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 \(\psi_n\):
Multiply both sides by \(\phi_k (t)\) and integrate over t:
Apply Fubini's theorem to swap integration order on the left side:
Substitute the eigenfunction expansion \(\psi_n (s) = \sum_{j = 1}^{\infty} c_{n, j} \phi_j (s)\):
Exchange summation and integration (justified by \(L^2\) convergence):
Recognize the kernel matrix elements:
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:
Divide both sides by \(\lambda_n\) (noting \(\lambda_n \neq 0\) for non-trivial eigenfunctions):
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 \(\psi_n\).\(\Box\)
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