For existence of Fourier coefficients of a function f is sufficient that f is absolutely integrable in [−π,π] but, is this condition necessary? that is, is there a function that is not absolutely integrable in [−π,π] so that its Fourier series Exists? Consider the usual trigonometric system.
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2$\begingroup$ How do you define $\int_{-\pi}^\pi f(x)dx$, if $\int |f|dx = \infty$? More concrete: What is then its Fourier Series? $\endgroup$– Dieter KadelkaCommented Jun 22, 2020 at 22:08
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$\begingroup$ This is the problem, because if $\int_{-\pi}^{\pi} |f(x)|dx$ converges then Fourier Series of $f$ exists always , but is possible that is there a functión that $\int_{-\pi}^{\pi} |f(x)|dx$ diverges (no necessarily $\int_{-\pi}^{\pi} f(x)dx$ diverges) and its Fourier series exists? if it is not possible why? $\endgroup$– SASCommented Jun 22, 2020 at 22:17
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$\begingroup$ If $\int |f|d\mu = \infty$, then always $\int fd\mu = \pm \infty$ (if either $\int f^+d\mu < \infty$ or $\int f^-d\mu < \infty$) or this integral does not exist. $\endgroup$– Dieter KadelkaCommented Jun 22, 2020 at 22:26
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2$\begingroup$ I consider that this is false, because $\int_{a}^{\infty} \dfrac{sin(x)}{x}dx$ converges $\forall a>0$ but $\int_{a}^{\infty} \dfrac{|sin(x)|}{x} dx$ diverges $\endgroup$– SASCommented Jun 22, 2020 at 22:46
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$\begingroup$ @PieroD'Ancona yes, I know. In this theory all is very nice because $\mathcal{L}_2$ is a Hilbert space. However, Serie Fourier also can be defined for another type of function (Function $f$ such that $f \notin \mathcal{L_2}$), and in these cases is this problem. $\endgroup$– SASCommented Jun 23, 2020 at 7:55
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All $2π$ periodic distributions are temperate and since all temperate distributions have a Fourier transform, you have plenty of examples. Note that the previous statements prove the power of abstract nonsense in Mathematics: you define for $u\in \mathscr S(\mathbb R)$ the Fourier transform $$ \hat u(\xi)=\int_{\mathbb R} e^{-2iπ x \xi } u(x) dx, $$ and you can prove that it is an isomorphism of $\mathscr S(\mathbb R)$, with the inverse given by $ u(x)=\int_{\mathbb R} e^{2iπ x \xi } \hat u(\xi) d\xi. $ Well, not trivial but very standard with direct proofs. Then you dramatically increase the generality by defining the Fourier transform of a temperate distribution $T$, as $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R)} =\langle T,\hat \phi\rangle_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R)}, $$ and with $\mathcal F$ standing for the Fourier transform and $\mathcal C$ for the mapping $T(x)\mapsto T(-x)$ (clearly defined for a function and also by duality for distributions). Then (and this is the power referred to above), you get trivially that for any tempered distribution $T$ $$ \mathcal C\mathcal F \mathcal F T=T. $$ For instance you get with $H=\mathbf 1_{\mathbb R_+}$, $ \hat H=\frac12\delta_0+\frac{1}{2π i}\textrm{pv}\frac1x. $
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1$\begingroup$ Yes, of course you can extend the domain of definition and get the above mentioned results. But I think a beginner (?) is misguided by not answering the original question. $\endgroup$ Commented Jun 22, 2020 at 22:56
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1$\begingroup$ The OP is working on the torus $[-\pi,\pi]$, not on the whole real line. $\endgroup$– PhoemueXCommented Jun 23, 2020 at 4:56
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1$\begingroup$ @PhoemueX Periodic functions are defined everywhere and a classical way to obtain the classical expansion in Fourier series for a periodic distribution is to use the Fourier inversion formula above on the real line. $\endgroup$– BazinCommented Jun 23, 2020 at 9:32
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1$\begingroup$ @Dieter Kadelka I tried to answer the question which was "Can we define the Fourier transform on a larger set than $L^1$?". The distributional framework with temperate distribution is providing a very large space on which the Fourier transform makes sense. $\endgroup$– BazinCommented Jun 23, 2020 at 9:35
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How about $1/x$, for distributional Fourier transform