If $f: \mathbb R \to \mathbb R$ is continuous and $2\pi$-periodic, then $f \in L^2[-\pi, \pi]$

Question

I'm learning about Fourier series, specifically $L^2$ convergence, and need help with the following problem:

Let $f: \mathbb R \to \mathbb R$ be continuous and $2\pi$-periodic. Show that $f \in L^2[-\pi, \pi]$.

This exercise looks rather simple but I'm stuck. I think the fact that $f$ is continuous is key here but I don't know how to use it. Also, since the function $f$ is $2\pi$-periodic, then $f$ expands as a Fourier series of the form

$$f(x) \sim \frac{a_0}{2} + \sum_{k = 1}^{\infty}(a_k\cos{kx} + b_k\sin{kx}).$$

Moreover, $f \in L^2[-\pi, \pi]$ if and only if

$$\int_{-\pi}^{\pi} \left| f(x) \right|^2 dx < \infty$$

but I still don't know how to prove the above statement.

Answer 1

If it's continuous, it's bounded (image of the compact set $[-\pi,\pi]$ is a compact hence bounded set), so $0\le |f(x)|^2\le M$. Then by Lebesgue's DCT or Fatou's lemma or whatever you want,

$$\int_{-\pi}^\pi |f(x)|^2\,dx\le \int_{-\pi}^\pi M\,dx = 2\pi M < \infty$$

which is exactly what it means to be in $L^2([-\pi,\pi])$