightarrow& 0 \end{eqnarray}

So the sequence $f_n$ converges to a function $f$ not continuous.

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As remarked by fedja in the comments under the question, this sequence converges to the zero function in the given norm. In fact that is exactly what you proved, the function $f$ does not appear in your proof! And your $f$ is not an element of the vector space $X$ considered (it is discontinuous) so saying that $f_n\to f$ in the norm on $X$ just makes no sense. –  Marc van Leeuwen Nov 21 '12 at 13:56
You are right. Im gonnna change it. –  Tomás Nov 21 '12 at 14:06
Is it right now @MarcvanLeeuwen ? –  Tomás Nov 21 '12 at 14:23
I didn't check the integrals, but the example looks OK now (and a lot like the examples in the other answers). –  Marc van Leeuwen Nov 21 '12 at 15:00

I replace $[0,1]$ by $I:=[-1,1]$. We have to produce a sequence $f_n:\ I\to{\mathbb R}$ of continuous functions which is a Cauchy sequence in $L^2(I)$ but does not converge in $L^2(I)$ to an $f\in X$.

Put $$f_n(t):=\cases{n t\quad &(|t|\leq{1\over n}) \cr {\rm sgn}\ t & (|t|\geq{1\over n}) \cr}\ .$$ Let an $\epsilon >0$ be given. When $m$, $n>\frac1\epsilon$ then $|f_m(t)-f_n(t)|\leq 1$ when $|t|\leq\epsilon$, and $=0$ when $|t|>\epsilon$. Therefore $\|f_m-f_n\|^2\leq 2\epsilon$, which shows that $(f_n)_{n\geq1}$ is a Cauchy sequence in $L^2(I)$. Therefore the $f_n$ converge in $L^2(I)$ to a certain $f\in L^2(I)$.

Assume that this $f$ has a continuous representant, again denoted by $f$, and that $f(0)=:c\leq0$. Then there is an $\epsilon_0>0$ with $$f(t)\leq{1\over 2}\qquad(0\leq t\leq\epsilon_0)\ .$$ When $n>{2\over\epsilon_0}$ then $$f_n(t)=1\qquad({\epsilon_0\over2}\leq t\leq\epsilon_0)\ .$$ It follows that $|f_n(t)-f(t)|\geq{1\over2}$ on an interval of length ${\epsilon_0\over2}$, and this implies $\|f_n-f\|^2\geq{\epsilon_0\over 8}$ for all these $n$. This contradicts the already established fact $\lim_{n\to\infty} f_n=f$ in $L^2(I)$.

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Let $$f_n(t) := \left(1-n \cdot dist\left(\{t\}, \left[\frac{1}{4},\frac{3}{4} \right] \right) \right) \vee 0$$

Then $(f_n)_n$ is a Cauchy-sequence,

$$f_n \to 1_{\left[\frac{1}{4},\frac{3}{4} \right]} \notin C[0,1]$$

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