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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 $x=1$?

Bana
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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 incompleteness – Thomas Andrews Oct 19 '11 at 20:07
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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. – Thomas Andrews Oct 19 '11 at 20:13
  • 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.... – Bana Oct 19 '11 at 20:14
  • Ow that's it!!! – Bana Oct 19 '11 at 20:14
  • Ok good thnx. I can also do it piecewise in three parts. – Bana Oct 19 '11 at 20:16

1 Answers1

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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 $\frac12<x\leqslant1$ but $g$ is not in $C[0,1]$ hence $(g_n)$ diverges in $(C[0,1],\|\cdot\|)$. But $\|g_n-g\|\leqslant\frac1{2\sqrt{n}}$ hence $(g_n)$ is a Cauchy sequence in $(C[0,1],\|\cdot\|)$.

Did
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  • I do not see how that last sentence demonstrates that $(g_n)$ is a Cauchy sequence. –  Sep 05 '16 at 19:25
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    @JacobGross Use $|g_n-g_k|\leqslant|g_n-g|+ |g_k-g|$ and rejoice. – Did Sep 05 '16 at 19:30