Are the norms $\Vert f\Vert_1=\Vert f\Vert _\infty+\Vert f'\Vert _\infty$ and $\Vert f\Vert _2=\vert f(a)\vert +\Vert f'\Vert _\infty$ equivalent?

Question

We have the norms $\Vert f\Vert_1=\Vert f\Vert _\infty+\Vert f'\Vert _\infty$ and $\Vert f\Vert _2=\vert f(a)\vert +\Vert f'\Vert _\infty$ where $f\in C^1[a,b]$. Are they equivalent and how shoud I prove/disprove this.

Answer 1

Well, obviously $$ \|f\|_2 \leq \|f\|_1 $$ because $|f(a)| \leq \|f\|_\infty$ by definition of $\|\cdot\|_\infty$. So the question is, is there a $C$ such that $$ \|f\|_1 \leq C\cdot \|f\|_2 \text{.} $$ To prove that, it suffices to show that there's a $C' > 0$ with $$ \|f\|_\infty \leq |f(a)| + C'\|f'\|_\infty \quad(\star) $$ because then (for $C = C'+1 > 1$)$$ \|f\|_1 = \|f\|_\infty + \|f'\|_\infty \leq \underbrace{|f(a)|}_{\leq C|f(a)|} + \underbrace{C'\|f'\|_\infty + \|f'\|_\infty}_{=C\|f'\|_\infty} \leq C|f(a)| + C\|f'\|_\infty = C\|f\|_2 \text{.} $$

So what's left is to prove $(\star)$. For that, you can either use the fundamental theorem of calculus, or the intermediate value theorem.