Let $X$ be a locally convex space, and let $\mathcal{E}(X)$ denote the cylindrical $\sigma$-field. Bogachev defines a measure $\mu$ on $(X, \mathcal{E}(X))$ to be Gaussian if the pushforward, $\mu \circ f^{-1}$ for any $f \in X^\ast$ is Gaussian.
He then claims: if $\mu$ is Gaussian, then by definition $X^\ast \subset L^2(\mu)$.
I want to make sure I am understanding why this is the case.
Here is my reasoning: say $f \in X^\ast$ and $\mu$ on $(X, \mathcal{E}(X))$ is Gaussian. Then $\mu \circ f^{-1}$ is Gaussian, with mean $a_\mu(f)$ and variance $\sigma^2(f)$. Then $$ \int f^2(x) \, \mu(dx) = \int t^2 \, [\mu \circ f^{-1}](dt) $$ Writing $t^2 = (t - a_\mu(f))^2 +2(t-a_\mu(f))a_\mu(f) + (a_\mu(f))^2$, we find that $$ \|f\|_{L^2(\mu)}^2 = \sigma^2(f) + (a_\mu(f))^2 < \infty. $$ Hence $f \in L^2(\mu)$.
The basic idea seems to be that $f \in X^\ast$ implies that the projection $\mu \circ f^{-1}$ is Gaussian, in particular that it has finite variance, which is essentially the same as the $L^2$-norm of $f$ under the original measure.
(The secondary manipulation is the standard formula $E[(X- E X)^2] = E[X^2] - E[X]^2$, just re-arranged, but I wanted to just write down the full details to keep it complete.)