Fréchet differentiability from Gâteaux differentiability Let $X$ be a Banach space and $\Omega \subset X$ be open.

The functional $f$ has a Gâteaux derivative $g \in X'$ at $u \in \Omega$ if, $\forall h\in X,$ $$\lim_{t \rightarrow 0}[f(u+th)-f(u)- \langle g,th \rangle]=0$$ 

How can I prove the following:

If $f$ has a continuous Gâteaux derivative on $\Omega$, then $f \in C^1(\Omega,\mathbb R)$.
It seems worth expanding user31373's answer above for the benefit of those who could come to this page by chance (as I did). 

The Mean Value theorem discussed above is the following statement: 

Let $f : X \to Y$ be a map of normed linear spaces. Given $x_0 \in X$ and a direction $h \in X$ such that the directional derivative $\delta f(x_0+th)(h)$ exists for all $t \in [0,1],$ we have that
$$
\|f(x_0+h)-f(x_0)\| \leqslant \|\delta f(x_0+th)(h)\| 
$$
for a suitable $t \in [0,1]$ 


(P.Drabek, J.Milota, Methods of Nonlinear Analysis, 2007; Th. 3.2.7).  It follows that 
\begin{equation} \tag{$*$}
\|f(x_0+h)-f(x_0)-\delta f(x_0)h\| \leqslant \|\delta f(x_0+th)(h)-\delta f(x_0)(h)\| 
\end{equation}
(the latter statement is left without the proof in Drabek--Milota, and, unfortunately, the hint that is given there is misleading; however the book is one of the best sources on Fr&#233;chet and G&#226;teaux differentiation). In fact, one can use the idea from user31373's answer. We consider the map $g : X \to Y$ such that
$$
g(x)=f(x)-\delta f(x_0)(x-x_0)
$$
for all $x \in X.$ It is easy to see that the Mean Value theorem applied to $g$ gives us $(*).$ Let us verify, for instance, the key fact of existense of the directional derivatives $\delta g(x_0+th)(h).$ First, recall that the G&#226;teaux differential preserves the scalar multiplication: $\delta f(x)(r h)=r \delta f(x)(h)$ for all $x,h \in X$ and for all $r \in \mathbf R$ (the homogeneity of $\delta f(x)$). We have that
\begin{eqnarray}
\delta g(x_0+th)(h) &amp;=&amp; \lim_{u \to 0} \frac{g(x_0+th+uh)-g(x_0+th)}u  \\
&amp;=&amp;\lim_{u \to 0} \frac{f(x_0+th+uh)-\delta f(x_0)( (t+u)h) - f(x_0+th)+\delta f(x_0)(th)}u  \\
&amp;=&amp;\delta f(x_0+th)(h)-\delta f(x_0)(h)
\end{eqnarray}
where we have used the homogeneity to justify the last equation.

Now to address the question under what conditions existense of G&#226;teaux differentials implies existense of Fr&#233;chet differential. The standard result is the following one (quoting again Drabek--Milota): 

if $f : X \to Y$ is a map of linear normed spaces such that in a certain open neighbourhood $O_{x_0}$ of $x_0 \in X$ the G&#226;teaux differential $\delta f(x)$ exists and is linear and continuous for all $x \in O_{x_0}$ and the map
$$
x \mapsto \delta f(x)
$$
from $X$ into $L(X,Y) $ is continuous at $x_0,$ then $f$ is Fr&#233;chet differentiable at $x_0$  

(Prop. 3.2.15 in Drabek--Milota); $L(X,Y)$ is the normed space of continuous linear maps from $X$ to $Y$ equipped the uniform norm
$$
\|\varphi\| =\mathrm{sup}_{x \ne 0} \frac{\|\varphi(x)\|}{\|x\|} \qquad (\varphi \in L(X,Y)).
$$
The proof is based on the equality $(*)$ above.
If $\phi$ is continuous at $x_0$ and ∂ϕ(x0) contains a singleton element,
then ϕ is Gâteaux differentiable at x0 and ϕ(x0) = ∂ϕ(x0).