Let $f\in C^1([a,b])$. Prove that $\|f\|_{C^1} = \|f\|_{Lip}$. By definition of Lip norm and $C^1$ norm, it is equivalent to prove that $\|f'\|_{\infty}=Lip(f,(a,b))$, where the second member is the Lipschitz constant of $f$ on the considered interval.
I proved the inequality $\|f'\|_{\infty}\geq Lip(f(a,b))$ using the Lagrange mean value theorem, but I have a problem to prove the inequality $(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \|f'\|_{\infty}\leq Lip(f(a,b)) \ \ \ \ \ \ \ \ \ \ \ \forall f \in C^1([a,b]).$ Well, since $f' \in C^0([a,b])$, $\exists \ x_0 \in [a,b]$ such that $|f'(x_0)|=\|f'\|_{\infty}$. So, in order to prove $(1)$, we only need to prove that $(2) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |f'(x_0)|\leq Lip(f,(a,b))$ for a given $f\in C^1([a,b])$. Now, recalling the definition of limit, we have that $\forall \ \epsilon >0 \ \ \exists \ \delta =\delta (\epsilon ,x_0 )>0$ such that $(3) \ \ \ \ \ \ \ \ \ \ \ |f'(x_0)|-\epsilon < \frac {|f(x)-f(x_0)|}{|x-x_0|} \ \ \ \ \forall \ x\in (x_o-\delta ,x_0 + \delta)\cap [a,b]-\{x_o\}.$
By definition on Lipschitz constant we can increase "uniformly" the second term of $(3)$ with the Lipschitz constant itself: $\frac {|f(x)-f(x_0)|}{|x-x_0|}\leq Lip(f,(a,b)).$ So we have, $\forall \ \epsilon >0$, $(4) \ \ \ \ \ \ \ \ \ |f'(x_0)|-\epsilon < Lip(f,(a,b)).$
Here is my questions:
how can I justify the implication $(4)\Rightarrow (2)$? Or, to put it simply, how can I gain also the equality in $(4)$? (I should have $\leq$ there). I think that the arbitrariness of $\epsilon$ is not enough.
Does "uniformly" means $\forall x$?
Thank you for your attention.