Norm of a functional given by difference of values How to calculate the norm of such functional?
$$\phi : \mathcal{C} ([0,1]) \ni f \rightarrow 3 f (\frac{1}{2}) - 5 f (\frac{2}{3}) \in \mathbb{K} (= \mathbb{R}, \mathbb{C})$$
If we equip $\mathcal{C} ([0,1]) $ with supremum norm, we have that $|3 f (\frac{1}{2}) - 5 f (\frac{2}{3})| \le 8 \sup_{[0,1]} |f(t)|$
And we can construct a function which is equal to one for $x \in [0, \frac{1}{2}]$  and $f(x) = -1$ for $x \in [\frac{2}{3}, 1]$ and we link $(1/2, 1)$ with $(2/3, -1)$.
This function has sup norm equal one and $\phi(f) = 3 - (-5)=8$. So $|| \phi || = 8$
But I don't know how to calculate the norm of the functional if we equip $\mathcal{C} ([0,1]) $ with $||f|| = \int_0^1 |f(t)|dt$ or $||f|| = |f(0)| + \sup_{t \in [0,1]} |f'(t)|$

Hint: they might be unbounded with respect to those norms.
For the $L^1$ norm, you want a sequence of functions that have constant (absolute) integral, but whose values grow higher and higher near either $1/2$ or $2/3$. Think of "tent" functions, where the tent grows taller and narrower as $n \to \infty$.
For the other norm, note that
$$
f(t) = f(0) + \int_0^t f'(s) \,ds 
$$
so
$$
|f(t)| \leq |f(0)| + \int_0^t \sup|f'| \,dt = |f(0)| + t\sup_{t\in[0,1]}|f'(t)| \leq |f(0)| + \sup_{t\in[0,1]}|f'(t)|
$$
