Who are you? The next Newton or something?: Eigenfunction Expansions for Mercer Kernels

Wednesday, December 25, 2024

Eigenfunction Expansions for Mercer Kernels

Consider an integral covariance operator with Mercer kernel $R (s, t)$

$\displaystyle T f (t) = \int_0^{\infty} R (s, t) f (s) \hspace{0.17em} ds$

The eigenfunctions satisfy the equation:

$\displaystyle T \psi (s) = \int_0^{\infty} R (s, t) \psi (s) \hspace{0.17em} ds = \lambda \psi (t)$

where $\{\psi_n \}_{n = 1}^{\infty}$ are the eigenfunctions with corresponding eigenvalues $\{\lambda_n \}_{n = 1}^{\infty}$

Let $\{\phi_j \}_{j = 1}^{\infty}$ be a complete orthonormal basis of $L^2 [0, \infty)$ and define the kernel matrix elements:

$\displaystyle K_{kj} = \int_0^{\infty} \int_0^{\infty} R (s, t) \phi_k (t) \phi_j (s) \hspace{0.17em} dt \hspace{0.17em} ds$

If $\psi_n (t) = \sum_{j = 1}^{\infty} c_{n, j} \phi_j (t)$ is an eigenfunction expansion, then:

$\displaystyle c_{n, k} = \int_0^{\infty} \phi_k (t) \psi_n (t) \hspace{0.17em} dt$

Proof.

Begin with the eigenfunction equation for $\psi_n$ :
$\displaystyle \int_0^{\infty} R (s, t) \psi_n (s) \hspace{0.17em} ds = \lambda_n \psi_n (t)$
Multiply both sides by $\phi_k (t)$ and integrate over t:
$\displaystyle \int_0^{\infty} \phi_k (t) \int_0^{\infty} R (s, t) \psi_n (s) \hspace{0.17em} ds \hspace{0.17em} dt = \lambda_n \int_0^{\infty} \phi_k (t) \psi_n (t) \hspace{0.17em} dt$
Apply Fubini's theorem to swap integration order on the left side:
$\displaystyle \int_0^{\infty} \int_0^{\infty} R (s, t) \phi_k (t) \hspace{0.17em} dt \hspace{0.17em} \psi_n (s) \hspace{0.17em} ds = \lambda_n \int_0^{\infty} \phi_k (t) \psi_n (t) \hspace{0.17em} dt$
Substitute the eigenfunction expansion $\psi_n (s) = \sum_{j = 1}^{\infty} c_{n, j} \phi_j (s)$ :
$\displaystyle \int_0^{\infty} \int_0^{\infty} R (s, t) \phi_k (t) \hspace{0.17em} dt \hspace{0.17em} \sum_{j = 1}^{\infty} c_{n, j} \phi_j (s) \hspace{0.17em} ds = \lambda_n \int_0^{\infty} \phi_k (t) \psi_n (t) \hspace{0.17em} dt$
Exchange summation and integration (justified by $L^2$ convergence):
$\displaystyle \sum_{j = 1}^{\infty} c_{n, j} \int_0^{\infty} \int_0^{\infty} R (s, t) \phi_k (t) \phi_j (s) \hspace{0.17em} dt \hspace{0.17em} ds = \lambda_n \int_0^{\infty} \phi_k (t) \psi_n (t) \hspace{0.17em} dt$
Recognize the kernel matrix elements:
$\displaystyle \sum_{j = 1}^{\infty} c_{n, j} K_{kj} = \lambda_n \int_0^{\infty} \phi_k (t) \psi_n (t) \hspace{0.17em} dt$
Note that $\sum_{j = 1}^{\infty} c_{n, j} K_{kj}$ is the $k$ -th component of $K \textbf{c}_n$ . Since $\psi_n$ is an eigenfunction, $\textbf{c}_n$ must satisfy $K \textbf{c}_n = \lambda_n \textbf{c}_n$ , thus:
$\displaystyle \lambda_n c_{n, k} = \lambda_n \int_0^{\infty} \phi_k (t) \psi_n (t) \hspace{0.17em} dt$
Divide both sides by $\lambda_n$ (noting $\lambda_n \neq 0$ for non-trivial eigenfunctions):
$\displaystyle c_{n, k} = \int_0^{\infty} \phi_k (t) \psi_n (t) \hspace{0.17em} dt$

This establishes that the coefficient $c_{n, k}$ in the eigenfunction expansion equals the inner product of the basis function $\phi_k$ with the eigenfunction $\psi_n$ . $\Box$

Who are you? The next Newton or something?

Wednesday, December 25, 2024

Eigenfunction Expansions for Mercer Kernels

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