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Let $X,Y$ be Banach spaces over $\mathbb{C}$ and let $U \subset X$ be open. If $f:U \to Y$ is Fréchet differentiable at every point of $U$, can we locally expand $f$ as a "power series"?

To be more precise (and I hope I'm making this precise in the right way), given $x \in U$ must there exist $A_0,A_1,A_2,\ldots$ with $A_n$ a bounded, symmetric $n$-linear map $X^n \to Y$ such that the expansion $ f(y) = \sum_{n = 0}^\infty A_n (y-x)^n $ is valid for all $y$ in some neighbourhood of $x$? To explain the above notation:

  • Let $X^0 := \mathbb{C}$ by definition so that $A_0$ is essentially an element of $Y$ (namely $f(x)$).
  • Let $(y-x)^n : = (y-x,y-x,\ldots,y-x) \in X^n$ for $n \geq 1$ and let $(y-x)^0 := 1 \in \mathbb{C}$.

Wikipedia's entry on infinite dimensional holomorphy seems to indicate the answer is yes. I found it pretty incredible that a classical result should hold at this level of generality so I checked out the reference they give which is Kadison & Ringrose Fundamentals of the theory of operator algebras, vol. 1, sect. 3.3 but it turns out that only the case $X = \mathbb{C}$ is treated — pretty unsatisfactory!

I'd be interested in to know a) whether Wikipedia has this right, even though the reference is faulty, and b) how difficult is this to prove?

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Yes, this is true, you should read the book J. Mujica, Complex analysis in Banach Spaces, Elsevier, 1986. In particular Theorem 13.16 on page 107.

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    Ok thanks, I will try to track down a copy.2011-10-02