Remark: Even when this question is about stochastic PDEs, it can be answered by someone who has no knowledge about probability theory or PDEs.

I'm reading Stochastic Differential Equations in Infinite Dimensions and try to understand the relationship between the Itō formula for a Q-Wiener process and the Itō formula for a cylindrical Wiener process.

Let me introduce the necessary objects and please stay with me even when it's a bit lengthy:

Let

- $U$ be a real separable Hilbert space
- $Q\in\mathfrak L(U)$ be nonnegative and symmetric
- $U_0:=Q^{1/2}U$ be equipped with $$\langle u,v\rangle_{U_0}:=\langle Q^{-1/2}u,Q^{-1/2}v\rangle_U\;\;\;\text{for }u,v\in U_0$$
- $V$ be a real separable Hilbert space and $\iota\in\operatorname{HS}(U_0,V)$ be an embedding
- $C:=\iota\iota^\ast$

It's easy to show that $C$ is a symmetric and nonnegative element of $\mathfrak L(V)$ with finite trace.

Let

- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $\left\{B^{(n)}:n\in\mathbb N\right\}$ be a family of independent real Brownian motions on $(\Omega,\mathcal A,\operatorname P)$
- $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U_0$

Then, $$W^{(n)}:=\sum_{i=1}^nB^{(i)}\iota e_i\;\;\;\text{for }n\in\mathbb N$$ converges in $L^2\left(\operatorname P,C^0\left(\left[0,\infty\right),V\right)\right)$ and its limit $W$ is a $C$-Wiener process on $(\Omega,\mathcal A,\operatorname P)$.

$\color{red}{\text{Warning:}}$ $W$ is called a **cylindrical $C$-Wiener process** on $(\Omega,\mathcal A,\operatorname P)$ by many authors. My reference is A Concise Course on Stochastic Partial Differential Equations where you can find the definition in Proposition 2.5.2.

However, the authors of the book mentioned at the beginning of the question use a different definition of a cylindrical Wiener process and, of course, they use this definition in their statement of the corresponding Itō formula.

**The whole reason for this question is**: I want to use the definition of $W$ given here, apply the Itō formula for a $C$-Wiener process and hope that I end with a formula which is equivalent to the Itō formula for a cylindrical Wiener process. That should be the case, right?

Let

- $V_0:=C^{1/2}V=\iota U_0$ (and note that $\iota$ is an isometry between $U_0$ and $V_0$)
- $\Phi:\Omega\times[0,\infty)\to\operatorname{HS}(V_0,H)$ and $\varphi:\Omega\times[0,\infty)\to H$
- $f:[0,\infty)\times H\to H$, ${\rm D}^2f:[0,\infty)\times H\to\mathfrak L\left(H,\mathfrak L\left(H\right)\right)$ denote the second Fréchet derivative of $f$ with respect to the second variable and $$f^{(n)}:=\langle f,x_n\rangle_H\;\;\;\text{for }n\in\mathbb N$$ for some orthonormal basis $(x_n)_{n\in\mathbb N}$ of $H$

We consider the equation $${\rm d}X_t=\varphi_t{\rm d}t+\Phi_t{\rm d}W_t\;\;\;\text{for all }t\ge 0\;.$$

The crucial part is the trace term in the Itō formula, which we want to apply to find an expression for ${\rm d}f^{(n)}(t,X_t)$. It is equal to $$\operatorname{tr}\left[{\rm D}^2f^{(n)}(t,X_t)\left(\Phi_tC^{\frac 12}\right)\left(\Phi_tC^{\frac 12}\right)^\ast\right]=\langle\sum_{k\in\mathbb N}{\rm D}^2f(t,X_t)\left(\Phi_tC^{\frac 12}v_k\right)\left(\Phi_tC^{\frac 12}v_k\right),x_n\rangle_H\;,\tag 1$$ where $(v_k)_{k\in\mathbb N}$ is any orthonormal basis of $V$ and ${\rm D}^2f^{(n)}(t,X_t)$ on the left-hand side of $(1)$ is considered as an element of $\mathfrak L(H)$ (which is possible by Riesz' representation theorem).

**Question:** Is the right-hand side of $(1)$ actually equal to the trace term in the mentioned Itō formula for a cylindrical Wiener process? Maybe we can further simplify the right-hand side. I'm not sure how I can use the fact that $C^{1/2}=C$ is the orthogonal projection. I could imagine that the sum on the right-hand side is equal to $$\sum_{k\in\mathbb N}{\rm D}^2f(t,X_t)\left(\Phi_t\tilde v_k\right)\left(\Phi_t\tilde v_k\right)\;,$$ where $(\tilde v_k)_{k\in\mathbb N}$ is any orthonormal basis of $V_0$.

$\mathfrak L(A,B)$ and $\operatorname{HS}(A,B)$ denote the space of bounded, linear operators and Hilbert-Schmidt operators from $A$ to $B$, respectively. Moreover, $\mathfrak L(A):=\mathfrak L(A,A)$ and $L^\ast$ denotes the adjoint of a bounded, linear operator $L$.