I am studying a proof, where the following computation is done:
$$ \int_0^T \int_0^T Y_t^2 |\rho(t-s)|dtds \leq 2 \left(\int_0^T(Y_t^2 - EY_0^2)dt+\rho(0)\right) \int_0^{+\infty}|\rho(t)|dt,$$
where $Y_t$ is Gaussian, and $E[Y_t Y_s] = \rho(|t-s|)$ is the covariance function. Could someone explain the analytical step to get the bound?
(1) $Y_t^2=Y_t^2-\mathsf{E}Y_0^2+\rho(0)$ ($\because\mathsf{E}Y_0^2=\rho(0)$).
(2) Changing the order of integration (the integrand is nonnegative) yields $$ \int_0^T\left(Y_t^2-\mathsf{E}Y_0^2+\rho(0)\right)\left(\int_0^T|\rho(|t-s|)|\,ds\right)dt. $$ (3) For each $t\in[0,T]$, $$ \int_0^T|\rho(|t-s|)|\,ds=\int_0^t|\rho(t-s)|\,ds+\int_t^T|\rho(s-t)|\,ds\le 2\int_0^\infty |\rho(z)|\,dz. $$