Necessary and sufficient condition differentiability

Question

I need to prove the equivalence that $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$ is differentiable at a given point $a$ if and only if, for every $h$, there exists a linear transformation $A(h)$ such that $$f(a+h)-f(a)=A(h)h$$with $A$ continuous at $0$. For the forward I tried putting $A(h)$ such that $A(h)h=df(a)h+r(h)$ (definition of derivative) put I'm having a hard time proving that is continuous at $0$... Any tips?

Answer 1

I found this answer: For the implication:
Define $A(0) = df(a)$ and $A(h)h = df(a)h +r(h)$. Then $r(h) = (A(h)-A(0))h$, so that $$0=\lim_{h\rightarrow 0}\frac{|r(h)|}{|h|}=\lim_{h\rightarrow0}\frac{|(A(h)-A(0))h|}{|h|}=\lim_{h\rightarrow0}\left|(A(h)-A(0))\frac{h}{|h|}\right|\iff ?A(h)\rightarrow A(0)$$ The converse is easier:
We have that: $$f(a+h)-f(a)= A(0)h+(A(h)-A(0))h$$ Notice that $A(0)h$ is linear with $h$, it remains to show that the rest of the RHS tends to zero "faster than $h$". In fact: $$0\leq\frac{|(A(h)-A(0))h|}{|h|}\leq||(A(h)-A(0))||$$ Now clearly $||A(h)-A(0)||\rightarrow 0$ since $A$ is continuous there, so the result follows.