I want to show that $$(\int _A|\omega (z)|²|dz|²)^{1/2}$$ is a norm on the space of square integrable differential 1-forms on a Riemann surface $M$ where $A$ is an open set such that $\omega|_{M-A}=0$ almost everywhere.
I only need to prove the triangle innequality but I'm stuck. Maybe I need to use a certain innequality.
I know that $|w_1(z)+w_2(z)| \leq |w_1(z)| + |w_2(z)|$ but I don't know how to get from there to the innequality involving the squares.
Why do you want to prove the triangle inequality? If you want to show that: $\forall \omega, \eta \in \Omega^1(M;\mathbb{C})$ the inner product $$<\omega, \eta>:= \int_{M} \omega \wedge \overline{*\eta}, $$ is a norm on the $\Omega^1(M;\mathbb{C})$, it is sufficient to prove the inner product is Hermitian, we let the $\omega= adx+bdy$, then $$<\omega, \omega>= \int_{M} \omega \wedge \overline{*\omega}= \int_{M} (\vert a \vert ^2+ \vert b \vert ^2)dx\wedge dy \geq 0,$$ and $<*\omega, *\eta>=<\omega, \eta>$.Hence, we can define the norm: $$\Vert \omega \Vert= <\omega, \omega>^\frac{1}{2}.$$