2
$\begingroup$

Let $k: [0,T] \times [0,T] \to \mathbb{R}$ symmetric, square-integrable and define $$(Kf)(t) := \int_0^T k(s,t) \cdot f(s) \, ds \qquad (f \in L^2([0,T])$$

Since $K$ is compact, by the spectral theorem, there exists a orthonormal system $(\varphi_j)_j$ in $L^2([0,T])$ of eigenfunctions of $K$ with corresponding eigenvalues $\lambda_j$, and

$$Kf = \sum_{j} \lambda_j \cdot (f|\varphi_j)_{L^2([0,T])} \cdot \varphi_j \tag{1}$$ (the series converges in $L^2([0,T])$).

My question is the following: What are sufficient conditions to conclude that the kernel $k$ admits the representation

$$k(s,t) = \sum_{j} \lambda_j \cdot \varphi_j(s) \cdot \varphi_j(t)$$

Mercer's theorem applies to continuous, non-negative definite kernels, but in my case, $k$ is not continuous (and also not in $L^{\infty}([0,T] \times [0,T])$).


I would like to apply this in the following framework: Given a Gaussian process $(X_t)_{t \in [0,T]}$, mean $0$, covariance $k(s,t) := \mathbb{E}(X_s \cdot X_t)$,

$$(Kf)(t) := \int_0^T k(s,t) \cdot f(s) \, ds \qquad (f \in L^2([0,T]))$$

and it is assumed that $K$ has finite trace. It is known that the paths $t \mapsto X_t(w)$ are in $L^2([0,T])$, $K$ is non-negative definit and admits a (non-negative symmetric) square root $K^{\frac{1}{2}}$.

$\endgroup$
1
$\begingroup$

As far as I can see, the square-integrability and symmetry of the kernel are sufficient conditions:

Denote by $\varphi_j \otimes \varphi_j(s,t) := \varphi_j(s) \cdot \varphi_j(t)$. By definition, $\varphi_j$ is an eigenfunction of $K$ with corresponding eigenvalue $\lambda_j$, so

$$(k|\varphi_j \otimes \varphi_j)_{L^2([0,T] \times [0,T]} = \int_0^T \int_0^T k(s,t) \cdot \varphi_j(s) \cdot \varphi_j(t) \, ds \, dt = \lambda_j \cdot \int_0^T \varphi_j(t) \cdot \varphi_j(t) \, dt = \lambda_j$$

Since $(\varphi_j \otimes \varphi_j)_k$ is orthonormal in $L^2([0,T] \times [0,T])$ and $k \in L^2([0,T] \times [0,T])$ we obtain

$$\sum_{j} \lambda_j^2 = \sum_j |(k|\varphi_j \otimes \varphi_j)_{L^2([0,T] \times [0,T]}|^2 \leq \|k\|_{L^2}^2<\infty$$

by applying Bessel's inequality. This implies that the series

$$\tilde{k} := \sum_{j} \lambda_j \cdot \varphi_j \otimes \varphi_j \tag{2}$$

converges in $L^2([0,T] \times [0,T])$. Consequently,

$$\begin{align*} (Kf,g)_{L^2([0,T]} &\stackrel{(1)}{=} \sum_{j} \lambda_j \cdot (f|\varphi_j)_{L^2([0,T])} \cdot (\varphi_j|g)_{L^2([0,T])} \\ &= \sum_j \lambda_j \cdot (f \otimes g|\varphi_j \otimes \varphi_j) \\ &\stackrel{(2)}{=} (f \otimes g|\sum_j \lambda_j \cdot \varphi_j \otimes \varphi_j) \\ &\stackrel{\text{def}}{=} \int_0^T \int_0^T \sum_j \lambda_j \cdot f(s) \cdot \varphi_j(s) \cdot \varphi_j(t) \cdot g(t) \, ds \, dt \\ &= \left( \int_0^T \tilde{k}(s,\cdot) \cdot f(s) \, ds, g \right)_{L^2([0,T])} \end{align*}$$

for any $f,g \in L^2([0,T])$. Thus,

$$(k|f \otimes g)_{L^2([0,T] \times [0,T])} = (\tilde{k}|f \otimes g)_{L^2([0,T] \times [0,T])}$$

Since $\{f \otimes g; f,g \in L^2\}$ is dense in $L^2([0,T] \times [0,T])$, we conclude $k=\tilde{k}$ almost everywhere.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.