<s>[INST] 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.
 [/INST] 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.
</s>