Formula involving double integral over a disc?

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SUMMARY

The discussion centers on proving that for a uniformly continuous function \( g: \mathbb{R}^{2} \rightarrow [0,\infty) \) satisfying the condition \( \sup_{r> 0}\iint_{{x^{2}+y^{2}\leq r^{2}}}g(x,y)dxdy< \infty \), it follows that \( \lim_{|(x,y)| \to \infty}g(x,y)=0 \). The approach involves using polar coordinates and a contradiction method to establish that if \( g \) does not approach zero, it leads to a non-zero limit along some angle, contradicting the bounded integral condition. The discussion emphasizes the importance of uniform continuity in this context.

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b.krom
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Problem:

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

I tried to use polar coordinates instead of rectangular ones, but it didn't work out. Any help?
 
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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 g(x,y)=g(x(r,\theta),y(r,\theta)), then there must be some constant angle theta along which g_{\theta}(x(r),y(r)) has nonzero limit, right?

If you try to use polar coordinates, you get

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

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 \int_0^\infty f(x) dx &lt; \infty, f(x)&gt;0 \iff f(x)\rightarrow0, which is easier than what you had first.

Hope that helps a bit!
 
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