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 $||f'(τ)-f'(t)+\vec{o}(t,δ)δ|| <ε$.
$\tiny{It's\space quite\space late\space here,\space so \space I \space might \space be \space missing \space a \space piece.}$