I'm trying to find a cauchy sequence in $C[0,1]$ that converges under $\|\cdot\|_2$ to a limit which isn't continuous.
Any ideas?
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1Instead of "converges under $|\cdot|_2$ to a limit which isn't continuous," you probably want "does not converge under $|\cdot|_2$ to a limit which is continuous." – Jonas Meyer May 11 '12 at 19:19
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Let $$f_n(x) = \left\{ \begin{array}{rl} 0 & \text{if } x \leq 1/2,\\ 1 & \text{if } x \geq 1/2+1/n,\\ n(x-1/2) & \text{if } 1/2\leq x\leq 1/2+1/n. \end{array} \right.$$
Michael Greinecker
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Does this sequence work for showing C[0,1] is not Banach for any $p<\infty$? I can see it works for $p=1$ & $p=2$. – rk101 May 11 '12 at 13:23
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I'm not completely ure. I think for simple functions you can use the corresponding reult for finite dimensioal spaces and then take limits. – Michael Greinecker May 11 '12 at 14:23
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@rk101: $|f|1=\int_0^1|f(t)|\cdot 1dt \leq \sqrt{\int_0^1 |f(t)|^2dt}\cdot\sqrt{\int_0^1 1dt}=|f|_2=\sqrt{\int_0^1|f(t)|^2dt}\leq\sqrt{\int_0^1|f|\infty^2 dt}=|f|_\infty$. – Jonas Meyer Nov 06 '12 at 06:24
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Another one. Think of a discontinuous (bounded measurable) function. Say: $f(x) = 0$ on $[0,1/2]$ and $f(x) = 1$ on $(1/2,1]$. Write down its Fourier series. The partial sums are continuous. They converge in $L_2$ norm (to $f$) but do not converge to any element of $C[0,1]$.
GEdgar
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2Note that $f$ should be not only discontinuous, but not a.e. equal to any continuous function. E.g. $f(1/2) = 1$, $f(x) = 0$ otherwise is discontinuous, but its Fourier series converges uniformly to the continuous function 0. – Nate Eldredge May 11 '12 at 13:10
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Think of the function that is $0$ on the interval $\left[0,1-\frac{1}{n}\right]$ and then is $y=n(x-1)+1$ on the remainder of the interval. As $n$ increases the "spike" gets sharper. The limit function is not continuous at $x=1$.
J126
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However, the limit function in the $L^2$ norm is $0$, which is in $C[0,1]$. These aren't the pointwise limits you're looking for... – robjohn May 11 '12 at 15:33