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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 $\alpha,\beta \in \mathbb{R}$, $$ f'(x;\alpha d + \beta d') =\alpha f'(x; d )+\beta f'(x; d' ) $$

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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.

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  • $\begingroup$ 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? $\endgroup$ – Saeed Nov 9 '18 at 14:45
  • $\begingroup$ use the search in this site, you have a lot of examples an discussions about that, by example: math.stackexchange.com/… $\endgroup$ – Masacroso Nov 9 '18 at 18:42
  • $\begingroup$ Appreciate that. $\endgroup$ – Saeed Nov 9 '18 at 20:22

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