real analysis - 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

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

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

Any hints would be appreciated.

Answer

Let's build off of Tomas' last remark, slightly modified:

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

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

$$\begin{eqnarray*} \|f(x+h) - f(x) + f(x+k) - f(x+h+k)\| &\leq& \|f'(x)h + f'(x)k - f'(x)(h+k)\|_2 + 3\epsilon(\|h\|+\|k\|)\\ &=& 3\epsilon(\|h\|+\|k\|) \end{eqnarray*}$$