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Let $f_n \rightarrow f$ be a sequence of $2\pi$-periodic functions, where the convergence is in $L^1({\mathbb R}/2\pi{\mathbb Z})$. Then the Fourier-coefficients satisfy $|F(f_n) -F(f)| \rightarrow 0 $ uniformly.

Now, I was wondering. Does this imply that the Fourier series $$h_k(x):=\sum_{n \in \mathbb{Z}} F(f_k)(n)e^{-inx} $$ converges to $$h(x):=\sum_{n \in \mathbb{Z}} F(f)(n)e^{-inx} $$ pointwise, where we assume that everything exists( so we assume that the sums $h_n,h$ converge for all $x$.)

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migrated from mathoverflow.net Jun 1 '15 at 5:36

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    $\begingroup$ My feeling is, this should be false, but a counterexample might be tricky to come by. The condition of convergence of the Fourier series is complicated and to my feeling unnatural. $\endgroup$ – Zero May 31 '15 at 13:16
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    $\begingroup$ Just to clarify, the hypotheses seems to be: (i) $f_n\to f$ in $L^1({\bf T})$; (ii) $f_n$ has pointwise convergent Fourier series, and $f$ has pointwise convergent Fourier series. The desired conclusion is that $f_n \to f$ pointwise. Is that correct? $\endgroup$ – Yemon Choi May 31 '15 at 13:58
  • $\begingroup$ For $0<= k < m$ consider the intervals $I_{k,m}=[k/m,(k+1)/m]$ (on [0,1] and make it periodic). Form a sequence with the characteristic functions. $\endgroup$ – juan May 31 '15 at 18:30
  • $\begingroup$ If this were true and $f_n\to f$ in $L^2$, then it would follow that $f_n\to f$ a.e. (since $L^2$ functions are represented pointwise by their Fourier series). So the claim is wrong. $\endgroup$ – user138530 May 31 '15 at 23:25
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Let $\{f_n\}$ be the typewriter sequence. $f_n$ converges to $f\equiv0$ in $L^1$ but not pointwise (not even pointwise almost everywhere). The Fourier coefficients of $f$ are all equal to $0$, so its Fourier series converges to $0$. The Fourier series of $f_n$ converges to $f_n$ at all points except two.

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