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 $U$.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.