(Generalization) Holomorphic vector-valued functions

Question

I am trying to generalize this result: Holomorphic Banach-valued functions and composition to the unit ball of Hilbert space $E$. With @EricWofsey's help I proved the following:

Let $X$ be a complex Banach space and let $f\in\mathcal{H}(\mathbb{D},X)$. Then there exists a function $F\in\mathcal{H}(\mathbb{D},X)$ with $F(0)=0$ such that $F'=f$.

Proof. Define $F\colon\mathbb{D}\to X$ by $$ F(z)=\int_0^z f(s)\ ds\qquad \forall z\in\mathbb{D}, $$ where we I am using the Bochner integral of $f\colon\mathbb{D}\to X$ along some path in $\mathbb{D}$ from $0$ to $z$. It is easy to prove that $F$ does not depend on the path chosen by classical Cauchy's theorem for star-shaped domains, the properties of the Bochner integral and since $X^*$ separates points of $X$.

$F(0)=0$ and given $z\in\mathbb{D}$ and $\phi\in X^*$, we have
$$ (\phi\circ F)(z)=\phi\left(\int_0^z f(s)\ ds\right)=\int_0^z (\phi\circ f)(s)\ ds, $$ and again Cauchy's theorem for star-shaped domains gives that $F'=f$. $\hspace{17cm} \blacksquare$

Now, I was wondering if the preceding result could be extended as follows:

Let $X$ be a complex Banach space, $E$ a Hilbert space and let $f\in\mathcal{H}(B_E,L(E,X))$. Then there exists a function $F\in\mathcal{H}(B_E,X)$ with $F(0)=0$ such that $DF=f$.

I think that the definition of $F: B_E \to X$ must be changed. Maybe by $$F(z) = \int_{A_z} f(s) \; ds, \qquad \forall z \in B_E,$$

with $A_z$ a subset of $B_E$ satisfying some properties? In this case, if we use the Bochner integral (or the Gelfand-Pettis integral), it is necessary to show that $F$ does not depend of the subset selected, but I don't know how to do it because I don't have some similar result of Cauchy's theorem for star-shaped domains in $B_E$ (and $E$ could be infinite-dimensional). Moreover, I think that prove $DF=f$ would not be easy.

Answer 1

No, this is not true in general. This is just the fact that unlike ODEs, PDEs do not generally have solutions. For instance, consider the simplest case where you are looking for a function $\mathbb{C}^2\to\mathbb{C}$ which has a prescribed derivative at each point. In other words, you are prescribing the partial derivatives $f$ and $g$ with respect to each variable of your function $\mathbb{C}^2\to\mathbb{C}$. But because mixed partial derivatives are equal, this is not possible unless $\partial f/\partial z_2=\partial g/\partial z_1$.

When $E$ is finite dimensional, at least, then equality of the mixed partial derivatives is a sufficient condition for the existence of an antiderivative. That is, suppose $B\subset\mathbb{C}^n$ is a polydisk and $f_1,\dots,f_n:B\to X$ are holomorphic such that $D_jf_i=D_if_j$ for all $i,j$, where $D_i$ is the partial derivative with respect to the $i$th variable. Then there exists $F:B\to X$ holomorphic such that $D_iF=f_i$ for each $i$. To prove this, start by integrating $f_1$ with respect to the first variable to get $g$ such that $D_1g=f_1$. Then for each $i\neq 1$, $D_1(f_i-D_ig)=D_1f_i-D_if_1=0$. That is, $f_i-D_ig$ is constant with respect to the first variable. By induction, then, we can find a function $h$ depending on only the last $n-1$ variables such that $D_ih=f_i-D_ig$ for each $i\neq 1$. The function $F=g+h$ then has $D_if=f_i$ for all $i$.

I don't know whether equality of the mixed partials is sufficient when $E$ is infinite-dimensional.