Prove that there is a constant $c >0$ such that $\int_0^1f(x)^2 dx \leq c \int_0^1 f'(x)^2 dx$ for all continuously differentiable function $f: \mathbb R \to \mathbb R$ with $f(x+1) = f(x)$ and $\int_0^1f(x) dx = 0$.
I am not sure how to go about this problem. I was thinking of expressing $f$ as a Fourier series, especially as we have the periodicity condition. I was also thinking of using the Dominated Convergence Theorem, but it was not immediate how we could use the DCT.
I would appreciate some hints. Thank you
$\int_0^1f(x) dx = 0$ implies there exists $x_0\in [0,1]$ such that $f(x_0)=0$.
Let $g(x)$ be the translation of $f(x)$ by $x_0$, i.e., $g(x)= f(x+x_0)$.
Since $f(x+1)=f(x)$, we have $\int_0^1f(x)^2 dx =\int_0^1g(x)^2 dx$ and $ \int_0^1 f'(x)^2 dx=\int_0^1 g'(x)^2 dx$.
It is enough to prove the same conclusion for $g(x)$, which is also continuously differentiable on $[0,1]$.
Since $g(0)=0$, we have the first equality below. The first inequality is Cauchy–Schwarz inequality. $$g(x)^2=\left(\int_0^x1\cdot g'(t)\,dt\right)^2 \le \int_0^x1^2\,dt \int_0^x g'(t)^2\,dt \le x \int_0^1 g'(t)^2\,dt,$$
$$\int_0^1g(x)^2\,dx \le \int_0^1 x\,dx \int_0^1g'(t)^2\,dt = \frac{1}{2}\int_0^1g'(t)^2\,dt.$$
We obtain constant $\frac12$.
Since $g(0)=g(1)=0$, we can apply the second version of Wirtinger's inequality to $g(x)$.
$$\int_0^1g(x)^2dx \le \frac{1}{\pi^2}\int_0^1g'(t)^2 dt.$$
We obtain constant $\frac1{\pi^2}$, which is the best possible.
Here are proofs for the Wirtinger's inequality.