If $f:R^n\to R$ is differentiable,then $df=D_1f.dx^1+...+D_nf.dx^n$

Question

From Calculus on Manifold Spivak,chapter 4.

If $f:R^n\to R$ is differentiable,then $df=D_1f.dx^1+...+D_nf.dx^n$

Proof:

$df(p)(v_p)$

=$Df(p)(v)$

=$\sum_{i=1}^{n}v^i.D_if(p)$

=$\sum_{i=1}^{n}dx^i(p)(v_p).D_if(p)$

I understood the proof,but my question is how from $df(p)(v_p)=\sum_{i=1}^{n}dx^i(p)(v_p).D_if(p)$ ,we conclude $df=D_1f.dx^1+...+D_nf.dx^n$.My question may be very basic. please help.

Answer 1

If $f$ and $g$ are two functions with $f(x)=g(x)$ for all $x$ then $f=g$. In the case at hand we have $$df(p).v=\sum_{i=1}^n D_if(p)\> dx^i.v\quad\forall v\in T_p\ .$$ It follows that the maps $$df(p):\ T_p\to{\mathbb R},\qquad \sum_{i=1}^n D_if(p)\>dx^i :\ T_p\to{\mathbb R}$$ are equal. Here the $D_if(p)={\partial f\over\partial x_i}(p)$ are numbers.