2

Let $f: \mathbb{R} \to \mathbb{R}$ be a $2\pi$ -periodic Riemann integrable function such that $1 \leq f(x) \leq 2$ for all $ x \in \mathbb{R}$. Let $f(x)$ ~ $\sum_{n \in \mathbb{Z}} c_n e^{inx}$ be the Fourier expansion of $f$. Prove that:

A) $\sum_{n \geq 1} |c_n|^2 \leq \frac {1}{8}$

B) There exists a continuous $2\pi$-periodic function $g: \mathbb{R} \to \mathbb{R}$ with Fourier expansion $g(x)$ ~ $\sum_{n \geq 1} \frac {|c_n|}{n} \cos {nx}$.

I know that the Fourier coefficients are defined as $c_n=\frac {1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx}dx$. I am also familiar with the Dirichlet kernel as an alternate way of finding partial sums. This is a test review problem so a full solution would be appreciated.

mathqueen459
  • 1,193

1 Answers1

3

For A), notice that $f(x) - c_0 \sim \sum_{n \neq 0} c_n e^{inx}$. Then by the Parseval's identity and the identity $c_{-n} = \overline{c_n}$, we have

$$ 2 \sum_{n \geq 1} |c_n|^2 = \sum_{n \neq 0} |c_n|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |f(x) - c_0|^2 \, dx. $$

So it suffices to show that the LHS is $\leq \frac{1}{4}$. To this end, we write

\begin{align*} \frac{1}{2\pi} \int_{-\pi}^{\pi} |f(x) - c_0|^2 \, dx &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(\left(f(x) - \tfrac{3}{2}\right) - \left(c_0 - \tfrac{3}{2}\right) \right)^2 \, dx \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(f(x) - \tfrac{3}{2}\right)^2 \, dx - \left(c_0 - \tfrac{3}{2}\right)^2 \\ &\leq \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(f(x) - \tfrac{3}{2}\right)^2 \, dx. \end{align*}

Since $1 \leq f(x) \leq 2$, it follows that $|f(x) - \frac{3}{2}| \leq \frac{1}{2}$. Plugging this inequality gives the desired bound.

For B), by Cauchy-Schwarz inequality we have

$$\sum_{n\geq 1} \frac{|c_n|}{n} \leq \left( \sum_{n\geq 1} |c_n|^2 \right)^{1/2}\left( \sum_{n\geq 1} \frac{1}{n^2} \right)^{1/2} < \infty. $$

Therefore by the Weierstrass M-test, the series defining $g(x)$ converges uniformly and so $g$ is continuous.

Sangchul Lee
  • 167,468
  • For part B, how do we know that $$\sum _{n\geq 1} \frac {|c_n|}{n}\cos {nx}<\infty$$ You have shown that $$\sum _{n\geq 1} \frac {|c_n|}{n}<\infty$$but this is not the whole sum. – mathqueen459 Apr 21 '17 at 16:34
  • @britgirl5, Have you heard of the Weierstrass M-test? – Sangchul Lee Apr 21 '17 at 17:14
  • Yes! It says that if $|f_n(x)| \leq M_n$, then $\sum f_n$ converges uniformly if $\sum M_n$ converges, correct? – mathqueen459 Apr 21 '17 at 17:24
  • @britgirl5, Exactly! Now can you see how this theorem applies to our case? – Sangchul Lee Apr 21 '17 at 17:24
  • I'm not sure that I completely understand. If I'm following correctly, we are letting $\sum _{n \geq 1} \frac {|c_n|}{n}$ be our $M_n$ and claiming that $\sum _{n \geq 1} \frac {|c_n|}{n} \cos {nx} \leq M_n$, which we have shown converges. Is this accurate? – mathqueen459 Apr 21 '17 at 17:28
  • @britgirl5, The heart of Weierstrass M-test is that term-wise bounds together with some extra condition guarantee uniform convergence, which saves lots of effort compared to other approaches. So you need to produce a term-wise bound. In this case, $$ \left| \frac{|c_n|}{n}\cos(nx) \right| \leq \frac{|c_n|}{n} =: M_n $$ and $\sum_n M_n < \infty$, which is shown in my answer. (I also suggest you to review your comment to see what makes your choice of $M_n$ inapplicable. For instance, does your $M_n$ satisfy $\sum_{n\geq 1} M_n < \infty$?) And finally, how does it guarantee the continuity of $g$? – Sangchul Lee Apr 21 '17 at 17:34
  • @Sanchul Lee I should let $M_n=\frac {|c_n|}{n}$, not the sum. Can you explain why this guarantees the continuity of $g$? – mathqueen459 Apr 21 '17 at 17:46
  • @britgirl5, Here comes another theorem related to uniform convergence: if $g_n \to g$ uniformly and each $g_n$ is continuous, then $g$ is also continuous. Now by the Weierstrass M-test, we already know that $g_n(x) = \sum_{k=1}^{n} \frac{|c_k|}{k}\cos(kx)$ converges uniformly to $g$. And of course, each $g_n$ is continuous. So $g$ is also continuous. (Notice that, if the continuity of $g$ were not required, then we could have simply appealed to the absolute convergence of the series by comparison test.) – Sangchul Lee Apr 21 '17 at 17:49
  • Got it! Thanks for your help! – mathqueen459 Apr 21 '17 at 17:52
  • @britgirl5 You're welcome :) – Sangchul Lee Apr 21 '17 at 17:53