I Uniform convergence of difference quotient in higher dimensions

[psie]

TL;DR Summary

In $\mathbb R$ with $f:[a,b]\to\mathbb R$, one can show that the difference quotient in the definition of the derivative of $f$ converges uniformly on $[a,b]$ to the derivative of $f$. I wonder, is this true for $\mathbb R^d$ with $d>1$ too?

Let $f:\mathbb R^d\to\mathbb R^d$ and let $f$ be differentiable on an open subset $U\subset\mathbb R^d$. We then have $$\lim_{h\to0}\frac{|f(h+a)-f(a)-Df(a)(h)|}{|h|}=0,\quad \forall a\in U.$$ $Df(a)$ is the linear transformation at $a$, i.e. the derivative of $f$, acting on $h$. Let now $K\subset U$ be compact. Under what conditions on $f$ does the fraction above converge uniformly to $0$ on $K$?

I'm partly reading this answer (and also my own textbook, where this claim seems to be made) and it seems to hold for $d=1$ with the help of the mean value theorem and the assumption that $f$ is continuously differentiable, but I have no clue how one would go about it for $d>1$, since I don't think any mean value theorem/inequality holds for vector-valued functions of a vector variable. If you know a proof, I'd be happy to hear about it.

 

[Gavran]

since I don't think any mean value theorem/inequality holds for vector-valued functions of a vector variable.

There is the mean value theorem for vector valued functions. See https://mathresearch.utsa.edu/wiki/index.php?title=Mean-Value_Theorems_for_Vector_Valued_Functions.

If you look at their proof you will see that you can, in similar way, define a new function $F(t)=\textbf{a}\textbf{f}(\textbf{x}+t\textbf{J})$ where $\textbf{a}\in\mathbb{R}^m$ and $\textbf{J}\in\mathbb{R}^n$ is a vector of ones. $F’(t)$ will be $\textbf{a}\textbf{f}’(\textbf{x}+t\textbf{J})\textbf{J}$. By using $$ |\frac{F(t+h)-F(t)}{h}-F’(t)|\lt\epsilon $$ for $t=0$ you can get the next $$ |\frac{\textbf{a}\textbf{f}(\textbf{x}+h\textbf{J})-\textbf{a}\textbf{f}(\textbf{x})}{h}-\textbf{a}\textbf{f}'(\textbf{x})\textbf{J}|\lt\epsilon $$.