Taylor expansion for Gâteaux derivative Let $\mathbb{X}$ be a normed Space and 
$
f: \mathbb{X}\mapsto\mathbb{R}
$ 
is twice Gâteaux differentiable (not necessary Fréchet differentiable). Is it possible to build a Taylorexpansion for $f$ in the following sense
$
f(u)= f(\bar{u}) + f'(\bar{u})(u-\bar{u})+\frac{1}{2}f
''(\bar{u}+\theta(u-\bar{u}))(u-\bar{u})^2,
$
with $\theta\in(0,1)$?
 A: Yes, this holds. It is basically a consequence of Darboux's theorem and the mean value theorem.
As indicated by Giuseppe Negro, it is sufficient to discuss the case of real arguments. Further, by simple transformation, it is sufficient to consider the situation
$$f(0) = f'(0) = 0, \quad f(1) = 1$$
and we will show the existence of $t \in (0,1)$ with $f''(t) = 2$.
First, we claim that there exists $t_1 \in (0,1)$ with $f''(t_1) \ge 2$.
We argue by contradiction and assume $f''(t) < 2$ for all $t \in (0,1)$.
The function $f'$ is differentiable, hence it has the mean value property. For every $t \in (0,1)$ we have $\hat t \in (0,t)$ with
$$ \frac{f'(t)-f'(0)}{t} = f''(\hat t) < 2,$$
i.e., $f'(t) < 2 \, t$. Now, the fundamental theorem of calculus yields
$$f(1) - f(0) = \int_0^1 f'(t) \, \mathrm{d}t < \int_0^1 2 \, t \,\mathrm{d}t < 1$$
which is a contradiction. This shows the existence of $t_1$.
Similarly, we can show the existence of $t_2 \in (0,1)$ with $f''(t_2) \le 2$.
If $f''(t_1) = 2$ or $f''(t_2) = 2$, we are finished. Otherwise,
$f''(t_1) > 2$ and $f''(t_2) < 2$. Hence, Darboux's theorem implies the existence of $t$ between $t_1$ and $t_2$ such that $f''(t) = 2$.
