On differentiability and Fourier coefficients (Vretblad's text)

[psie]

TL;DR

I'm reading Vretblad's Fourier Analysis and its Applications. In the chapter on Fourier series, there is a section on differentiable functions and an exercise to prove variants of a theorem that relates smoothness to an upper bound on the Fourier coefficients.

Let $\mathbb T$ be the unit circle and denote the complex Fourier coefficient of $f$ by $c_n$. Then there is the following theorem;

Theorem 4.4 If $f\in C^k(\mathbb T)$, then $|c_n|\leq M/|n|^k$ for some constant $M$.

This theorem is not really proved in the book, but if $f$ is (Riemann) integrable over $\mathbb T$, then the Fourier coefficients are bounded. This follows from the definition of $c_n$, namely $$|c_n|=\left|\frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)e^{-int}dt\right|\leq \frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f(t)\right|\left|e^{-int}\right|dt=\frac{1}{2\pi}\int_{-\pi}^{\pi} |f(t)|dt=M,$$ since $f$ is integrable on $\mathbb T$. If $b_n$ denotes the Fourier coefficient of $f^{(k)}$, then by recursively applying partial integration, and noting that if $g$ is continuous on $\mathbb T$, then $g(\pi)=g(-\pi)$, so \begin{align} b_n &= \frac{1}{2\pi}\int_{\mathbb T}f^{(k)}(t)e^{-int}dt \nonumber \\ &= \frac{1}{2\pi}[f^{(k-1)}(t)e^{-int}]^{\pi}_{-\pi}+\frac{1}{2\pi}in\int_{\mathbb T}f^{(k-1)}(t)e^{-int} dt \nonumber \\ &= \ldots \nonumber \\ &=(in)^k \frac{1}{2\pi}\int_{\mathbb T}f(t)e^{-int}dt \nonumber \\ &= (in)^kc_n \nonumber.\end{align}
Since $f^{(k)}$ is continuous (and thus integrable), we have $|b_n|\leq M$ for some $M$, i.e. $|n^k c_n|\leq M$, and the claim of the theorem follows.

Then there is the following exercise in the book;

Try to prove the following partial improvements of Theorem 4.4:
(a) If $f'$ is continuous and differentiable on $\mathbb T$ except possibly for a finite number of jump discontinuities, then $|c_n|\leq M/|n|$ for some constant $M$.
(b) If $f$ is continuous on $\mathbb T$ and has a second derivative everywhere except possibly for a finite number of points, where there are "corners" (i.e., the left-hand and right-hand first derivative exist but are different from each other), then $|c_n|\leq M/n^2$ for some constant $M$.

1. I struggle with seeing the difference in the assumptions of these two statements. Is (a) not assuming the same as (b)?
2. Consider statement (a) and the assumptions on $f'$. What does this tell us about $f$? I've been trying to compute the Fourier coefficients of $f'$ as above for the "proof" of theorem 4.4, i.e. via partial integration, but I'm not sure what properties $f$ has.
3. Any hints for (b)?

Grateful for any help.

 

[pasmith]

In (a), f does not have to be continuous, but can have a jump discontinuity wherever f&#039; does. An example would be the periodic function defined by <br /> f: x \mapsto \begin{cases}<br /> 0 &amp; -\pi &lt; x \leq 0 \\<br /> 1 + \sin x &amp; 0 &lt; x \leq \pi<br /> \end{cases} with derivative <br /> f&#039;(x) = \begin{cases}<br /> 0 &amp; -\pi &lt; x &lt; 0 \\<br /> \cos x &amp; 0 &lt; x &lt; \pi. \end{cases} In (b) it is expressly stated that f is continuous.

 

[psie]

In (a), f does not have to be continuous, but can have a jump discontinuity wherever f&#039; does.

Maybe $f'$ in (a) is a typo for $f$? Otherwise I do not see how to obtain the bound $|c_n|\leq M/|n|$.

 

[pasmith]

The key is that we must define <br /> c_n = \frac{1}{2\pi} \sum_{j=1}^N \int_{x_j}^{x_{j+1}} f(x)e^{-inx}\,dx where -\pi = x_1 &lt; \dots &lt; x_j&lt; \dots &lt; x_{N+1} = \pi with x_2, \dots, x_{N} being points of discontinuity of f&#039; or f. Then integrating by parts gives <br /> c_n = \frac{i}{2n\pi} \sum_{j=1}^N \left(\left[ f(x)e^{-inx} \right]_{x_j}^{x_{j+1}} - \int_{x_j}^{x_{j+1}} f&#039;(x)e^{-inx}\,dx\right). Now <br /> \sum_{j=1}^N \left[ f(x)e^{-inx} \right]_{x_j}^{x_{j+1}} = (-1)^n(f(\pi^{-}) - f(-\pi^{+})) - \sum_{j=2}^{N} (f(x_j^{+}) - f(x_j^{-}))e^{-inx_j} and if f is not continuous then either f(\pi^{-}) \neq f(-\pi^{+}) or f(x_j^{+}) \neq f(x_j^{-}) for some j.

 

[psie]

The key is that we must define <br /> c_n = \frac{1}{2\pi} \sum_{j=1}^N \int_{x_j}^{x_{j+1}} f(x)e^{-inx}\,dx where -\pi = x_1 &lt; \dots &lt; x_j&lt; \dots &lt; x_{N+1} = \pi with x_2, \dots, x_{N} being points of discontinuity of f&#039; or f. Then integrating by parts gives <br /> c_n = \frac{i}{2n\pi} \sum_{j=1}^N \left(\left[ f(x)e^{-inx} \right]_{x_j}^{x_{j+1}} - \int_{x_j}^{x_{j+1}} f&#039;(x)e^{-inx}\,dx\right). Now <br /> \sum_{j=1}^N \left[ f(x)e^{-inx} \right]_{x_j}^{x_{j+1}} = (-1)^n(f(\pi^{-}) - f(-\pi^{+})) - \sum_{j=2}^{N} (f(x_j^{+}) - f(x_j^{-}))e^{-inx_j} and if f is not continuous then either f(\pi^{-}) \neq f(-\pi^{+}) or f(x_j^{+}) \neq f(x_j^{-}) for some j.

I think this shows that we need $f$ to be piecewise $C^1$ in (a) to arrive at $$|c_n|=\left|\frac{1}{n2\pi}\sum_{j=1}^N \int_{x_j}^{x_{j+1}} f'(x)e^{-inx}dx\right|\leq \frac{M}{|n|},$$ where $M=\frac{1}{2\pi}\int_{-\pi}^\pi |f'(x)|dx$ if $f$ is assumed to be piecewise $C^1$ (meaning it is continuous and has a piecewise continuous derivative).

(b) seems to also have some missing assumptions. If we repeat your argument, we require integrability of the second derivative, which is not stated in the exercise. Here probably too we require piecewise $C^2$. Then the exercise is just the same as (a) basically.