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