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I know there's a theorem that states that for $f\in L^{1}(\mathbb{R})$ we have $f(x)= \frac{1}{2 \pi} \int \hat{f} (k) e^{ikx} dk$ (or something similar, it depends on how you define the Fourier transform). I think the proof I have seen uses $C^{\infty}_0 $ functions (smooth functions with compact support), and even though it doesn't use them, I'm sure it uses a dense subspace of $L^1$. Now, if I consider the Fourier series, I have that for $f\in L^1([0,2 \pi])$ and periodic, then $f(x)= \frac{1}{2\pi} \sum_{k\in \mathbb{Z}} \hat{f}(k) e^{ikx}$. But this isn't true for every $f$, for instance the Fourier series of some continuous functions doesn't converge (maybe it even diverges?).

The question is: given the fact that for $f\in C^1$ the convergence is uniform (total in fact), why can't I conclude that for all functions in $L^1([0,2 \pi])$? Why can't I use the same density argument used for the Fourier inverse theorem?

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  • $\begingroup$ Practically everything you want goes for $C^\infty_C$ functions. They are dense in $L^1(\mathbb{R})$ but you cannot conclude that $L^1$ functions get some of them good properties by some limiting machinery. $\endgroup$ – Ranc Jun 22 '17 at 17:54
  • $\begingroup$ @Ranc but why in the first theorem I can use the argument of density (which in fact I've seen used many times when dealing with functions in $L^{p} (\mathbb{R})$ ) while in the second I cannot? Is it because of the periodicity or what? $\endgroup$ – tommy1996q Jun 22 '17 at 19:23
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First of, it is not true that the fourier inversion formula holds for any $f\in L^1(\mathbb{R}$. To see this, you have $f\in L^1$ such that $\widehat{f} \notin L^1$: Just take indicator function of finite symmetric interval, the fourier transform is (up to a constant) just an inflation of $\frac{\sin x}{x}$, which is not summable, and so the inversion formula is not meaningful in any ordinary (function-wise) way.

Second, Yes, some fourier series diverge (and infact, once you find one that diverges then you can make it diverge quite alot). But to see why the formula doesn't work, or rather, why knowing it works for very good functions does not imply it works for other functions, we must see where it should fail more carefully.

Denote a sequence of functions $f_n \in L^1([-\pi ,\pi])$ for $n=1,2,\ldots$ and also suppose they are very good, meaning they are $C^\infty$. Then they equal their fourier series, we have

$$f_n (x)= \sum_{k\in \mathbb{Z}} c^{(n)}_k \mathrm{e} ^{ikx}$$ where $c^{(n)}_k = \widehat{f_n} (k)$. Further suppose we have $f_n \rightarrow f$ in some topology. We will try to figure out what is this topology later. Then when can we say that $f$ equals its fourier series?

First suppose we have just $f_n \rightarrow f$ in $L^1$. This is the first reasonable thing to say, since all we know is $f\in L^1$. Then we can get that $\widehat{f_n}(k) \rightarrow \widehat{f}(n)$ uniformally, that is, independent of $n$, indeed:

$$\left|\int_\mathbb{T} f_n(x)\mathrm{e}^{-inx} \mathrm{d}x - \int_\mathbb{T} f(x)\mathrm{e}^{-inx} \mathrm{d}x \right| \leq \int_\mathbb{T} |f_n(x)-f(x)| \mathrm{d}x \rightarrow 0 $$ So the series converge, but $f_n(x)$ does not necessarily converge to $f(x)$! So equality could not hold unless further details are given. Then we must improve our assumptions on the convergence of $f_n$ to $f$, but this will obviously improve our conditions of $f$, that is to say that not every function $f\in L^1([-\pi,\pi])$ can be the limit of such good functions.

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  • $\begingroup$ I misread the theorem, it states that the inversion holds "if $f$ and $\hat{f}$ are in $L^1(\mathbb{R})$". Anyway, I don't get why you can use a dense subspace to prove the theorem in the first case while you cannot in the second. Also I don't know if I get this thing: you said that $f_n \to f$ but $f_n (x) \to f(x)$ is not necessarily true. I understand this when the non-convergence is about a subset of measure zero, but I know there are fourier serieses that diverge almost everywhere, so I don't see how can $f_n \to f$ be true while it's not true that $f_n(x) \to f(x)$ foe almost all x. $\endgroup$ – tommy1996q Jun 23 '17 at 9:38
  • $\begingroup$ You need to distinguish the different modes of convergence. $\endgroup$ – Ranc Jun 23 '17 at 12:05
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    $\begingroup$ To ammend my laconic answer: (1) The inversion theorem holds almost everywhere if both $f,\widehat{f} \in L^1$. (2) To further see what might go wrong: The fourier series of a $C^\infty$ function is very good, the coefficients decay very rapidly. Then if a sequence of such functions converge to a $L^1$ function, the coefficients decay rate must break somwehere, although every sequence of coefficients (corresponding to $f_n$: $(c^(n)_k)_{k\geq 1}$ decays faster than any polynomial decay, the limiting sequence $(\widehat{f}(k))_{k\geq 1}$ would generally not. $\endgroup$ – Ranc Jun 23 '17 at 12:21

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