2

For existence of Fourier coefficients of a function $f$ is sufficient that $f$ is absolutely integrable in $[-\pi,\pi]$ but, is this condition necessary? that is, is there a function that is not absolutely integrable in $[-\pi,\pi]$ so that its Fourier series Exists?
Consider the usual trigonometric system.

SAS
  • 49
  • You can even consider distributions such as Dirac delta with the FT being constant, but these are not functions, its up to you what spaces you admit. – VojtaK Jun 22 '20 at 21:54
  • The Fourier series of $f(x)=\frac{\sin(e^{1/x})}{x}1_{[0,1]}$ should converge to $f$ everywhere due to the Dirichlet kernel and that $f$ oscillates very fast. – reuns Jun 22 '20 at 22:05
  • There are convergent trigonometric series that are not integrable (absolutely integrable in your jargon) Is this what you mean? – Mittens Jun 22 '20 at 23:14
  • Do you maybe mean functions $f$ for which $\int_{-\pi}^\pi f(x) e^{-ikx} dx$ exists for all $k \in \mathbb{Z}$, but for which $f$ is not absolutely integrable on $[-\pi , \pi]$? – Charles Hudgins Jun 22 '20 at 23:37
  • The integral $\int_{-\pi}^{\pi}f(x) e^{-ikx}\ dx$ exists if and only if $f$ is (absolutely) integrable, so the answer is no, if $f$ is not integrable then it doesn't have a Fourier series. –  Jun 22 '20 at 23:45
  • @Bungo $\int_{-\pi}^{\pi}f(x) e^{-ikx}\ dx$ exists of course for many more functions (and distributions) than integrable ones. – reuns Jun 23 '20 at 00:09
  • @reuns OK, that's fair, I retract my comment. Should have qualified with "assuming we are talking about ordinary Lebesgue integrals..." –  Jun 23 '20 at 00:30
  • Is correct @CharlesHudgins this is what we need: Do you maybe mean functions f for which $\int_{-\pi}^{\pi}f(x)e^{-ikx}dx$ exists for all $k\in \mathbb{Z}$, but for which $f$ is not absolutely integrable on [−π,π]?
    Some function?     – SAS Jun 23 '20 at 02:05

1 Answers1

1

Fourier coefficients (and thus Fourier series) are defined for integrable functions. There are however trigonometric series that are not integrable (elements of $L_1(\mathbb{S}^1)$. Is this what you mean?

Here is an example: let $\{c_n:n\in\mathbb{Z}\}\subset\mathbb{R}_+$ be such that both $c_n\searrow0$ and $c_{-n}\searrow0$ as $n\rightarrow\infty$. Dirichlet's convergence test shows that $f(x)=\sum_n\operatorname{sign}(n)c_ne^{inx}$ is a pointwise convergent series. In particular, for $c_n=\frac{1}{\log |n|}\mathbb{1}(|n|\geq2)$ we have that \begin{align*} f(x)=\sum_{|n|\geq2}\frac{e^{inx}}{\operatorname{sign}(n)\log |n|}=2i\sum^\infty_{n=2}\frac{\sin nx}{\log n} \end{align*} is pointwise convergent. However, as $\sum^\infty_{n=2}\frac{1}{n\log n}$ diverges, $f\notin\mathcal{L}_1(\mathbb{S}^1)$.

Here I have used a well know fact that states that if $f\in L_1(\mathbb{S}^1)$, then $\sum^\infty_{n=1}\frac{\hat{f}(n)-\hat{f}(-n)}{n}$ converges, where $\hat{f}(n)=\frac{1}{2\pi}\int^{\pi}_{-\pi}f(t)e^{-int}\,dt$.

Mittens
  • 39,145
  • Dear Oliver what is explicitly the function $f$ so that his Fourier series is $\sum_{n=2}^{\infty}\dfrac{\sin nx}{\log n}$? I consider that is sufficient find a function $f$ odd in $[-\pi,\pi]$ so that $\int_{-\pi}^{\pi}f(x)\sin nx dx= \frac{1}{\log n}$ – SAS Jun 23 '20 at 06:43
  • I understand you but, How is the existence of this function proven? that is, always exists $f$ such that $\int_{-\pi}^{\pi}f(x)\sin nx dx = \frac{1}{\ln n}$ – SAS Jun 23 '20 at 20:03
  • That $f$ is a well defined function follows by showing that the series $f(x)=\sum^\infty_{n=1}\frac{\sin nx}{\log n}$ converges for any $x$. This can be done by standard arguments such as the Dirichlet convergence test. Whether $\frac{1}{2\pi}\int^{\pi}{-\pi}f(t)e^{-nit},dt=\frac{1}{\log n}$ is a different matter. Tis has to be a prroached as a proper integral since $f$ is not integrable. – Mittens Jun 23 '20 at 20:18