Bogachev's "Gaussian measures" contains the following theorem:
Theorem 3.3.1: Let $\mathcal{X}$ be a locally convex space and $G\subset \mathcal{X}^\ast$ a linear subspace separating the points of $\mathcal{X}$. Consider two nonnegative quadratic forms $V,W$ on $G$ and suppose that the function $g\mapsto \exp\left(-\frac{1}{2}V(g)\right)$ coincides with the Fourier transform of a Radon Gaussian measure $\gamma$ on $\mathcal{X}$. If $W(g)\leq V(g)$ for all $g\in G$, then $g\mapsto \exp\left(-\frac{1}{2}W(g)\right)$ also coincides on $G$ with the Fourier transform of a Radon Gaussian measure.
The definition of the Fourier transform is classically given by $$\widetilde{\gamma}(f)=\int_{\mathcal{X}}\exp(if(x))\, d\gamma(x),\ f\in \mathcal{X}^\ast.$$ Later in the book, in the proof of Theorem 3.5.1, that theorem is applied to a measure whose Fourier transformation is equal to $$\widetilde{\gamma}(f)=\exp\left(-\frac{1}{2}\sum_{i=1}^{\infty} f(e_n)^2\right),\ f\in \mathcal{X}^\ast.$$ where $\{e_n\}$ is an orthonormal basis of the Cameron-Martin space. The claim is that, because of the above theorem, there exists a measure $\lambda$ whose Fourier transformation is equal to $$\widetilde{\lambda}(f)=\exp\left(-\frac{1}{2}\sum_{i=1}^{\infty} f(Ae_n)^2\right), f\in \mathcal{X}^\ast,$$ where $A$ is a bounded operator from the Cameron-Martin space to itself.
Hence, it looks like $\sum_{i=1}^{\infty} f(e_n)^2$ should dominate $\sum_{i=1}^{\infty} f(Ae_n)^2$ on the whole $\mathcal{X}^\ast$, or at least on some subspace separating the points of $\mathcal{X}$. But I cannot figure out why this is the case.
Does anyone have any hints or suggestions? Thank you very much in advance!