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Let $f\in L^2(0,1)$. I was wondering if the Fourier series of $f$ is a linear map $L^2(0,1)\to L^2(0,1)$. The linearity is obvious, but if $f\in L^2(0,1)$ does $S(f)\in L^2$ or not ? I tried as follow,

$$\int_0^1 S(f)^2(x)\mathrm d x=\int_0^1\left(\sum_{k=-\infty }^\infty c_ne^{inx}\right)^2\mathrm d x.$$

But first, does $\sum_{k=-\infty }^\infty c_ne^{inx}$ always converge ? And now, how can I compute the previous integral ? It looks complicated.

user349449
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  • You might want to look at Parseval's Theorem. https://en.wikipedia.org/wiki/Parseval%27s_theorem – John Hughes Jul 02 '16 at 12:25
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    there is a huge difference between the pointwise convergence $|f_n(x) - f(x)| \to 0$ and the convergence in $L^2$ norm $\sqrt{\int_0^{2\pi} |f_n(x)-f(x)|^2 dx} = |f_n - f |{L^2([0,2\pi])} \to 0$. With $f_n = S_n(f)$ the Parseval theorem is that $|f_n - f |{L^2([0,2\pi])} \to 0$ whenever $f \in L^2([0,2\pi])$ – reuns Jul 02 '16 at 13:42

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You look to make same confusion as your previous post about Fourier transformation. Let $S_N(f)$ the partial Fourier series. The convergence of $S_N(f)$ (if it converge to something, it converge to $f$) needs more condition that only $L^2$. In my memories, there is a something with Holder continuity. But, by a theorem, $S_n(f)$ will converge to $f$ in the $L^2-$sense. This mean that $$\lim_{N\to \infty }\int_0^1|S_N(f)-f|^2=0,$$ which is not the same. In particular, using Minkovski inequality, you will indeed have that the limit of $S_N(f)\in L^2$, since Minkovski tels you that $$\int_0^1|S_N(f)|^2\leq \int_0^1|S_N(f)-f|^2+\int_0^1|f|^2,$$ (notice that it's not an obvious result). So, all we can say it that the limit of $S_N(f)$ (in the $L^2$ sense) will be $L^2$ (but this is nothing surprising since it's $f$ which is in $L^2$). But you can't say more a priori.

Here there is an interesting trap

If you can find the subtlety, I think you will understand more those $L^2$ space. We know that $L^2$ are Hilbert space. An orthonormal basis of $L^2(0,1)$ is given by $\{e^{inx}\}_{n\in\mathbb Z}$. Then, $$f(x)=\sum_{k=-\infty }^\infty \left<f,e^{inx}\right>e^{inx}=\sum_{k=-\infty }^\infty \int_0^1 f(x)e^{-iny}\mathrm d ye^{inx}=\sum_{n=-\infty }^\infty c_ne^{-inx}$$ $$=\lim_{N\to \infty }\sum_{n=-N}^N c_ne^{inx}=\lim_{N\to \infty }S_N(x),$$ where $c_n=\int_0^1 f(x)e^{-inx}\mathrm d x$. As we can see, if $S_N$ converge in the $L^2-$sense to $f$, then it converge in the usual sense to $f$.

Question : What's wrong here ?

Surb
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  • To me you have that $\lim_{N\to \infty }S_N(f)=f(x)$, there is no mistakes. What is the problem ? I don't see what you mean. – user349449 Jul 02 '16 at 13:30
  • Look at the definition of $f(x)=\sum_{k=-\infty }^\infty \left<f,e^{inx}\right>e^{inx}$. – Surb Jul 02 '16 at 13:32
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    @MathBeginner Hint: consider the sequence $g_m$ given by enumerating the elements of $f_{n,k}=\chi_{[k/n,(k+1)/n]}(x)$ where $n=1,2,\dots$ and $k=0,1,\dots,n-1$. Does it converge in $L^2$? Does it converge pointwise? Does it converge pointwise a.e.? (This example has the same issues as the Fourier series example but the analysis is far simpler.) – Ian Jul 02 '16 at 13:48
  • @Ian: I don't understand your hint (I'm sorry). Thanks for trying to help me, but this example (given by Surb) muddles me, I'm confuse. May be I should give up for the moment... (but thank you to all of you) – user349449 Jul 02 '16 at 14:04
  • @MathBeginner In my example when $n$ is large, $f_{n,k}$ is $1$ only on a small set; everywhere else it is zero. This means $g_m$ converges in measure and in $L^p$ for $1 \leq p<\infty$. Yet because of the variation in $k$, $g_m$ can't converge pointwise, because for any $x$ and any $n$ there is a $k$ such that $f_{n,k}(x)=1$. – Ian Jul 02 '16 at 14:14
  • @Ian: Ok thank you. But could you please tell me why are we intrested in $L^2$ convergence of Fourier series ? It looks really unnecessary to me (I don't see in what this convergence is interesting). Thank you :) – user349449 Jul 02 '16 at 14:51
  • @MathBeginner That depends too much on the application that you have in mind to give a universal answer. A decent a posteriori answer is "it's the kind of convergence that we always have whenever Fourier series make sense". – Ian Jul 02 '16 at 15:01
  • @MathBeginner: For your 2nd comment, in $(L^2,|\cdot |{L^2})$, $f(x)=\sum{k=-\infty }^\infty \left<f,e^{nix}\right>e^{nix}$ mean that $$\forall \varepsilon>0, \exists N\in \mathbb N: \forall n\in \mathbb N, n\geq N\implies \underbrace{\left|\sum_{k=-n}^n\left<f,e^{nix}\right>e^{nix}-f(x)\right|{L^2}}{=:\int_0^1|...|}<\varepsilon.$$ In $(L^2,|\cdot |_{L^2})$ the norm of $\mathbb R$ doesn't exist ! – Surb Jul 02 '16 at 15:09
  • @Ian: Thank you for your active and constructive participation on my both answer on Fourier. – Surb Jul 02 '16 at 15:29