Let $L$ denote the space of piecewise linear functions $f: \mathbb R \to \mathbb C$ that is, $f(x) = \begin{cases} A - |x| & x \in [-1,1] \\ 0 & \text{ otherwise} \end{cases}$ where $A \in \mathbb R_{>0}$.
Define $p(f) := \sqrt{\int_{-\infty}^\infty |f(x)|^2 dx}$.
Claim: $p: L \to [0,\infty)$ defines a norm on $L$.
Proof:
(i) $p \geq 0$ is clear. If $f=0$ then clearly $p(f) = 0$. Now assume $p(f)=0$ and show that $f=0$. By contradiction, if $f$ was non-zero at some point then since it is continuous it would have to be non-zero on an open interval and then the integral would also be non-zero. Contradiction. So $f$ is the zero function.
(ii) $ p(af) = \sqrt{\int_{-\infty}^\infty |af(x)|^2 dx} = |a|^2 \sqrt{\int_{-\infty}^\infty |f(x)|^2 dx}= |a|^2 p(f)$
(iii) Here you want to show $p(f+g) \leq p(f) + p(g)$.