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

• Yes. Just Taylor expand the function of one real variable $g(r)=f(\overline{u}+r u),$ where $r\in\mathbb R$. – Giuseppe Negro Jan 10 at 13:56
• I am not really sure why this should be the same, since the second Gâteaux dirivative f'' could be discontinuosly. – Bara Jan 10 at 14:07
• 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. – Giuseppe Negro Jan 11 at 14:37

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 $$f''(t) = 2$$.