Gateaux differentiability of univariate function? (Chain rule is true for Gateaux differentiable function)

Let the function $$f: U \subseteq \mathbb{R}^n \rightarrow \mathbb{R}$$ be G-differentiable on open set $$U$$. Let $$x,y \in U$$ be such that the line segment $$[x,y] \subset U$$. Define the function $$g : [0,1]\rightarrow \mathbb{R}$$ as $$g(t):=f(x+t(y-x))$$ Show that for any $$t_* \in [0,1]$$, $$g'(t_*) = \langle \nabla f(x+t_*(y-x)),y-x\rangle$$. Simply, it says, that when the function is Gateaux differentiable chain rule can only be applied on the line segment that is laid onto $$U$$.

Remark:

When the function is G-differentiable at point $$x$$, directional derivative $$f'(x;d)$$ exists for all $$d \in \mathbb{R}^n$$, and $$f'(x;.)$$ is linear in the second argument, i.e., for all $$d,d' \in \mathbb{R}^n$$ and $$\alpha,\beta \in \mathbb{R}$$, $$f'(x;\alpha d + \beta d') =\alpha f'(x; d )+\beta f'(x; d' )$$

Let $$e_1,\ldots, e_n$$ the (standard) orthonormal basis of $$\Bbb R^n$$, and $$v\in\Bbb R^n$$ such that $$v:=v_1e_1+\ldots+v_ne_n$$ where the $$v_k\in\Bbb R$$, then from the remark you quoted you have that

$$f'(x,v)=\sum_{k=1}^n v_k f'(x,e_k)=\langle \nabla f(x),v\rangle\tag1$$

And for $$v:=y-x$$ we have that

\begin{align}g'(t_0)&=\lim_{h\to 0}\frac{g(t_0+h)-g(t_0)}h\\&=\lim_{h\to 0}\frac{f(x+t_0v+hv)-f(x+t_0v)}{h}\\&=f'(x+t_0v,v)\end{align}\tag2

Then from $$(1)$$ and $$(2)$$ together we have the desired identity.

• Can you help me to understand why we need derivative in just linear direction when we have Gateaux differentiability? The statement says that for any two point in the $U$ and all points on the line joining them, derivative exists so we have the whole $U$ which is differentiable. In such a situation, can we have a point where we have G-diff but we don't have F-diff? – Saeed Nov 9 '18 at 14:45
• use the search in this site, you have a lot of examples an discussions about that, by example: math.stackexchange.com/… – Masacroso Nov 9 '18 at 18:42
• Appreciate that. – Saeed Nov 9 '18 at 20:22