Let $F: X \to Y$ be continuously differentiable around $\bar{x} \in X$. where $X$ and $Y$ are Banach normed spaces. Then prove that F can be represented as follow
$$ F(y) = F(x) + \nabla F(\bar{x}) (y -x) + o(y-x)$$
where $\lim_{(x,y) \to (\bar{x} , \bar{x} )} \frac{o(y-x)}{\|y-x \|} = 0$
In another word I want to prove that
$$ \lim_{(x,y) \to (\bar{x} , \bar{x} )} \frac{F(y) - F(x) + \nabla F(\bar{x}) (y -x) } {\|y-x\|} =0 $$
Yes, this is true. The property you are asking for is known as 'strict differentiability'.
It can be proven as follows. We define $$g(x) := f(x) - f'(\bar x)(x - \bar x).$$ From the mean value theorem (Necessity of the Hahn Banach Theorem for the Gateaux Mean Value Theorem) it follows \begin{align*} \frac{\| f(x) - f(y) - f'(\bar x)(x-y)\|}{\|x-y\|}&= \frac{\| g(x) - g(y) \|}{\|x-y\|}\\ &\le \sup_{t \in [0,1]} \| g'(x + t \, (y-x))\| = \sup_{t \in [0,1]} \| f'(x + t \, (y-x)) - f'(\bar x)\| \to 0 \end{align*} as $x,y \to \bar x$.
