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I'm reading Stochastic Differential Equations in Infinite Dimensions and don't understand what the authors do in Chapter 2.3.1. Let me introduce the necessary objects: Let

  • $K$ and $H$ be real Hilbert spaces
  • $Q\in\mathfrak L(K)$ be nonnegative and symmetric
  • $K_Q:=Q^{1/2}K$
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\Phi:\Omega\times[0,\infty)\to\operatorname{HS}(K_Q,H)$ and $\varphi:\Omega\times[0,\infty)\to H$
  • $F:[0,\infty)\times H\to\mathbb R$ and $F_{xx}$ be the second Fréchet derivative of $F$ with respect to the second variable

I don't understand the term $$\operatorname{tr}\left[F_{xx}(t,x)\left(\Phi_tQ^{\frac 12}\right)\left(\Phi_tQ^{\frac 12}\right)^\ast\right]\tag 1$$ which occurs in equation (2.53).

By definition, $F_{xx}$ is an element of $\mathfrak L(H,\mathfrak L(H,\mathbb R))$. However, the authors obviously use the fact that $\mathfrak L(H,\mathbb R)\cong H$ and hence $F_{xx}$ can be identified with an element of $\mathfrak L(H)$. With this interpretation, we've got $$\underbrace{F_{xx}(t,x)}_{\in\mathfrak L(H)}\underbrace{\underbrace{\left(\Phi_tQ^{\frac 12}\right)}_{\in\mathfrak L(K_Q,H)}\underbrace{\left(\Phi_tQ^{\frac 12}\right)^\ast}_{\in\mathfrak L(H,K_Q)}}_{\in\mathfrak L(H)}\in\mathfrak L(H)\;.$$ Thus, at least it makes sense to talk about the trace of this expression. However, can we rewrite the expression $(1)$ without the identification?


Above $\mathfrak L(A,B)$ 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 $L$.

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1 Answer 1

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The answer is: Yes, we can.

It's not important that we talk about real Hilbert spaces. So, let's replace $\mathbb R$ by $\mathbb K\in\left\{\mathbb C,\mathbb R\right\}$ in the question.

Let

  • $(t,x)\in[0,\infty)\times H$
  • $(e_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ be orthonormal bases of $K$ and $H$, respectively

Note that $$\tilde\Phi:=\Phi_tQ^\frac12\in\mathfrak L(K,H)$$ is trace class and that ${\rm D}^2F(t,x):=F_{xx}(t,x)\in\mathfrak L(H,\mathfrak L(H,\mathbb K))$ can be identified with $L\in\mathfrak L(H)$, where $$Lu:=\sum_{n\in\mathbb N}\overline{{\rm D}^2F(t,x)uf_n}f_n\;\;\;\text{for }u\in H\;.$$ It's easy to check that $$\langle v,Lu\rangle={\rm D}^2F(t,x)uv\;\;\;\text{for all }u,v\in H\;.$$ Thus, we can conclude that $$\operatorname{tr}L\tilde\Phi\tilde\Phi^\ast=\sum_{n\in\mathbb N}\langle\tilde\Phi e_n,L\tilde\Phi e_n\rangle_H=\sum_{n\in\mathbb N}{\rm D}^2F(t,x)\left(\tilde\Phi e_n\right)\left(\tilde\Phi e_n\right)\;.$$

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