0

I know that there exists a $2\pi$ periodic continuous function whose Fourier series diverges at a point.

However, I heard that for any $2\pi$ periodic continuous function $f(x)$, it is wrong that $\mid S_N(f,0)\mid \to \infty$ as $N$ goes to $\infty$.

Here $S_N(f,x)=1/2\pi\int_{-\pi}^{\pi}[\sin(N+1/2)(x-\theta)/\sin((x-\theta)/2)]f(\theta)d\theta$.

I tried with the $L^2$ convergence of $S_Nf$ to $f$. But this only shows that there exists 'some points' on which $S_Nf$ has a subsequence converging to $f$. Those points are not guaranteed to include $0$. So I am just stymied.

David C. Ullrich
  • 89,985
Keith
  • 7,673
  • currently the formula for $S_n(f,x)$ seems to have nothing to do with $f$. I suggest using \frac{a}{b} for $\frac{a}{b}$? But in any case, if you know the first result, then you can simply translate the point of divergence to 0, right? Or just follow the proof here http://www-users.math.umn.edu/~garrett/m/fun/notes_2012-13/05b_banach_fourier.pdf – Calvin Khor Oct 18 '19 at 12:16
  • Oops I will edit it. A typo – Keith Oct 18 '19 at 12:19
  • What do you mean by translate the point of divergence? Do I define another function $g(x)=f(x-a)$ if $a$ is the point of divergence for $f$? Then $g$ is a different function from $f$. – Keith Oct 18 '19 at 12:25
  • And the link does not say anything about the convergence of the whole series of partial sums.... – Keith Oct 18 '19 at 12:29
  • Oh, I see the distinction. You know that its divergent, but you want to know if it specifically diverges to plus/minus infinity. OK, sorry – Calvin Khor Oct 18 '19 at 12:31
  • Not that it says anything about $x=0$, but you seem to be unaware of Carleson's theorem: If $f\in L^2$ then $S_Nf\to f$ almost everywhere. – David C. Ullrich Oct 18 '19 at 14:58
  • I mentioned the fact in my question and that it does not seem to help much. – Keith Oct 18 '19 at 15:02
  • You cannot really know on which points the partial sums converge. – Keith Oct 18 '19 at 15:03
  • No, it doesn't help much, but no you didn't mention that fact! You alluded to the fact that some subsequence converges to $f$ almost everywhere. That's completely trivial - Carleson's theorem is not trivial, it's like the deepest theorem in the theory of Fourier series. – David C. Ullrich Oct 18 '19 at 15:15
  • Oh I see... there is a huge difference.... however that deep theorem does not seem to help either.... – Keith Oct 18 '19 at 15:58

1 Answers1

1

I don't know the answer in general. But if $f$ is real-valued it's very simple.

Easy Lemma. Suppose $(x_n)$ is a sequence of real numbers such that $|x_n|\to\infty$ and $|x_{n+1}-x_n|$ is bounded. Then $\left|\frac{x_1+\dots+x_n}n\right|\to\infty$.

(Hint: The hypothesis implies that either $x_n\to+\infty$ or $x_n\to-\infty$.)

Now set $$\sigma_n(f)=\frac{s_1(f)+\dots+s_n(f)}n$$as usual. If $f$ is a real-valued continuous function and $|s_n(f,0)|\to\infty$ then $|\sigma_n(f,0)|\to\infty$, which is impossible since $\sigma_n(f)=f*K_n\to f$ uniformly. (Where of course $K_n$ is the Fejer kernel.)

But the lemma is false for sequences of complex numbers...

(Say $C_j$ is the circle of radius $j$ centered at the origin. Let $(z_n)$ be a sequence consisting of one point on $C_1$, then two points evenly spaced on $C_2$, then three on $C_3$, etc. Then $(z_1+\dots+z_n)/n\to0$.)

David C. Ullrich
  • 89,985