Recently I'm studying something about Fourier series on the space $\mathbb{T}$ and I find this particular question:

Is possible to prove that the Fourier series of an integrable function, is summable by Abel at almost every point?

Now, I know that Abel summability is a method to regularize divergent series and that makes finite sums that would otherwise be infinite as the limit of partial sums and I also know that exist some theorems about the convergence according to Abel of the Fourier series. But here, for my question, how can I use these informations?



1 Answer 1


This was a comment that got too long which explains why the result required - Fourier series of integrable functions are Abel summable a.e. to the function - is true and how it can be derived in two ways; either way some non-trivial facts about the Lebesgue integral and Feijer or Poisson kernel are used so there is work involved and it's better to look it up in any standard text like Zygmund, Edwards (Trigonometric/Fourier series) or Duren ($H^p$ spaces) say.

For continuous functions, the proof is much easier (including uniform continuity) and uses only general stuff about the Feijer/Poisson kernel (positivity, normalization and the uniform convergence to zero of the Kernel on compact sets outside zero and its periods $2k\pi$)

Caesaro summability of the first order (arithmetic means) implies Abel summability (easy manipulations of series - Frobenius 1880's) so all the results of Feijer, Lebesgue etc apply; here the proof that for any integrable function, its Fourier series sums Caesaro a.e. to the function uses Lebesgue's theorem that the fundamental theorem of calculus holds a.e. for indefinite integrals of integrable functions and some standard manipulations of convolutions with the Feijer kernel, but it is not quite straightforward.

The proof that the Fourier series is Abel summable a.e. to the function also can be given directly using the Poisson Kernel and Fatou's result about the existence of radial limits a.e. of harmonic functions in $l^1(\mathbb D)$ and again the result uses some specific facts about the behavior of the Poisson Kernel near the boundary of the unit circle.

Edit later - as requested I will add a sketch of the direct proof assuming:

1 -Lebesgue's theorem (the generalization of the fundamental theorem of calculus) which states in the context here that if $f \in L^1([0, 2\pi]$ and $F(r)=\int_{0}^{r}f(t)dt, 0 \le r \le 2\pi$, then $F$ as a function on $[0, 2\pi)$ is absolutely continuous, differentiable a.e., $F'(t)=f(t)$ ae and we can integrate by parts $fg=F'g$ as usual when $g$ differentiable say. Instead of $[0,2\pi]$ we can use $[-\pi, \pi]$ (or any period length interval) too when $f$ is a function on the unit circle $T$ so is periodic with period $2\pi$

This result is basic real variable theory though non-trivial and should be known and understood before trying to study Fourier series

2 - Poisson formula for harmonic functions on the unit disc - if $f \in L^1(T)$ as above and $P(r,\theta)= \Re \frac{1+re^{i\theta}}{1-re^{i\theta}}=\frac{1-r^2}{1-2r\cos \theta +r^2}=1+2\sum_{n \ge 1} {r^n \cos n \theta}=1+\sum _{n <0}r^{-n}e^{in\theta}+\sum _{n >0}r^{n}e^{in\theta}$ is the Poisson Kernel, then $u(re^{i\theta})=\frac{1}{2\pi}\int_0^{2\pi}P(r,\theta-t)f(e^{it})dt$ is the harmonic extension of $f$ to the unit disc and it is precisely $\sum_{n<0}{a_nr^{-n}e^{in\theta}}+a_0+\sum_{n>0}{a_nr^{n}e^{in\theta}}$ the Abel mean of $\sum {a_ne^{int}}$ the Fourier series of $f$

Then Fatou's theorem claims that $u(re^{i\theta}) \to f(e^{i\theta}), r \to 1$ at all points where $F$ an indefinite integral of $f$ (where $f$ is now regarded as a periodic function on some given segment of length $2\pi$ on the real axis) is differentiable and has derivative $f$ and that happens almost everywhere by Lebesgue's theorem.

The above statements put together prove that the Abel mean of the Fourier series of $f$ converges almost everywhere to the function $f$.

(Fatou's theorem is true more generally for harmonic extensions of $fdt+d\mu, f \in L^1, d\mu$ singular measure and also with a slightly weaker condition that only the symmetric derivative of the indefinite integral of $fdt+d\mu$ needs to exist finite or allowing $\pm \infty$ for real functions $f$ and singular real measures $d\mu$ and then convergence happens to it whether that symmetric derivative is finite or infinite)

Given 1 and 2, we assume the derivative $F'(\theta)$ which we also write by a slight abuse of notation as $F'(e^{i\theta})$ exists at $\theta_0$ and for notational simplicity, we let wlog $\theta_0=0$ (by a translation that doesn't affect anything) and let $A$ the derivative.

$2\pi(u(r)-A)=\int_{-\pi}^{\pi}P(r,t)(F'(t)-A)dt=P(r,t)(F(t)-At)|_{\pi}^{\pi}-\int_{-\pi}^{\pi}\frac{\partial P(r,t)}{\partial t}(F(t)-At)dt$ as integration by parts by point 1 (Lebesgue)

But now it is clear that $P(r, \pm \pi) \to 0, r \to 1$ so only the integral term matters and there we do the usual decomposition in $|t| \le \delta, |t| \ge \delta$. The second term again is easy as $\frac{\partial P(r,t)}{\partial t} \to 0, r \to 1$ uniformly in $r$ on $ |t| \ge \delta$, while if we fix $\delta$ and write $((F(t)-F(-t))/2-At)=(\frac{F(t)-F(-t)}{2t}-A)t$ change the integral from $|t| \le \delta$ to $0 \le t \le \delta$ and make it symmetric with $t \to -t$ and given $\epsilon$ we find $\delta, |\frac{F(t)-F(-t)}{2t}-A| \le \epsilon, 0<t \le \delta$ by the defintion of $A$, while $-t\frac{\partial P(r,t)}{\partial t} >0, 0 \le t \le \delta$ so we can take the abolute value hence the $\epsilon$ in front, integrate again by parts $-t\frac{\partial P(r,t)}{\partial t}$ and show it is uniformly bounded near $0$ for $r \to 1$ so we are finally done and have proved Fatou's theorem, hence the statement about the Abel mean of a Fourier series.

Examining carefully the proof above shows how it extends to the more general statements mentioned (most notably the real infinite case).

(It is a good exercise to fill in the little details in the above which are standard stuff in kernel theory)

  • $\begingroup$ thanks you very much! So there are many and many ways of demonstrate it!! If you have time, can you report here one of them? $\endgroup$
    – Adam
    Apr 30, 2020 at 18:23

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