Prove that continuity in x of the Gateaux derivative, $f'(x;y)$, implies Frechet differentiability 
Prove that continuity in x of the Gateaux derivative implies Frechet differentiability

Let $x$ be te point, $y$ the direction and $f'(x;y)=y·a(x)$.
First, I considere the function $g(\varepsilon)=f(x+\varepsilon y)-f(x)-\varepsilon y·a(x)$.
Then I have $g(0)=0$, $g(1)=f(x+ y)-f(x)-y·a(x)$, and $g(\varepsilon)=o(\varepsilon)$.
Note: $o(h(x))$ is a function such that $$\lim\limits_{\|y\|\rightarrow 0}\dfrac{\|o(h(x))\|}{\|h(x)\|}=0.$$
And I want to show that $$f(x+y)-f(x)-y·a(x)=o(y)$$
So I think I have to show that $g(1)=o(y)$, but I don't know how can I do that.
 A: Notice: I changed the fundamental theorem of calculus into mean value theorem, as the fundamental theorem does not work in very Banach space.
I assume $f:U\to Y$ where $U\subseteq X$ is open and $X, Y$ are real Banach spaces. Let $x\in U$.
Notice that $a(x)$ is a continuous linear operator. By assumption $a$ is continuous. Thus, $a(x+y) - a(x)$ converges to $0$ in operator norm for $y\to 0$. 
Now let $B\subseteq U$ be a sufficiently small open ball around $x$. For $y\in B - x$ let $g(t) = f(x+ty)$ for $0\le t \le 1$. Then, $g$ is  differentiable by assumption and we have $g'(t) = f'(x+ty, y) = a(x+ty)y$.
Thus, the mean value theorem applies and yields
$$ g(1) - g(0) \in \operatorname{\overline{co}}\{ g'(t) \mid 0 < t < 1 \} = \operatorname{\overline{co}}\{ a(x + ty)y \mid 0 < t < 1 \} $$
where $\operatorname{\overline{co}}$ denotes the closure of the convex hull.
Thus, we have
\begin{align} 
f(x + y) - f(x) - a(x)y \in \mathcal C := \operatorname{\overline{co}}\{ a(x + ty)y - a(x)y \mid 0 < t < 1 \}.
\end{align}
For every $t_1,\dotsc,t_n\in (0,1)$ and $\lambda_1,\dotsc,\lambda_n\in (0,1)$ with $\lambda_1+\dotsb+\lambda_n = 1$ follows
$$ \left\| \sum_{i=1}^{n} \lambda_i a(x+t_i y)y - a(x)y \right\| \le \max_{0\le t \le 1} \| a(x + t y) - a(y) \| \|y \| = \mathcal o(1) \|y\|. $$
Thus, the diameter of $\mathcal C$ is $\mathcal o(1)\|y\|$ and
$$ \| f(x + y) - f(x) - a(x)y \| = \mathcal o(1)\|y\|. $$
