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· 2 min read
· by Xianbin
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.
\[\lVert \xi_1u_1+\ldots+\xi_n u_n\lVert^2 =\sum_{i=1}^n (\xi_i u_i)\cdot (\xi u_i)^\intercal + \sum_i \Pi_{i\neq j} \xi_i \xi_j u_i u_j = \sum_i^n\lVert \xi_i u_i\rVert^2\]