Fourier series and a specific identity

Question

Let $f \in C[0,2\pi]$ and where $a_k, b_k $ are its Fourier coefficients. Assume there are finitely many nonzero coefficients.

Then $||f||^2_2=2\pi a^2_0 \ + \pi\sum_{k=1}^{\infty}(a_k^2+b_k^2).$

I know that this identity holds for any function on interval $[0,2\pi]$ where $\int f^2 < \infty$.

How can I show the above identity holds?

Answer 1

Recall that $L_\mathbb{R}^2[0, 2\pi]$, the space of all real square-integrable functions on $[0, 2\pi]$, is a Hilbert space equipped with the inner product:

$$\langle f, g \rangle = \int_0^{2\pi} f(t)g(t)\,dt\quad \forall f, g \in L^2_\mathbb{R}[0, 2\pi]$$

For an orthonormal basis $(e_n)_{n=1}^\infty$ of a Hilbert space $X$ the Parseval identity holds:

$$\|x\|^2 = \sum_{n=1}^\infty |\langle x, e_n\rangle|^2, \quad\forall x\in X$$

In particular, the set of functions $$B = \left\{\frac1{\sqrt{2\pi}}\right\}\cup \left\{\frac{1}{\sqrt{\pi}}\sin nx : n \in \mathbb{N}\right\} \cup \left\{\frac{1}{\sqrt{\pi}}\cos nx : n \in \mathbb{N}\right\}$$ is an orthonormal basis for $L^2[0,2\pi]$ so we have:

\begin{align} \|f\|_2^2 &= \left\langle f, \frac1{\sqrt{2\pi}}\right\rangle^2 + \sum_{n=1}^\infty\left(\left\langle f, \frac{1}{\sqrt{\pi}}\sin nx\right\rangle^2 + \left\langle f,\frac{1}{\sqrt{\pi}}\cos nx\right\rangle^2\right)\\ &= \frac1{2\pi}\left(\int_0^{2\pi}f(t)\,dt\right)^2 + \frac{1}{\pi}\sum_{n=1}^\infty\left(\left(\int_0^{2\pi}f(t)\sin nt\,dt\right)^2 + \left(\int_0^{2\pi}f(t)\cos nt\,dt\right)^2\right)\\ \end{align}

Now, it seems your Fourier coefficients are defined as:

$$a_0 = \frac{1}{2\pi}\int_0^{2\pi} f(t)\,dt$$ $$a_n = \frac{1}{\pi}\int_0^{2\pi} f(t)\cos nt \,dt, \quad n\in\mathbb{N}$$ $$b_n = \frac{1}{\pi}\int_0^{2\pi} f(t)\sin nt \,dt, \quad n\in\mathbb{N}$$

so we can recognize them in the above formula:

\begin{align} \|f\|_2^2 &= 2\pi\left(\frac1{2\pi}\int_0^{2\pi}f(t)\,dt\right)^2 + \pi\sum_{n=1}^\infty\left(\left(\frac{1}{\pi}\int_0^{2\pi}f(t)\sin nt\,dt\right)^2 + \left(\frac{1}{\pi}\int_0^{2\pi}f(t)\cos nt\,dt\right)^2\right)\\ &= 2\pi a_0^2 + \sum_{n=1}^\infty \left(a_n^2 + b_n^2\right) \end{align}

We obtain the desired result.

This holds for any $f \in L_\mathbb{R}^2[0, 2\pi]$, so in particular it holds for $f \in C_\mathbb{R}[0, 2\pi] \subseteq L_\mathbb{R}^2[0, 2\pi]$, as continuous functions are square-integrable.

Answer 2

Correction: The first term on the RHS is "$\frac{\pi}{2}a_0^2$".

This is Parseval's identity (restricted to the case that only finitely many Fourier coefficients are nonzero). It explicitly appears as the last line of the statements of Parseval's theorem on the English Wikipedia. In both cases, your condition of finitely many nonzero coefficients is enforcing the hypothesis of square integrability.

Depending on what your goals are, either ... Use Parseval's theorem and observe that since the sum on the RHS is finite, the integral in $||\cdot||_2$ is finite and hence $f$ is square integrable, providing post hoc justification of the use of the theorem. (The theorem has a standard real analysis property: it holds even if both sides are $\infty$. So if, after applying the theorem, you can show either side is finite, you have shown both sides are.) Alternatively, re-prove Parseval's theorem using your new boundedness for the indices of nonzero coefficients as the hypothesis replacing square integrability.