Proving Integral Norms on C[0,1] for Continuous Functions

[cummings12332]

Homework Statement

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]

Homework Equations

(there are 3 conditions , i just don't know how to prove that ||v||>0,||v||=0 implies v=0)

 

[LCKurtz]

Homework Statement

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]

Homework Equations

(there are 3 conditions , i just don't know how to prove that ||v||>0,||v||=0 implies v=0)

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

 

[cummings12332]

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

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

 

[LCKurtz]

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

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 $\int_a^b f(x)\, dx > \int_a^b g(x)\, dx$

 

[cummings12332]

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 $\int_a^b f(x)\, dx > \int_a^b g(x)\, dx$

yes,i worked out the first one now,but how about the secound one for t f(t) i don't know how to construct the inequality that u have mentioned above..