Formula involving double integral over a disc?

b.krom

Problem:

Can anyone help me out with the following problem:
I am given a uniformly continuous function : [itex]g:\mathbb{R}^{2}\rightarrow [0,\infty )[/itex] such that the following condition is satisfied:
[tex]\sup_{r> 0}\iint_{{x^{2}+y^{2}\leq r^{2}}}g(x,y)dxdy< \infty [/tex]
The question is to prove that:[tex]\lim_{| (x,y)| \to \infty}g(x,y)=0[/tex]

I tried to use polar coordinates instead of rectangular ones, but it didn't work out. Any help?

jacobrhcp

First, note that the limit intuiteively means that if you go really far from the origin, g becomes almost zero.

if you first suppose by contradiction the limit does not hold and [itex]g(x,y)=g(x(r,\theta),y(r,\theta))[/itex], then there must be some constant angle theta along which [itex]g_{\theta}(x(r),y(r))[/itex] has nonzero limit, right?

If you try to use polar coordinates, you get

[itex]\iint r g(x(r,\theta),y(r,\theta)), \theta\in[0,2\pi], r\in[0,\infty)[/itex]

Because g is always positive and uniformly continous, I expect this is larger than the integral over the radiant with theta such that there is a nonzero limit (admittedly, this needs to worked out a bit), which exists as assumed by contradiction. So then you just need something like [itex]\int_0^\infty f(x) dx < \infty, f(x)>0 \iff f(x)\rightarrow0[/itex], which is easier than what you had first.

Hope that helps a bit!