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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.

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up vote 1 down vote accepted

It's just a general principle of analysis: $a\le b$ is equivalent to the following statement: for all $\varepsilon>0$, $a-\varepsilon < b$. The proof of this equivalence is very easy. And for the $x$-dependence: Your inequality (3) is valid for all $x$ inside an interval, but you only have to choose one $x$ from this interval to continue.

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Thank you, I didn't think that was so easy. –  the_elder Nov 15 '12 at 19:27

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