Can we apply an Itō formula to find an expression for $f(t,X_t)$, if $f$ is taking values in a Hilbert space? Let


*

*$U$ and $H$ be separable Hilbert spaces

*$Q\in\mathfrak L(U)$ be nonnegative and symmetric with finite trace

*$U_0:=Q^{1/2}U$

*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$(W_t)_{t\ge 0}$ be a $Q$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$

*$X_0$ be a $H$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$

*$v:[0,\infty)\times H\to H$ be continuously Fréchet differentiable with respect to the first argument and twice continuously Fréchet differentiable with respect to the second argument

*$\xi:\Omega\times[0,\infty)\times H\to\operatorname{HS}(U_0,H)$ suitable for the following



Assuming that $$X_t=X_0+\int_0^tv_s(X_s)\;{\rm d}s+\int_0^t\xi_s(X_s)\;{\rm d}W_s\;\;\;\text{for }t>0\;,\tag 1$$ I would like to use an Itō-like formula to find an expression for $$Y_t:=v_t(X_t)\;\;\;\text{for }t\ge 0\;.\tag 2$$ However, the Itō formula (see Da Prato, Theorem 4.32) for $(1)$ can only be used to find an expression for $f_t(X_t)$, if $f$ is a continuously partially Fréchet differentiable mapping $[0,\infty)\times H\to\color{red}{\mathbb R}$.

Since I want to derive a SPDE for $Y$, I'm unsure what I need to do. Maybe we can do the following: Let $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ and $$v^{(n)}_t(x):=\langle v_t(x),e_n\rangle_H\;\;\;\text{for }n\in\mathbb N\;.$$ Then, each $v^{(n)}$ is continuously partially Fréchet differentiable and we can apply the Itō formula to find an expression for $$Y^{(n)}_t:=v_t^{(n)}(X_t)\;\;\;\text{for }t\ge 0\;.$$ Is this a good idea? We should be able to obtain $(2)$ by $$Y=\sum_{n\in\mathbb N}Y^{(n)}e_n\;.$$ Later, I'm interested in numerically obtaining $Y$.

EDIT: By the referenced version of the Itō formula, we obtain 
\begin{equation}
\begin{split}
\langle v(t,X_t),e_n\rangle_H&=\langle v(0,X_0),e_n\rangle_H+\int_0^t\langle{\rm D}u(s,X_s)(\xi_s(X_s){\rm d}W_s),e_n\rangle_H\\
&+\int_0^t\langle\frac{\partial u}{\partial t}(s,X_s),e_n\rangle_H+\langle{\rm D}u(s,X_s)(u_s(X_s)),e_n\rangle_H\\
&+\frac 12\operatorname{tr}\langle{\rm D}^2u(s,X_s)\left(\tilde\xi_s(X_s)\tilde\xi_s^\ast(X_s)\right),e_n\rangle_H{\rm d}s
\end{split}\tag 3
\end{equation}
where $\tilde\xi:=\xi Q^{1/2}$. I've got two questions:


*

*Can we write the infinite system of equations $(3)$ as one equation, as it is possible in the case $H=\mathbb R^d$ and $e_n=n\text{-th standard basis vector}$? It seems like that's the case, cause almost all terms are simply projections to the $n$-th basis vector, but I don't know how I need to deal with the trace term.

*In my real application, I have another known expression for $v(t,X_t)$. Thus, my SPDE would be obtained by equating the SPDE of the question with that other expression. Now the question is: I want to solve that SPDE numerically. Is there any recommended method for such a type of equation? [I should note that I will consider something like $H=[L^2(\mathcal V)]^3$ or $H=[H_0^1(\mathcal V)]^3$ for some bounded domain $\mathcal V\subseteq\mathbb R^3$].

 A: This longer version of my comment addresses only the first question.
Yes, you can do that. Let, for simplicity, $v$ be independent of $t$, i.e. $v\colon H\to H$, and let it be twice continuously differentiable with bounded derivatives. Then $D^2 v(x)$ can be thought as a bilinear map $H\times H\mapsto H$. On the other hand, $\xi\xi^*(x)$ can be regarded as a trace class operator on $H$. In turn, the space of trace class operators is the closure of $H\otimes H$ with respect to the trace norm. Hence, the map $H\otimes H\mapsto H$ which sends $h\otimes g$ into $D^2v(x) (h,g)$ can be extended to the space of trace class operators by continuity; denote this by $\operatorname{tr}_H(D^2v(x)\cdot)$. This makes the writing $\operatorname{tr}_H\big(D^2v(X_s) \xi Q\xi^*(X_s)\big)$ meaningful as an element of $H$.
So you can write
$$
v(X_t) = v(X_0) + \int_0^t Dv(X_s)\Big(v(X_s)\mathrm{d}s + \xi_s(X_s) \mathrm{d}W_s\Big) \\
+ \frac12 \int_0^t \operatorname{tr}_H\big(D^2v(X_s) \xi Q\xi^*(X_s)\big)\mathrm{d}s.
$$
Long story short, it is just a matter of making some notation to write this in a coordinate-free form on $H$.
