I was trying to prove the following statement:
Let $U\subset\mathbb{R}^m$ be open and $f:U\to\mathbb{R}^n$. Show that $f$ is continuously differentiable if, and only if, for each $x\in U$, there exists a linear operator $A(x):\mathbb{R}^m\to\mathbb{R}^n$ such that
$$
\lim\frac{f(x+h)-f(x+k)-A(x)(h-k)}{\|h-k\|}=0
$$
whenever $(h,k)\to(0,0)$, with $h\ne k$.
If $f$ is continuously differentiable, then, for each $x\in U$, defining $g_x:=f-f'(x)$ (then $g_x'(y)=f'(y)-f'(x)$) and applying the Mean Value Theorem, using the continuity for $f'$, and for adequate closed ball $B=\bar B(x;\delta)\subset U$ such that $\|f'(a)-f'(b)\|<\epsilon/2$ if $a,b\in B(x;2\delta)\subset U$, and for any $h,k\in B,h\ne k$, we have
$$
\frac{\|f(x+k+h-k)-f(x+k)-f'(x)(h-k)\|}{\|h-k\|}=\frac{\|g_x(x+h)-g_x(x+k)\|}{\|h-k\|}\\\le\sup\limits_{h,k\in B}\|f'(x+k)-f'(x+h)\|<\epsilon
$$
so, the equality holds, where $A(x):=f'(x)$, for each $x\in U$.
Reciprocally, fixing $k=0$, we see that $f$ is differentiable in $U$ and $f'(x)=A(x),\forall x\in U$ (and hence continuous). But I couldn't prove that $f'$ is continuous…
Any hint? Thanks!
Best Answer
Pick $x\in U$. We can assume $A(x)=0$ (subtract a linear function to achieve this). For every $\epsilon>0$ there is a neighborhood $V$ of $x$ such that $|f(y)-f(z)|\le \epsilon |y-z|$ for all $y,z\in V$. Hence $\|f'\|\le \epsilon$ in $V$, as was to be proved.
(This is a mere sketch: you should write this out in detail, of course.)