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I am given the following problem statement:

"Let $f$ be a periodic function of class $C^1(\Bbb{R})$, of period $2L$. Show that the coefficients of the fourier series expansion of $f$,

$$\frac{a_0}{2} + \sum_{k=1}^\infty a_k \cos\left(\frac{k\pi x}{L}\right) + b_k \sin\left(\frac{k\pi x}{L}\right)$$

satisfy $\lim_{k\to\infty} a_k = 0$, $\lim_{k\to\infty} b_k = 0$."

I proceeded as follows:

Given that $f$ is $C^1(\Bbb{R})$, we know that the Fourier Series converges to the value of $f$ in every point, so that

$$f(x) = \frac{a_0}{2} + \sum_{k=1}^\infty a_k \cos\left(\frac{k\pi x}{L}\right) + b_k \sin\left(\frac{k\pi x}{L}\right),\ \forall x \in \Bbb{R} \tag{1}$$

In particular, let us consider $x = 0$. $f(0)$ must be some real value, say $f_0$. On the other hand, evaluating the Fourier series yields

$$f_0 = \frac{a_0}{2} + \sum_{k=1}^\infty a_k$$

and we are well aware that if a series converges, its terms go to $0$. (I can easily prove this, if it turns out to be necessary).

At this point I realized that, if I could get an $x$ such that all cosines vanished and all sines became $1$, I would get the same result for the $b_k$ with ease. Having realized I wouldn't be able to do so, I thought of taking the derivative on both sides of $(1)$.

Hence I get

$$f'(x) = \sum_{k=1}^\infty -a_k\frac{k\pi}{L}\sin\left(\frac{k\pi x}{L}\right) + b_k\frac{k\pi }{L}\cos\left(\frac{k \pi x}{L}\right)$$

Again, at $x=0$ I have that $f'(0) = f_0'$ and evaluating the series gives

$$f_0' = \sum_{k=1}^\infty b_k\frac{k\pi }{L} = \frac{\pi}{L}\sum_{k=1}^\infty kb_k$$

Because the series converges, $\lim_{k\to\infty} kb_k = 0 \implies \lim_{k\to\infty} b_k = 0$ and we proved what was asked.

I am not sure if I proved it correctly, in particular the part where I show $\lim_{k\to\infty} b_k = 0$. Please suggest alternative solutions if any come to mind, and point me out my mistakes. Thanks.

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1 Answer 1

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The claim is true for any $f$ Riemann integrable on $[-L, L]$. A simple proof relies on Bessel's inequality that, for $L=\pi$, reads $$ \frac{a_0^2}{2} + \sum_{k=1}^\infty (a_k^2 + b_k^2) \leq \frac{1}{\pi} \int_{-\pi}^\pi f(x)^2 dx. $$ Since $\sum a_k^2$ and $\sum b_k^2$ are convergent, then $a_k, b_k\to 0$.

(The inequality is actually an equality for any $f\in L^2(-\pi,\pi)$.)

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  • $\begingroup$ This seems like a nice alternative proof! Can you tell however if my proof is correct? $\endgroup$
    – RGS
    May 27, 2017 at 11:59
  • $\begingroup$ You are using the fact that the series of derivatives converges pointwise. In general this is not true. But you can take integrals instead of derivatives. $\endgroup$
    – Rigel
    May 27, 2017 at 12:09

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