Similar question to this one but I could not understand the answer.
In Rudin, Principles of Mathematical Analysis, theorem $9.21, Rudin provides a proof of this result:
$f: E (\subseteq \mathbb R^n) \rightarrow \mathbb R^m$ is continuously differentiable, if and only if, it's partial derivatives exist and are continuous on $E$.
The $\implies$ way is no problem. But in the $\impliedby$ way, Rudin does the proof considering just the $m=1$ case.
Here is my attempt at understanding why Rudin only treats the $m=1$ case:
Let $f_1 ,..., f_m$ be the components of $f$, so that $f= (f_1, ... , f_m)$
Knowing that all the partial derivatives of $f$ exist and are continuous, we can conclude that $\forall i, 1\leq i \leq m , f_i$ is continuously differentiable.
Now remains to prove that:
- $\forall i, 1\leq i \leq m , f_i$ differentiable $\implies$ $f$ differentiable
- $\forall i, 1\leq i \leq m , f_i'$ continuous $\implies$ $f'$ continuous.
I've tried a bit to prove the second bullet, but I am not sure if $f'= (f_1 ', ..., f_m ')$ hold in multivariable calculus? My idea would be to use $|f'(x)-f'(y)| = \sqrt{\sum |f_i '(x) - f_i '(y)|^2}$
Is my approach correct ?
Thanks In Advance.
$\frac{|f(x+h)-f(x)-(f_1'(h),\dots,f_m'(h))|}{|h|}=\frac{|((f_1(x+h)-f_1(x)-f_1'(h)),\dots,(f_m(x+h)-f_m(x)-f_m'(h)) )|}{|h|}\le \sum\limits_{j=1}^m \frac{|f_j(x+h)-f_j(x)-f_j'(h)|}{|h|}<\sum\limits_{j=1}^m \frac{\epsilon}{m}$
(Relevant derivatives at $x$, evaluated at $h$)