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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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    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."2012-05-11

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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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    @rk101: $\|f\|_1=\int_0^1|f(t)|\cdot 1dt \leq \sqrt{\int_0^1 |f(t)|^2dt}\cdot\sqrt{\int_0^1 1dt}=\|f\|_2=\sqrt{\int_0^1|f(t)|^2dt}\leq\sqrt{\int_0^1\|f\|_\infty^2 dt}=\|f\|_\infty$.2012-11-06
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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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    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 0.2012-05-11
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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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    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...*2012-05-11