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Formula involving double integral over a disc?

by b.krom
Tags: analysis, calculus
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Apr26-12, 08:42 AM
P: 1

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?
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Apr26-12, 06:01 PM
P: 169
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!

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