Show the following norms $$\|f\|_1 = \|f\|_\infty +\|f'\|_\infty $$ $$\|f\|_2 = |f(a)| +\|f'\|_\infty $$ on the linear space $\mathcal{C}([a,b],\mathbb{K})$ are equivalent.
I am trying to find a global $n,N>0$ such that
$$n\|f\|_1 \leq \|f\|_2 \leq N\|f\|_1$$
so
$$n|f(a)| \leq \|f\|_\infty \leq N|f(a)|$$
Since $\|f\|_{\infty}$ is the supremum of the value obtained on $[a,b]$, $n=1$ suffices.
Then what can I set $N$ to?
It is actually not the same as $\|f\|_{\infty}\leq C|f(a)|$ for some universal constant $C$; that would work if it were possible, but it isn't, even for the example $f(x) = x-a$.
It is enough to have $\|f\|_{\infty}\leq c_1|f(a)|+c_2\|f'\|_{\infty}$, because then you will have $\|f\|_1=\|f\|_{\infty}+\|f'\|_{\infty}\leq c_1|f(a)|+(c_2+1)\|f'\|_\infty\leq C\|f\|_2$ where $C=\max\{c_1,c_2+1\}$.
That estimate comes from $f(x) = f(a)+\int_a^x f'(t)\,dt$, because it implies that for all $x\in[a,b]$, $|f(x)|\leq |f(a)|+\int_a^b|f'(t)|\,dt\leq |f(a)|+(b-a)\|f'\|_\infty$. The right hand side doesn't depend on $x$, so the inequality will also hold with $|f(x)|$ replaced by $\|f\|_\infty$. This gives $c_1=1$ and $c_2=b-a$.