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 $\theta\in(0,1)$?

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    $\begingroup$ Yes. Just Taylor expand the function of one real variable $g(r)=f(\overline{u}+r u), $ where $r\in\mathbb R$. $\endgroup$ – Giuseppe Negro Jan 10 at 13:56
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    $\begingroup$ I am not really sure why this should be the same, since the second Gâteaux dirivative f'' could be discontinuosly. $\endgroup$ – Bara Jan 10 at 14:07
  • $\begingroup$ That could be a problem even if X is $\mathbb R$. My point is that this general case is actually exactly the same as the one-variable case. $\endgroup$ – Giuseppe Negro Jan 11 at 14:37

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 $f''(t) = 2$.


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