Uniform convergence integration

[samithie]

f is a continuous function on [0,infinity) such that 0<=f(x)<=Cx^(-1-p) where C,p >0 f_k(x) = kf(kx)

I want to show that lim k->infinity ∫from 0 to 1 of f_k(x) dx exists

so my idea is if I have that f_k(x) converges to f(x)=0 uniformly which I was able to show and that f_k(x) are all integrable, then the limit can be moved inside and so this thing will exist.

to show f_k(x) is integrable from 0 to 1, I replaced it with Cx^(-1-p) and ran into a problem, this is only integrable from 0 to 1 when the exponent (1+p) on the bottom now needs to be <1 which requires p to be less than 0 but p is given to be > 0. is my understanding of integrability of improper integrals correct or is it something else I'm doing wrong. Thanks a lot.

 

[Dick]

f is a continuous function on [0,infinity) such that 0<=f(x)<=Cx^(-1-p) where C,p >0 f_k(x) = kf(kx)

I want to show that lim k->infinity ∫from 0 to 1 of f_k(x) dx exists

so my idea is if I have that f_k(x) converges to f(x)=0 uniformly which I was able to show and that f_k(x) are all integrable, then the limit can be moved inside and so this thing will exist.

to show f_k(x) is integrable from 0 to 1, I replaced it with Cx^(-1-p) and ran into a problem, this is only integrable from 0 to 1 when the exponent (1+p) on the bottom now needs to be <1 which requires p to be less than 0 but p is given to be > 0. is my understanding of integrability of improper integrals correct or is it something else I'm doing wrong. Thanks a lot.

You don't have to worry much about what happens on [0,1] particularly near 0. You are given that f is continuous on [0,1]. It integrable on [0,1]. f is much better behaved there than your bounding function is.

 

[samithie]

thanks that's very helpful!