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.

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*}

This answer only the case $f:U\subset\mathbb{R}\to \mathbb{R}^n$. By hypothesis we have that: given $\epsilon>0$, there exist $\delta>0$ such that if $|x-y|<\delta$ $$\left|\frac{f(x)-f(y)}{x-y}-f'(y)\right|<\frac{\epsilon}{2}\tag{1}$$

and

$$\left|\frac{f(y)-f(x)}{y-x}-f'(x)\right|<\frac{\epsilon}{2}\tag{2}$$

From $(2)$ we get that $$\left|\frac{f(x)-f(y)}{x-y}-f'(x)\right|<\frac{\epsilon}{2}\tag{3}$$

From $(1)$ and $(3)$ $$|f'(x)-f'(y)|<\epsilon$$

I dont know how to tackle the case $U\subset \mathbb{R}^m$.

Update: I would like to add here some thoughts, maybe it can help someone to give a full answer. First note that

\begin{eqnarray} \|f'(x)-f'(y)\| &=& \sup_{\|w\|=1}\|\langle f'(x)-f'(y),w\rangle\| \nonumber \\ &=& \sup_{\|w\|=1}\|f(x+w)-f(x)-[f(y+w)-f(y)]+o(x,w)-o(y,w)\| \nonumber \end{eqnarray}

By hypothesis, $\|o(x,w)-o(y,w)\|\leq \epsilon\|w\|+\epsilon\|w\|$, hence for $\|w\|=1$ we have that $\sup_{\|w\|=1}\|o(x,w)-o(y,w)\|$ is small, independetly of $x,y$. Therefore, it remains to prove that $f$ is uniformly continuous. One way to prove it is for example showing that $f'$ is bounded, which I think is true when $U$ is bounded. When $U$ is unbounded, I think we need a direct argument to show that $f$ is uniformly continuous.

The difficulty here is that the obvious estimate (as in Tomás' answer) only gives a result along lines - i.e. it can't say anything about how $f'(x)h$ changes as you move in any direction but $h$. Thus I think we need some kind of "polarization argument" to get an isotropic result. When $f$ is $C^2$, this is literally polarizing the Hessian matrix $H(x) = f''(x)$ to show it is bounded from the fact that it is bounded on the diagonal. While we cannot use this literally for general $f$, hopefully we can use a finite-difference analogy and the analysis will go through. Here's the proof for the $C^2$ case and an optimistic starting point for the general case.

For any $\epsilon>0$ there is a $\delta>0$ such that for all $x,h$ with $|h|<\delta$, we have (cf. Tomás' answer)

$$ \begin{align} |f(x+h) - f(x) - f'(x)h| &< \epsilon |h| \\ |f(x) - f(x+h) + f'(x+h)h| &< \epsilon |h| \end{align} $$

which combine to give $$t|(f'(x+t v)-f'(x))v| < 2 \epsilon $$ for $v$ a unit vector, $t<\delta$.

In the $C^2$ case the left side is $t^2 |H(x)(v,v)| + o(t^2)$, so we have a uniform bound $$|H(x)(v,v)| < M \tag{1}.$$ The polarization formula is $$ H(x)(v,w) = \frac12 \left( H(x)(v+w,v+w) - H(x)(v,v) - H(x)(w,w) \right). \tag{2}$$

We want to show uniform continuity of $f'$. A line integral estimate gives $$ |f'(x)w-f'(y)w| \le |x-y| \sup_{z \in U} |H(z)\left(v,w\right)|$$ where $|x-y|v = x-y$; so it suffices to show that $\sup_{z \in U, |v| = 1} H(z)(v,w)$ is finite. Applying $(1)$ and $(2)$ we estimate

$$ \begin{align} |H(z)(v,w)| & \le \frac12 M \left( |v+w|^2 + |v|^2 + |w|^2 \right)\\ & \le M \left( 1 + |w| + |w|^2 \right) \end{align} $$

so we are done.

Now to try generalising this. Attempting to write down the polarization formula using finite difference terms gives

$$ \begin{align} &(f'(x+v+w) - f'(x+v))w + (f'(x+w+v)-f'(x+w))v \\ = &(f'(x+v+w) - f'(x))(v+w) - (f'(x+v)-f'(x))v-(f'(x+w) - f'(x))w \\ \end{align}$$

which (since $f$ is uniformly differentiable) is bounded by $2 \epsilon (|v+w| + |v| + |w|)$.

Now we would like to equate the two terms on the LHS, which is the reason polarization works for symmetric forms. Our expression is of course not symmetric but we can hope it is close enough for small $u,v$ - I have no idea whether or not this can be done without more assumptions, however. If anyone has any ideas (or has spotted any mistakes) please let me know.