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I am searching for integrable functions $(f_k)_{k=0}^\infty$ on the circle with $$ \lim_{k \rightarrow \infty} \frac 1 {2\pi} \int_0^{2 \pi} |f_k(x)|^2 dx = 0 $$ and s.t. $\lim_{k \rightarrow \infty} f_k(x)$ does not exist for all $x$.

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Do you mean "does not exist for all $x$" as "there is some $x$ for which it does not exist" (easy), or "does not exist _for all $x$_" as in "there is no $x$ for which it exists"? –  Clement C. Jun 25 '13 at 13:07
    
Ok I will reformulate. I mean pointwise convergence in $(\mathbb R, d_{l^1})$ and for all $x$ the limit does not exist. –  André Jun 25 '13 at 13:08
    
Take the usual counterexample on the unit interval: $\ \ f_1=\chi_{[0,1]}$, $f_2=\chi_{[0,{1\over2}]}$, $f_3=\chi_{[{1\over2},1]}$, $f_4=\chi_{[0,{1\over4}]}$, $f_5=\chi_{[{1\over4},{2\over4}]}$, $f_6=\chi_{[{2\over4},{3\over4}]}$, $f_7=\chi_{[{3\over4},1]}$, $\ldots$ and "wrap them around the circle". –  David Mitra Jun 25 '13 at 13:09
    
Ok yes. You can post this as an answer if u want to. Thank you. –  André Jun 25 '13 at 13:10

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Take the usual counterexample on the unit interval:

$\ \ f_1=\chi_{[0,1]}$, $f_2=\chi_{[0,{1\over2}]}$, $f_3=\chi_{[{1\over2},1]}$, $f_4=\chi_{[0,{1\over4}]}$, $f_5=\chi_{[{1\over4},{2\over4}]}$, $f_6=\chi_{[{2\over4},{3\over4}]}$, $f_7=\chi_{[{3\over4},1]}$, $\ldots$

and "wrap them around the circle".

So, $f_1$ is the characteristic function over the entire unit circle, $f_2$ is the characteristic function of the upper half of the unit circle, $f_3$ is the characteristic function of the lower half of the unit circle, $...\,\,$.

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