24 Jun 2018

Parseval's identity

Geometrically, it is the Pythagorean theorem for inner-product spaces. In its special case for a single vector \(x\) it can be written as:

\[\sum_{n} |\prec x,e_n \succ|^2 = ||x||^2\]


\[\prec *,* \succ\]

represents an inner product and

\[||x||^{2} = \prec x, x \succ\]

This is directly analogous to the Pythagorean theorem, which asserts that the sum of the squares of the components of a vector in an orthonormal basis is equal to the squared length of the vector. Here, we project vector \(x\) on each of the basis vectors \(e_i\).

In more general form, the Parseval’s identity is:

\[\prec u,v \succ = \sum_i \prec u,w_i \succ \prec w_i, v \succ\]

First, we prove that \(c_i = \prec u, w_i \succ\). The vectors \(w_i\) for \(i=1,...,n\) are the orthonormal basis for inner-product vector space V. Then for any \(u \in V\) there exists \(c_1,c_2, ..., c_n\) such that \(u = \sum_i c_i w_i\). The coefficients \(c_i\) are called Fourier coefficients. Let’s prove that \(c_i = \prec v, w_i \succ\) using the orthogonormality (orthogonal and normal) vectors \(w_i\).

\[\prec v, w_i \succ = \prec \sum_{i=1}^{n} c_i w_i, w_i \succ = \sum_{i=1}^{n} c_i \prec w_i, w_i \succ = c_i\] \[\prec w_i, w_j \succ = 1 \mbox{ if } i=j \mbox{; } 0, \mbox{ otherwise}\]

Now, we prove the general form of Parseval’s identity:

\[\prec u,v \succ = \prec \sum_{i=1}^{n} c_i w_i, v \succ = \sum_{i=1}^{n} c_i \prec w_i, v \succ = \sum_{i=1}^{n} \prec v, w_i \succ \prec w_i, v \succ\]