Task:
On the set $C^2[a, b] := \lbrace f: [a, b]\to R \text{ are twice continuous differentiable function} \rbrace$
are defined functions $F_i :C^2[a, b]\to R, i \in\{ 1, 2, 3\}$. We have 3 specific functions:
$$F_1(f)=\max_{a\le t\le b} |f(t)|$$ $$F_2(f)=\max_{a\le t\le b} |f'(t)|$$ $$F_3(f)=\max_{a\le t\le b} |f''(t)|$$
Is it a norm of $C^2[a, b]$:
(a) $\Vert f\Vert =\vert f(b)-f(a)\vert + F_2(f) + F_3(f)$
(b) $\Vert f\Vert =\vert f(b)\vert+\vert f(a)\vert + F_3(f)$
(c) $\Vert f\Vert = \int_{a}^ b \vert f(t)\vert dt + F_3(f)$
My solution for (a) is:
I need to check 3 axioms:
1) $\Vert f\Vert \ge 0$. (it is true because of absolute values) and
$\Vert f\Vert = 0 \iff f=0$ (here I have some problems to show it is true).
2) $\Vert \mathcal L f\Vert = \vert\mathcal L \vert \Vert f \Vert$
Proof:
\begin{align*} \Vert \mathcal L f\Vert & = \vert\mathcal L f(b) - \mathcal L f(a) \vert + F_2(\mathcal L f) + F_3(\mathcal L f)\\ & = \vert\mathcal L (f(b) - f(a)) \vert + \mathcal L F_2(f) + \mathcal L F_3(f) \\ &= \vert \mathcal L \vert ( \vert (f(b) - f(a)) \vert + F_2(f) + F_3(f)) \\ &= \vert\mathcal L \vert \Vert f \Vert \end{align*}
So it is true.
3) $\Vert f+ g \Vert \le \vert f \vert + \vert g \vert$.
True, because of $ F_2(f+g) \le F_2(f) + F_2(g)$.
Question: The same I did with (b) and (c) and my result is that all of them are norms. Is my proof correct? What about (b) and (c)? Maybe I cannot notice something and make mistake? It looks for my strange that I got 3 norms here.
The keypoint is the condition $$\|f\|=0\Rightarrow f=0.$$ Since all the proposed candidates for norms contain the term $F_3(f)$, from $$\|f\|_i\Rightarrow0$$ follows that $f$ is linear, i.e. $f(x)=kx+d$ for some constants $k,d$. Now for $\|\cdot\|_b,\|\cdot\|_c$ this leads to $f=0$, since a linear function with two zeros is zeros, and also it is easy to check that $$\int_a^b|kx+d|\,dx=0\Rightarrow k,d=0.$$
For $\|\cdot\|_a$ on the other hand as @Aniruddha Deshmukh pointed out in the comments for any function $f(x)=c$ we have $$\|f\|_a=|f(a)-f(b)|+F_2(f)+F_3(f)=0.$$