I know that, if $F:X\to Y$, where $X,Y$ are Banach spaces, is a map whose $n$-th Fréchet derivative $x\mapsto F^{(n)}(x)$ is continuous as a function of $x$ in a neighbourhood of $x_0\in X$, then the Taylor formula of $F$ in $x_0$ holds:$$F(x_0+h)=F(x_0)+F'(x_0)h+\frac{1}{2!}F''(x)(h,h)+...+\frac{1}{n!}F^{(n)}(x_0)(h,...,h)+\omega(x,h)$$where $\|\omega(x,h)\|=o(\|h\|^n),h\to 0$ and $F^{(n)}(x_0)$ is the $n$-linear form corresponding to the $n$-th derivative.
I have never see the condition of continuity of $F^{(n)}$ in $x_0$ relaxed either in the general case or when $X=\mathbb{R}^n$, $Y=\mathbb{R}$ (contrarily tom what happens when $X=\mathbb{R}=Y$, when De l'Hôpital's rule can be used and the assumption of the continuity of $F^{(n)}$ avoided). Can that assumption be relaxed?
I am asking that because Kolmogorov-Fomin's Элементы теории функций и функционального анализа proves the Taylor expansion under the continuity assumption (p. 491 here), but then proves the following theorem, which has the Hessian matrix test as a particular case, without saying that $F''$ must be continuous by using the Taylor expansion $F(x_0+h)=F(x_0)+F'(x_0)h+\frac{1}{2!}F''(x)(h,h)$ $+o(\|h\|^2)$: if functional $F:X\to\mathbb{R}$ where $X$ is a Banach space and (1) $F'(x_0)=0$ and (2) $F''(x_0)$ is strongly positive, i.e. $\exists c>0:\forall h\in X\quad F''(x_0)(h,h)\ge c\|h\|^2$, then $F$ has a minimum in $x_0$.