3

Let $U \subset \mathbb{R}^n$, $V \subset \mathbb{R}^k$ be open, $f\colon U \to V$ a smooth map, $x \in U$. For $h \in \mathbb{R}^n$, let $$ d_x(h) := \lim_{t \to 0} \frac{f(x+ th) - f(x)}{t} $$ Now, in "Topology from the differential viewpoint", Milnor writes that clearly, $d_x$ is a linear function of $h$. Why is that so? In other words, how could one see that $d_x$ is simply the Jacobian matrix?

Vincent
  • 2,064
  • I guess the author assumes you know about differentiability between vector spaces and so, the map is clearly linear as it is a derivative between such spaces. – William M. May 02 '17 at 20:16
  • http://teachingdm.unito.it/paginepersonali/sergio.console/Dispense/Milnor%20Topology%20from%20%23681EA.pdf – sav May 02 '17 at 20:20
  • It's straightforward; just compute $d_x(h_1 + h_2)$ and $d_x(ah)$. –  May 02 '17 at 21:24
  • @OpenBall This would solve the problem but it is not straight forward – sav May 02 '17 at 22:13

1 Answers1

1

Define $c_x(t) = x + th$ so $t \mapsto f(c(t))$ is differentiable at zero with derivative (by Chain Rule) $d_x(h) = f'(x)c'(0)=f'(x)h...$ linear in $h.$

William M.
  • 7,532