there are three conditions for a norm,
I am stuck on one of them which is : $N(f)=0$ iff $f=0$.
If i say $\int_{0}^{1} |f(t)|dt=0$ does this imply that $f=0$(zero function) ?
Ok the other sense of the equivalence relation is obvious but this sense is not(i mean by this sense: $\int_{0}^{1} |f(t)|dt=0$ then f=0)
This is true. Assume that $f$ is not zero at a point $x_0$. By continuity you can find a small ball $B_r(x_0)$ such that $|f(x)| \geq \delta > 0$ for all $x \in B_r(x_0) \cap [0,1]$. But this means that the norm must be greater than $\delta \cdot | B_r(x_0) \cap [0,1]|$ (where $| B_r(x_0) \cap [0,1]|$ is the size of the interval $B_r(x_0) \cap [0,1]$), i.e. it cannot be zero.
