Thread: integral norm proving

show that ||f||1 = ∫|f| (integral from 0 to 1) does define a norm on the subspace C[0,1] of continuous functions
(there are 3 conditions , i just dont know how to prove that ||v||>0,||v||=0 implies v=0)
and also the same for ||f||= ∫t|f(t)|dt is a norm on C[0,1]

Hey cummings123321.

For the ||v|| > 0 consider the definition of the integral and how you partition it up and add up these partitions. Each partition will be positive which means the norm will be positive and zero only if all elements are zero since the sum of all positive things can only be zero if all are zero.

You will need to specify what kind of integral you are using (because it won't be the Riemann integral for C[0,1]).

Originally Posted by chiro

Hey cummings123321.

For the ||v|| > 0 consider the definition of the integral and how you partition it up and add up these partitions. Each partition will be positive which means the norm will be positive and zero only if all elements are zero since the sum of all positive things can only be zero if all are zero.

You will need to specify what kind of integral you are using (because it won't be the Riemann integral for C[0,1]).

yes ,i worked out the first part, but how about the secound ∫t|f(t)| i should get an inequality that smaller than t|f(t）| and greater than 0,but i dont how to get it