Let
- $T>0$
- $I:=(0,T]$
- $\mathcal B(I)$ denote the Borel $\sigma$-algebra on $I$
- $E$ be a $\mathbb R$-Banach space
- $\mu:\mathcal B(I)\to E$ be a vector measure
- $X:I\to E$ be bounded and $(\mathcal B(I),\mathcal B(E))$-measurable
Now, let $$f:E\to E\:\hat\otimes_\pi\:E\;,\;\;\;x\mapsto x\otimes x.$$ I want to show that $$\frac12\mu((0,t])=\int_0^t{\rm D}^2f(X)\:{\rm d}\mu\tag1\;\;\;\text{for all }t\in I.$$
If I didn't made any mistake, we should have $${\rm D}f(x)y=y\otimes x+x\otimes y\;\;\;\text{for all }x,y\in E\tag2$$ and $$({\rm D}^2f(x)y)z=y\otimes y+y\otimes y\;\;\;\text{for all }x,y,z\in E\tag3.$$ However, if $u=\sum_{n=1}^\infty y_n\otimes z_n\in E\:\hat\otimes_\pi\:E$, then (identifying ${\rm D}f(x)\in\mathfrak L(E,\mathfrak L(E,E\:\hat\otimes_\pi\:E))$ with an element of $\mathfrak L(E\:\hat\otimes_\pi\:E)$ on the left-hand side) $${\rm D}f(x)u=\sum_{n=1}^\infty({\rm D}^2f(x)y_n)z_n\tag4$$ and that doesn't seem to yield $(1)$.
So, what am I doing wrong and how can we show $(1)$?