Let $F : L^2(S^1) \to L^2(S^1)$ be a (nonlinear) mapping such that \begin{equation} \lVert F(f) \rVert \leq \lVert f \rVert \end{equation} for all $f \in L^2(S^1)$. For the space of smooth periodic functions $\mathcal{E}(S^1)$, we know that \begin{equation} \mathcal{E}(S^1) \subset L^2(S^1) \subset \mathcal{E}'(S^1). \end{equation}
By the Minlos theorem, there exists a unique Gaussian measure $d\mu$ on $\mathcal{E}'(S^1)$ such that \begin{equation} e^{-\lVert f \rVert^2}=\int_{\mathcal{E}'(S^1)}e^{iT(f)} d\mu(T) \end{equation} for all $f \in \mathcal{E}(S^1)$.
Let $P_N$ be the projection mapping defined by \begin{equation} (P_N g)(x):=\sum_{k=-N}^N e^{2\pi ikx}\int_{-1}^1 g(y) e^{-2\pi iky} dy \end{equation} for $g \in L^2(S^1)$. Then, $P_N$ may be extended to $\mathcal{E}'(S^1)$ by dual pairing.
It is clear that for each $N \in \mathbb{N}$, \begin{equation} \mathcal{F}_N:=\int_{\mathcal{E}'(S^1)} F\bigl(P_N T \bigr) d\mu(T) \end{equation} is an well-defined element of $L^2(S^1)$ due to the above assumption for $F$.
Now, if we set $F(T)=0$ for $T \in \mathcal{E}'(S^1)- L^2(S^1)$, we have \begin{equation} \mathcal{F}:=\int_{\mathcal{E}'(S^1)} F(T) d\mu(T)= \int_{L^2(S^1)} F(f) d\mu(f) \in L^2(S^1). \end{equation}
My question is:
Does $\mathcal{F}_N$ converges to $\mathcal{F}$ in any sense as $N \to \infty$?
This seems like quite a nontrivial problem...Could anyone please help me?
In a previous question
I was answered that $\{ P_N \}$ is equicontinuous even on $\mathcal{E}'(S^1)$. But I cannot see how to use this fact..
