Good morning. I'm having a hard time finding what's wrong with the following argument.
Let $f$ be any function in $C^{1}([0;1])$ and let $||f||$ and $N(f)$ be two norms defined as follows: $$||f|| = \sup_{x\in[0;1]}{|f(x)|} + \sup_{x\in[0;1]}{|f'(x)|}$$ $$N(f) = \int_0^1{|f'(x)|dx} + \sup_{x\in[0;1]}{f(x)} \text{.}$$ A priori, ignoring the fact that the two norms are not equivalent (as it could be seen considering the sequence of functions $\{x^n\}$ in $[0;1]$ and their different limits with respect to the two norms), if I try to find two positive constants $\alpha$, $\beta$ such that $\forall x \in [0;1]$ it is $$\alpha ||f|| \leq N(f) \leq \beta ||f||\text{,}$$ I can apparently succeed in it.
I'd like you to show me the pitfalls in my (pseudo)-demostration, since I cannot find them.
Let's begin with the second inequality, $\quad \forall x\in[0;1] \text{ }\exists \beta > 0 \text{ s. t. } N(f) \leq \beta||f||$: $$ \forall x \in [0;1] \text{: } |f'(x)| \leq \sup{|f'(x)|} \quad \Rightarrow \quad \int_0^1{|f'(x)|dx} \leq \int_0^1{\sup{|f'(x)|}}dx = \sup{|f'(x)|}\text{,} $$ therefore I could take $\beta = 1 > 0 $ and I'm finished.
As for the first inequality (and here's is the trouble, I guess), I proceed as follows: $$f\in C^1([0;1]) \quad \Rightarrow \quad f' \in C^0([0;1]) \quad \Rightarrow \quad |f'| \in C^0([0;1]) \quad \mathit{(right?),}$$ therefore, according to Weierstrass theorem, $$\exists x_m \in [0;1] \text{ s. t. } |f'(x)| \geq |f'(x_m)| = \mu \quad \forall x\in[0;1]\text{.}$$ The inequality $$ \alpha ||f|| \leq^{?} N(f) \Leftrightarrow \int_0^1{|f'(x)|dx} + \sup{|f(x)|} \geq^{?} \alpha(\sup{|f(x)|} + \sup{|f'(x)|}) \text{ ,} $$ for some positive $\alpha$ to be found (...), leads to the following: $$ \int_0^1{|f'(x)|dx} \geq \mu \geq^{?} (\alpha - 1)\sup{|f(x)|} + \alpha\sup{|f'(x)|}\text{;}$$ if I take $\quad 0 < \alpha < 1\text{,} \quad$ then $\quad (\alpha - 1 )\sup{|f(x)|} \leq 0\cdot \sup{|f(x)|} = 0\text{,}\quad$ therefore, for the inequality to be proved I have: $$\int_0^1{|f'(x)|dx} \geq \mu \geq^{?} \alpha\sup{|f'(x)|} \geq (\alpha - 1)\sup{|f(x)|} + \alpha\sup{|f'(x)|}$$ and I must find $\alpha$ such that $\quad \mu \geq \alpha\sup{|f'(x)|}\text{:}\quad$ if I then take $$\alpha \leq \dfrac{\mu}{M} \leq 1 \text{,}$$ with $M = \sup{|f'(x)|}$, I should be finished.
So what's wrong?! I'd appreciate any help from you. Thanks.
I believe the error is in the last inequality/step: first, why do you have $0 < \frac{\mu}{M}\leq 1$? (you need it for your argument to go through, as $\alpha$ was specifically assumed to be in $(0,1]$.)
Note also, crucially, that this $\alpha$ should not depend on $f$... and yours does.