Based on intuitive definitions of the differential of a function, it seems to me that for $f:\mathbb{R}^n \to \mathbb{R}^m$ something like $$ df(a) = \lim_{|r| \to 0} \frac{|f(a+r)-f(a)|}{|r|} $$ should hold for $a \in \mathbb{R}^n$, although this may be an abuse of notation. I haven't been able to prove it from the usual definition I've encountered that the differential is the linear map defined by $$ df(a)(v_1, \dots, v_n) = \left(\sum_{k=1}^n \frac{\partial f_1}{\partial x_k}(a)v_k, \dots, \sum_{k=1}^n \frac{\partial f_m}{\partial x_k}(a)v_k\right). $$ Does any such limit definition of the differential of a function exist?
My motivation for asking this is from Narasimhan's Analysis on Real and Complex Manifolds where he uses in a proof that for $g: \mathbb{R}^n \to \mathbb{R}^n$, $(dg)(0)=0$ implies that there exists a neighborhood $W$ of $0$ such that for any $x,y \in W$, $$ |g(x)-g(y)| \leq \frac{1}{2}|x-y|. $$ This would follow directly from the limit formula that I gave, but is there a more rigorous way to show this?
Edit: Of course, in the book $g$ is also specified to be continuously differentiable on an open subset $\Omega \subset \mathbb{R}^n$ containing the origin.
No such formula holds because the thing on the right is a (limit of) numbers, while $df(a)$ is a linear map. THese are different objects. If you remove the norm in the numerator, $\frac{f(a+r)-f(a)}{\|r\|}$, the numerator is not a linear map, so you can’t say much about its limit. See this question for some related discussions.
The definition of $df_a$ is that (when it exists) it’s the unique linear map such that \begin{align} \lim_{h\to 0}\frac{\|f(a+h)-f(a)-df_a(h)\|}{\|h\|}&=0.\tag{$*$} \end{align} So, if $df_a=0$, then plug in $0$ above to get \begin{align} \lim\limits_{h\to 0}\frac{\|f(a+h)-f(a)\|}{\|h\|}&=0. \end{align}
But, the claim you make is wrong; the Lipschitz estimate does not follow from merely the limit condition (even when $n=m=1$). Consider the function $f:\Bbb{R}\to\Bbb{R}$ defined as \begin{align} f(x)&= \begin{cases} x^2&\text{if $x\in\Bbb{Q}$}\\ 0&\text{else} \end{cases} \end{align} Then, $f$ is differentiable only at the origin and $f$ is not even continuous away from the origin, with $f’(0)=0$. However, if the Lipschitz estimate were true in a neighborhood of the origin, then $f$ must be continuous in a neighborhood of the origin, which we just said is not the case (if you wanted to be fancy, you could also invoke Rademacher’s theorem and get a contradiction).
If however you assume that $f$ is continuously differentiable in a neighborhood of the point $a$, then you can obtain this Lipschitz estimate using the mean-value inequality and the fact that $df_a=0$, and hence in a neighborhood, $\|df_x\|$ can be made small. Note that in general, if you’re trying to get information about the function at different points using only information about the derivative, then the mean-value inequality is often lurking in the shadows.
One very minor remark I’d make is that I’d rather not say that
I would prefer to say that this is a consequence of the definition $(*)$. Because if you define $df_a$ by this formula, then you’d have to define partial derivatives first, and one runs the very widespread and common risk of thinking that partial derivatives are the key concept, and $df_a$ is a secondary concept, when it is completely the other way around.