3
$\begingroup$

I found a problem that goes like this:

Suppose $f\in C^1(\mathbb{R}/\mathbb{Z})$ and $\int_0^1 fdx=0$. Show that $\int_0^1 (f')^2dx\ge4\pi^2\int_0^1 f^2dx$.

My first thought was to use Fourier series, by writing $$ f(x) = \sum_{n=1}^\infty \left(a_n\sin(2\pi nx)+b_n\cos(2\pi nx)\right) $$

Then we have:

  1. $f^2(x) = \left(\sum_{n=1}^\infty \left(a_n\sin(2\pi nx)+b_n\cos(2\pi nx)\right) \right)^2$
  2. $\int_0^1 f^2dx=\sum_{n=1}^\infty \frac{a_n^2+b_n^2}{2\pi n}$
  3. $f'(x)=\sum_{n=1}^\infty \left(2\pi na_n\cos(2\pi nx)-2\pi nb_n\sin(2\pi nx)\right)$
  4. $(f')^2(x)=\left(\sum_{n=1}^\infty \left(2\pi na_n\cos(2\pi nx)-2\pi nb_n\sin(2\pi nx)\right)\right)^2$
  5. $\int_0^1 (f')^2dx=\sum_{n=1}^\infty 2\pi n(a_n^2+b_n^2)$

And then the inequality of the problem follows easily. My problem is I'm not sure if each step above is valid for the infinite series. I'm pretty sure the first two are fine since $f$ is continuously differentiable and the domain is compact, which implies the Fourier series converges uniformly. However, I'm not sure if differentiating term-by-term in step 3 is valid, and I have no idea how to justify the last two. Are these infinite series manipulations valid?

Finally, although I'm more interested in knowing whether the above approach is mathematically sound, I'm also interested to hear about other ways to solve the original problem.

$\endgroup$
1
  • $\begingroup$ I can't be sure of your notation, but if the function is continuous in its periodic extensions, and such that its derivative is continuous almost always, then its Fourier series converges uniformly (and absolutely) to the function, and you can thus termwise differentiate the series and the Fourier seris of $\;f\;$ . To integrate the series termwise it is enough the function is a.a. continuous in its periodic extension. $\endgroup$ – DonAntonio Dec 22 '20 at 21:54
3
$\begingroup$

I would be nice if series $3$ converges uniformly; however, this is not always the case. Examples of continuous function with divergent Fourier series are not difficult to find. To tackle this, note that whether or not we can differentiate term by term (in classical sense), the Fourier series of $f'$ is always given by series $3$. Since $f'\in C^0,$ the Fourier series converges almost everywhere to $f'$. This is sufficient for us to perform Riemann or Lebesgue integral on $f'$, which is what we need to prove.

It is then a simple application of Plancherel theorem to go from 3 to 5. Or, if you want to square the series directly, then you can use Mertens' theorem (see also Cauchy product).

As for more ways of proving this, search for Poincaré inequality. The thing you prove generalises to higher dimensions. This is a very interesting topic.

$\endgroup$
3
  • $\begingroup$ Thank you so much for the links! I don't see how uniform convergence of the Fourier series for $f$ gives term-by-term differentiation - that's not true in general, but maybe it works for Fourier series? In fact, I think I can show 3-5 if I can prove $\sum_{n=1}^\infty \left(2\pi na_n\cos(2\pi nx)-2\pi nb_n\sin(2\pi nx)\right)$ is uniformly convergent. $\endgroup$ – Hempelicious Dec 23 '20 at 7:30
  • $\begingroup$ @Hempelicious I have updated my answers. Please have a look. $\endgroup$ – Ma Joad Dec 23 '20 at 8:57
  • $\begingroup$ Ah, now I get it. We have the Fourier series for $f$ and $f'$, and two applications of Plancherel's theorem gives steps 2 and 5. Thanks again! $\endgroup$ – Hempelicious Dec 24 '20 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.