If $f$ is uniformly differentiable then $f '$ is uniformly continuous? The following theorem is true?

Theorem. Let $U\subset \mathbb{R}^m$ (open set) and $f:U\longrightarrow \mathbb{R}^n$ a differentiable function.

If $f$  is uniformly differentiable  $ \Longrightarrow$ $f':U\longrightarrow \mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$ is uniformly continuous.

Note that  $f$ is uniformly differentiable if

$\forall \epsilon&gt;0\,,\exists \delta&gt;0:|\!|h|\!|&lt;\delta,\color{blue}{[x,x+h]\subset U} \Longrightarrow |\!|f(x+h)-f(x)-f'(x)(h)|\!|&lt;\epsilon |\!|h|\!| $ (edited)

$\forall \epsilon&gt;0\,,\exists \delta&gt;0:|\!|h|\!|&lt;\delta,\color{blue}{x,x+h\in U} \Longrightarrow |\!|f(x+h)-f(x)-f'(x)(h)|\!|&lt;\epsilon |\!|h|\!|\qquad  \checkmark$

Any hints would be appreciated.
Let's build off of Tomas' last remark, slightly modified:

Let $t&gt;0$ be small. Then
\begin{eqnarray}
 \|f'(x)-f'(y)\| &amp;=&amp; \frac{1}{t}\sup_{\|w\|=1}\|\langle f'(x)-f'(y),tw\rangle\|      \nonumber \\
   &amp;\leq&amp; \frac{1}{t}\sup_{\|w\|=1}\|f(x+tw)-f(x)-[f(y+tw)-f(y)]\| + 2\epsilon \nonumber 
\end{eqnarray}

It suffices to show that this weighted combination of four close points on a parallelogram can be bounded by $C\epsilon t$.

Let us bound $\|f(x+h) - f(x) + f(x+k) - f(x+h+k)\|_2 \leq C\epsilon(\|h\|+\|k\|)$, and then in this case $\|h\|=t$ and $\|k\|\leq \delta$, so if $t=\delta$ the whole expression is bounded by a constant times $\epsilon$.

Note applying uniform differentiability three times in directions $h,k,$ and $h+k$, for small $\|h\|,\|k\|$ we have

\begin{eqnarray*}
\|f(x+h) - f(x) + f(x+k) - f(x+h+k)\| &amp;\leq&amp; \|f'(x)h + f'(x)k - f'(x)(h+k)\|_2 + 3\epsilon(\|h\|+\|k\|)\\
 &amp;=&amp; 3\epsilon(\|h\|+\|k\|)
\end{eqnarray*}
Let us start with the simplest case f:$R^1 \rightarrow R^1$.  The discussion below will generalize nicely to more dimensions. 

Then f is differentiable at $x_0$ iff $ \lim_{x \rightarrow x_0} {\frac{f(x)-f(x_0)}{x-x_0}}$ exists and is finite.  We call that limit $f'(x_0)$ .  So if f is differentiable at $x_0$  then we may write

f(x) = $f(x_0) + f'(x_0)(x-x_0) + R(x)$ where  $\lim_{x \rightarrow x_0} R(x)/(x-x_0)$ = 0.  &nbsp; &nbsp; (1)

If f is uniformly differentiable that means 

$\lim_{x \rightarrow x_0} R(x)/(x-x_0)$ converges uniformly to zero. &nbsp; &nbsp; (2)

From (1) we have $f'(x)$ = $f'(x_0)$ + R'(x).  [We know R is differentiable because it is the difference between f which is differentiable and a linear polynomial.] 

So $\lim_{x \rightarrow x_0} (f'(x)-f'(x_0))$ = $\lim_{x \rightarrow x_0} R'(x)$  = $\lim_{x \rightarrow x_0} R(x)/(x-x_0)$ = 0.  And this convergence is uniform as per our hypothesis as stated in (2). 

Moving on to the case where $f:R^n \rightarrow R^m$. For this part of the discussion capital letters will be used to represent vectors (in whatever dimension). 

According to many sources (and I will state specifically Robert Sealey "Calculus of Several Variables") f is differentiable at $X_0$ iff there exists a linear function g (X) = $f(X_0)$ + $(M,X-X_0)$  [where M is a constant vector and (a,b) means the inner product], such that

$ \lim_{X \rightarrow X_0} {\frac{f(X)-g(X)}{|X-X_0|}}$ = 0. &nbsp;&nbsp;(3)

If f is uniformly differentiable that means simply that this limit converges uniformly.We could have stated differentiability this way in the $R^1 \rightarrow R^1$ case, but I stated it as I did in that case to throw some light on what g actually is.  

Let R(X) = f(X) - g(X) $\Rightarrow$  f(X) = g(X) + R(X). Since R = f - g then $R/|x-x_0| \rightarrow 0$ uniformly. Notice also that $R(X_0)$ = 0, since it is $f(X_0) - g(X_0)$ and by (3) this must be 0.  Since g is linear and f is differentiable R is also differentiable and from the above $R'(X_0)$ = 0.

Looking at $f'$ we have
$f'(X) = g'(X) + R'(X) = M + R'(X)$;
$f'(X_0) = M + R'(X_0)$ and thus $f'(X_0) = M$.

Finally $\lim_{X \rightarrow X_0} (f'(X)-f'(X_0))$ = $\lim_{X \rightarrow X_0} R'(X)$ = $\lim_{X \rightarrow X_0}{\frac{R(X)-R(X_0)}{X-X_0}}$ = $\lim_{X \rightarrow X_0}{\frac{R(X)}{X-X_0}}$  and this goes to 0 uniformly because f is uniformly differentiable, R = f - g, and $R(X_0)$ is 0.