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Consider the vector space $C([0,1])$ of real-valued continuous functions on $[0,1]$ endowed with the standard norm: $$ \Vert f\Vert_2 = \sqrt{\int_0^1 f(x)^2 dx}.$$ I know that this normed space is not complete.

Is this because the function $f_(x) = x^n$ converges to the discontinuous function which is zero on $[0,1)$ and $1$ at $x=1$?

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    Via that norm, $f_n(x)=x^n$ converges to $f(x)=0$. So that's not the source of the incompleteness2011-10-19
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    A quick way to do it is to let $f_n(x)=(2x)^n i$f $0\leq x \leq \frac{1}{2}$, and 1 if $x>\frac{1}{2}$. Then the sequence doesn't converge in $C[0,1]$. Now show it is Cauchy in that norm.2011-10-19
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    Ow I see. You're right! Ok so I should have sticked with the example I didn't write down. I can draw a picture of it, but I can't do that on the computer. It's a sequence of functions which converges to the step function: 1 if $0 \leq x <1/2$ and $0$ if $1/2 < x \leq 1$. I just can't seem to write down the sequence that does this explicitly....2011-10-19
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    Ow that's it!!!2011-10-19
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    Ok good thnx. I can also do it piecewise in three parts.2011-10-19

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The sequence $(f_n)$ converges in $(C[0,1],\|\cdot\|)$ to the function $0$ so this does not prove that $(C[0,1],\|\cdot\|)$ is not complete.

An example is the sequence $(g_n)$ defined by $g_n(x)=0$ if $0\leqslant x\leqslant\frac12-\frac1{2n}$, $g(x)=nx-\frac12n+\frac12$ if $\frac12-\frac1{2n}\leqslant x\leqslant\frac12+\frac1{2n}$ and $g(x)=1$ if $\frac12+\frac1{2n}\leqslant x\leqslant1$. Then $(g_n)$ converges in $(L^2[0,1],\|\cdot\|)$ to the function $g$ defined by $g(x)=0$ if $0\leqslant x\leqslant\frac12$ and $g(x)=1$ if $\frac12

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    I do not see how that last sentence demonstrates that $(g_n)$ is a Cauchy sequence.2016-09-05
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    @JacobGross Use $\|g_n-g_k\|\leqslant\|g_n-g\|+ \|g_k-g\|$ and rejoice.2016-09-05