Let $||f||_2=\sqrt{\int_{-\pi}^{\pi} f^2(x) dx}$
$||f||_{\infty}=\sup \{ |f(x)| \mid x \in [-\pi,\pi]\}$.
Suppose $f: \mathbb{R} \to \mathbb{R}$ an in the space of piecewise continuous functions with period $2 \pi$.
Prove that $||f||_2 \le \sqrt{2 \pi} ||f || _{\infty}$.
$\textbf{My Attempt:}$
I know that $a_0 = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) dx$
It seems that I could try using something like this, $2 \pi a_0 = \int_{-\pi}^{\pi} f(x) dx$ so $\sqrt{2 \pi} \sqrt{a_0} = \sqrt{ \int_{-\pi}^{\pi} f(x) dx}$.
Of course I run in to the problem that the values are not necessarily nonnegative.
It seems that using the fourier coefficient $a_0$ would be useful to complete the proof. However, I am having a hard time with this.
Help is appreciated! Thank you!!
Since $|f| \leq \|f\|_\infty$, $$ \|f\|_2^2 = \int_{-\pi}^\pi |f|^2 \leq \int_{-\pi}^\pi \|f\|_\infty^2 = 2\pi \|f\|_\infty^2 . $$