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Let $f: (0; 1) \rightarrow \mathbb{E}^n$. Does $f'$ being Lipschitz imply that $f$ is uniformly differentiable?

In my professor's textbook (yet unreviewed professionally, thus I'm coming here for an answer) this implication is a part of a proof of a theorem:

$(\forall t,τ\in (0,1))||f'(t)-f'(τ)||\leq L|t-τ|,\space L<\infty.$

implies

$(\forallε>0)(\exists\overline{δ}>0)(\forall t\in (0,1))(\forall δ\in (-\overline{δ}, \overline{δ})) \frac{||f(t+δ)-f(t)-f'(t)δ||}{δ}\leq ε$

The comment that is given to this statement is exactly the title.

First, I tried to apply the triangle inequality ($||x+y||\geq||x||+||y||$), with $f(t+δ)-f(t)-f'(t)δ$ and $-f(t+δ)+f(t)-f'(τ)δ$ as operands, but it seems to me that the inequality would not actually follow through.

Eariler in the work, a following equality is constructed:

$f(t+δ)-f(t)=f'(t')δ+\vec{ο}(t,δ)$, where $t'=t+\frac{δ}{2}$

It seems to me like the logic is flawed once again, with $||f'(t)-f'(τ)||\leq L|t-τ|$ not implying $||f'(τ)-f'(t)+\vec{o}(t,δ)δ|| <ε$.

$\tiny{It's\space quite\space late\space here,\space so \space I \space might \space be \space missing \space a \space piece.}$

michael15346
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Apr 10 '23 at 21:28
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    Thanks! Will do, I'm sorry I made such an impression. – michael15346 Apr 10 '23 at 21:53
  • Can you use the mean value theorem for the first part? You'd get $f(t+\delta) - f(t) = f'(\tau) \delta$ for some $\tau \in (t, t+\delta)$. – User203940 Apr 10 '23 at 22:27
  • @User203940 the textbook states that this theorem is not applicable to multi-dimensional spaces (for an unstated reason). I am inclined to believe this at this time, but I'd be eager to apply this theorem if it is actually applicable, as it actually saves about a page worth of proof. – michael15346 Apr 10 '23 at 22:37
  • Ah, I didn't see that this was not just on $\mathbb{R}$. Maybe you can work with the generalization? https://en.m.wikipedia.org/wiki/Mean_value_theorem – User203940 Apr 10 '23 at 23:07
  • For further reference, here's the case $n=1$: https://math.stackexchange.com/questions/2556707/a-function-is-uniformly-differentiable-if-its-derivative-is-uniformly-continuous – User203940 Apr 10 '23 at 23:28

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