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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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    $\begingroup$ Instead 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." $\endgroup$ – Jonas Meyer May 11 '12 at 19:19
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Let $$f_n(x) = \left\{ \begin{array}{rl} 0 & \text{if } x \leq 1/2,\\ 1 & \text{if } x \geq 1/2+1/n,\\ n(x-1/2) & \text{if } 1/2\leq x\leq 1/2+1/n. \end{array} \right.$$

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  • $\begingroup$ Does this sequence work for showing C[0,1] is not Banach for any $p<\infty$? I can see it works for $p=1$ & $p=2$. $\endgroup$ – rk101 May 11 '12 at 13:23
  • $\begingroup$ Yes, I think so. $\endgroup$ – Michael Greinecker May 11 '12 at 13:25
  • $\begingroup$ Yes it does actually, I have just checked it. $\endgroup$ – rk101 May 11 '12 at 13:38
  • $\begingroup$ A quick question: Why $\|\cdot\|_1 \leq \|\cdot\|_2 \leq \|\cdot\|_{\infty}$ ? $\endgroup$ – rk101 May 11 '12 at 13:56
  • $\begingroup$ I'm not completely ure. I think for simple functions you can use the corresponding reult for finite dimensioal spaces and then take limits. $\endgroup$ – Michael Greinecker May 11 '12 at 14:23
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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 $C[0,1]$.

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    $\begingroup$ Note 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. $\endgroup$ – 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 $x=1$.

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  • $\begingroup$ 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... $\endgroup$ – robjohn May 11 '12 at 15:33

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