# Sufficient and necessary condition for continuously differentiable

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!

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+1 for showing what you've tried. – Rasmus Jun 6 '12 at 19:08
I'm confused. Are you sure the that there is no assumptions like "$A$ is continuous in $x$"? – user20266 Jun 6 '12 at 19:10
@Thomas: the question says nothing about $A$... only that there exists $A(x)$ for each $x$, and nothing more... but truly, if we assume this, we have the equivalence... ow, i'm confused now! =p – Yuki Jun 6 '12 at 19:16
Well, if it says nothing about continuity of $A$, then this probably the whole point of the exercise. Did not see this before, I have to admit. – user20266 Jun 6 '12 at 19:18

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.
Fix $x\in U$ and consider $g_x:=f-f'(x)$ (certainly $g$ is differentiable, since is the sum of a differentiable function and a linear operator). By hypothesis, for $h',k'\in B(0;\delta)$ we have $$\frac{\|g_x(x+h')-g_x(x+k')\|}{\|h'-k'\|}=\frac{\|f(x+h')-f(x+k')-f'(x)(h'-k')\|}{\|h'-k'\|}<\epsilon$$ in particular, taking $h'=x_0+th-x$ and $k'=x_0-x$, with $x_0\in B(x;\delta)$ and $th\in B(0;\delta)$, we have $$\frac{\|g_x(x_0+th)-g_x(x_0)\|}{|t|\|h\|}<\epsilon$$ Since this holds for all $t$ small, we can take the limit and we have $\displaystyle\frac{\|g'_x(x_0)\cdot h\|}{\|h\|}\le\epsilon$. Since $h$ is arbitrary (in sense that given $u\in\mathbb{R}^m$, with $\|u\|=1$, we can find an adequate $h$ with $\frac{h}{\|h\|}=u$), we have the norm of the operator $\|g'_x(x_0)\|\le\epsilon$. But $g'_x(x_0)=f'(x_0)-f'(x)$, so, if $x_0,y_0\in B(x;\delta)$, then $$\|f'(x_0)-f'(y_0)\|\le\|f'(x_0)-f'(x)\|+\|f'(x)-f'(y_0)\|\le2\epsilon$$ and this shows that $f'$ is continuous in $x$, but since $x$ is arbitrary, we have $f$ continuously differentiable.