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Let $C([0, 1])$ be the space of all continuous real-valued functions on the interval $[0, 1]$, with norm

$\displaystyle\lVert f\rVert_2 = \sqrt{\int_0^1 \lvert f(x)\rvert^2 \,dx}$.

I need to show that $C([0, 1])$ is not a complete space (i.e. a Banach space) with respect to this norm.

Yiorgos S. Smyrlis
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Raio
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    Let $f_n$ be the function whose graph consists of the straight line segments joining $(0,1)$, $(1/n,0)$, and $(1,0)$. Show that $(f_n)$ is Cauchy under your $2$-norm. Note that its only possible limit is not in $C[0,1]$. – David Mitra Dec 24 '13 at 09:13
  • @DavidMitra: "The only possible limit is not in $C[0,1]$"? - How do you define a limit if it is not in the space of reference? – Yiorgos S. Smyrlis Dec 24 '13 at 09:39
  • @YiorgosS.Smyrlis Perhaps I should have said: "If it converged in $C[0,1]$, it would have to converge to its pointwise limit. But the pointwise limit is not in $C[0,1]$, so..." – David Mitra Dec 24 '13 at 09:44

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This is an interesting question, which is not always answered rigorously. You should construct a Cauchy sequence which is not convergent. One such is $$ f_n(x)=\left\{ \begin{array}{lll} 1 & \text{if} & x\in \left[0,\tfrac{1}{2}-\tfrac{1}{n}\right), \\ \tfrac{1}{2}+\tfrac{n}{2}(\tfrac{1}{2}-x) & \text{if} & x\in \left[\tfrac{1}{2}-\tfrac{1}{n},\tfrac{1}{2}+\tfrac{1}{n}\right],\\ 0 & \text{if} & x\in \left(\tfrac{1}{2}+\tfrac{1}{n},1\right]. \end{array} \right. $$ Clearly, all the $f_n$'s are continuous, and is it not hard to show that $\{f_n\}_{n\in\mathbb N}$ is a Cauchy sequence.

The delicate part (which I am not answering in detail) is how to show that it does not converge, with respect the this norm, in $C[0,1]$.

Hint. If it did converge, say to $f$, then show that $f(x)=1$, for $x<½$ and $f(x)=0$, for $x>½$, and hence discontinuous at $x=½$.

Yiorgos S. Smyrlis
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