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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.2012-11-21
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    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 $x=0$.2012-11-21
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    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 $[-1,0]$ ;).2012-11-21
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    So you mean, "why is this space not complete *in the norm induced by this inner product*?"2012-12-16

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