I can't seem to be able to prove why the expected value below is finite since I have no knowledge of what's the L2-norm of a stochastic process is:
Let $ \{X(t), t\in[0, 1]\}$ be a stochastic process with trajectories in $L²[0,1]$ and continuous covariance function. Denote by $\left\lVert . \right\rVert$ the $L²$-norm.
Prove that $E\left\lVert X \right\rVert$² < $\infty$
As you stated it I would suppose that $\vert\vert X\vert\vert^2 = \int_0^1 (X(t))^2 \text d t $.
But then this statements seems to be false. Consider any random variable $Z$ with $\Bbb E [Z^2] = \infty$, but $Z<\infty $ a.s., for example $\Bbb P (Z = n^2) = \frac 6 {(\pi n)^2} $, $n\in \Bbb N$. Then we can define a stochastic process as $X(t) := Z\cdot t$. The trajectories of $X$ lie in $L^2$ because they are continuous (thus measurable) and $$\vert\vert X\vert\vert^2 (\omega) = Z^2 \int_0^1 t^2 \text d t= Z^2 \frac 1 3.$$ But $\Bbb E [\vert\vert X\vert\vert^2] = \frac 1 3 \Bbb E[Z^2] =\infty.$