# Integral norm proving

1. Nov 16, 2012

### cummings12332

1. The problem statement, all variables and given/known data
show that ||f||1 = ∫|f| (integral from 0 to 1) does define a norm on the subspace C[0,1] of continuous functions

and also the same for ||f||= ∫t|f(t)|dt is a norm on C[0,1]

2. Relevant equations

(there are 3 conditions , i just dont know how to prove that ||v||>0,||v||=0 implies v=0)

2. Nov 16, 2012

### LCKurtz

If a continuous function is nonzero (or positive) at a point, what about its value nearby that point, and why?

3. Nov 16, 2012

### cummings12332

yes if it is not zero ,then |f(x)-f(c)|<esillope then choose esillope to be f(c)/2 then it will get |f(x)|>sth..... but how can u ensure that f(c) >0 for esillope has to be >0

4. Nov 16, 2012

### LCKurtz

You have this property of integrals to work with: If $f,g\in C[a,b]$ and $f(x)> g(x)$ on [a,b] then $\int_a^b f(x)\, dx > \int_a^b g(x)\, dx$

5. Nov 17, 2012

### cummings12332

yes,i worked out the first one now,but how about the secound one for t f(t) i dont know how to construct the inequality that u have mentioned above..