Proving $\int_{-\pi}^{\pi}|f(x)|^2dx$ $\le$ $\int_{-\pi}^{\pi}|g(x)|^2dx$

Question

I have this question in my homework.

I've been trying to solve it for a while now, But couldn't get much further.

Let $f(x)$ be a function with period of $2 \pi$.

Define:

$f''(x) + f(x+\frac{\pi}{2}) = g(x)$

Prove: $\int_{-\pi}^{\pi}|f(x)|^2dx$ $\le$ $\int_{-\pi}^{\pi}|g(x)|^2dx$

Any hints/Assistance will be useful.

Answer 1

Under the assumption that $f''$ is continuous (weaker assumptions may be made), if $f(x)= \sum_{n\in\mathbb{Z}}c_n e^{-inx}$, where $$c_n=\frac{1}{2\pi}\int^{2\pi}_0 f(s)e^{-ins}\,ds,$$ then $$f'' (x)=\sum_{n\in\mathbb{Z}}(in)^2c_n e^{-i n x}=-\sum_{n\in\mathbb{Z}}n^2c_ne^{-inx}$$ This can be done by integration by parts and the use of the fundamental theorem of Calculus. On the other hand, if $\tau_{a}f(x):=f(x-a)$, then $$\tau_af(x)=\sum_{n\in\mathbb{Z}}e^{-ina}c_n e^{-inx}$$ This can be seen using the periodicity of $f$ and a translation.

Then $$g(x)=\sum_{n\in\mathbb{Z}}((i)^n-n^2)c_n e^{-inx}$$

• For $n\in\{1,3\}\mod 4$, $|((i)^n-n^2)c_n|^2=(1+n^2)|c_n|^2>|c_n|^2$,
• For $n\equiv 2\mod 4$, $|((i)^n-n^2)c_n|^2=(n^2+1)^2|c_n|^2>|c_n|^2$,
• For $n\equiv0\mod 4$, $|((i)^n-n^2)c_n|^2=(n^2-1)^2|c_n|^2\geq|c_n|^2$.