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I would like to known if the following converse of the taylor's theorem is true:

Let $E$, $F$ Banach spaces, and $f:E\rightarrow F$ continuous. Suppose there are $k$ continuous functions $T_i: E \rightarrow L_i(E,F)$ such that for every $x \in E$ $$ f(x+h) = f(x) + T_1(x)h + T_2(x)h^2 + ... + T_k(x)h^k + r(x,h), $$ with $\lim_{h\to 0} \frac{r(x,h)}{\|h\|^k} = 0$. Then $f$ is $C^k$ and $f^{(k)}(x) = k! T_k(x)$

I found an analogue result with the same hypotesis on S. Dayal, "A converse of Taylor's Theorem on Banach Spaces", Proc. of Amercian Math. Soc., vol 68, but the conclusion is just that $f$ has Frechet derivative of order $k$.

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1 Answer 1

Hey, sorry to answer my own question but... I found an answer on F. Albrecht and H. Diamond, "A Converse of Taylor's Theorem", Indiana Univ. Math. J. 21 No. 4.

The version I posted is not true in general. The problem is the non uniform limit of $\frac{r(x,h)}{\|h\|^k}$ . If we suppose $\lim_{(x,h) \to (x_0,0)} \frac{r(x,h)}{\|h\|^k} = 0$, then the conclusion is true.

The refferences on mathscinet are MR0448394 (56 #6701) and MR0320744 (47 #9278)

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When $f$ is defined on a closed subset of $E=\mathbb{R}^n$, and $F=\mathbb{R}$, this result is known as Whitney's extension theorem, proved in 1934: jstor.org/stable/1989708 –  Margaret Friedland Feb 29 '12 at 15:57

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