T1: A Simple Question 2: Parseval's Inequality

July 18, 2024

• Parseval’s Inequality

Parseval’s Inequality

Let $A = \{u_1,\ldots,u_n\}$ be an orthonormal basis for an inner-product space $S$, and let $\textbf{x} = \sum_i \xi_i u_i$ be the Fourier expansion of $\textbf{x} \in S$. Prove that

$$\sum_{i=1}^n \lvert \xi_i\rvert^2 = \lVert \textbf{x} \rVert^2$$

$\textbf{Proof}$.

$$\begin{align} \lVert \textbf{x} \rVert^2 = \lVert \xi_1u_1+\ldots+\xi_n u_n\lVert^2 \\ =\sum_i^n\lVert \xi_i u_i\rVert^2 \\= \sum_i^n \lvert \xi_i \rvert \end{align}$$

Now, let us prove Equality 2.

$$\lVert \xi_1u_1+\ldots+\xi_n u_n\lVert^2 =\sum_{i=1}^n (\xi_i u_i)\cdot (\xi u_i)^\intercal + \sum_i \Pi_{i\neq j} \xi_i \xi_j u_i u_j = \sum_i^n\lVert \xi_i u_i\rVert^2$$