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Given a $\textbf{square integrable stochastic process}$ $X$ with $E\left(X\left(t\right)\right)=0$ $\forall t $ the $\textbf{Covariance Operator}$ is defined by
\begin{align} C_X: L^2 \rightarrow L^2 \end{align} with \begin{align} C_X\left(f\right)\left(t\right)&= \int Cov\left(X\left(s\right),X\left(t\right)\right)f\left(s\right) ds \\ &= \int E\left(X\left(s\right)X\left(t\right)\right)f\left(s\right) ds. \end{align} The operator $C_X$ is a $\textbf{Hilbert-Schmidt Operator}$ on the Hilbert-Space $L^2$. I have read several times, that \begin{align} \langle C_X\left(f\right),f \rangle_{L^2} = \int \int E\left(X\left(s\right)X\left(t\right)\right)f\left(s\right) f\left(t\right) ds dt \overset{!}{\geq} 0 \,\,\,\,\, \forall f\in L^2. \tag1 \end{align} Thus $C_X$ is a $\textbf{non-negative operator}$ and has $\textbf{non-negative eigenvalues}$. I don't understand why (1) is greater than 0. Can anyone give a $\textbf{proof}$ or a $\textbf{counter example}$?

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Observe that $$ C_X(f)(t)f(t)=f(t)\int E\left(X\left(s\right)X\left(t\right)\right)f\left(s\right) ds $$ and integrating, we get $$ \langle C_X\left(f\right),f \rangle_{L^2} =\int C_X(f)(t)f(t) dt=\int f(t)\left(\int E\left(X\left(s\right)X\left(t\right)\right)f\left(s\right) ds\right)dt $$ hence inserting $f(t)$ in the inner integral, we get (1). This can be rewritten as $$ \langle C_X\left(f\right),f \rangle_{L^2} =\iint E\left(f\left(s\right)X\left(s\right)f(t)X\left(t\right)\right) dsdt $$ and assuming that we can switch the expectation and the integral, we get $$ \langle C_X\left(f\right),f \rangle_{L^2} = E\left(\iint f\left(s\right)X\left(s\right)f(t)X\left(t\right)dsdt\right) $$ and the double integral in the right hand side is $\left(\int f(s)X(s)ds\right)^2$.

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