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Please help to improve/correct the following arguments.

I want to show that the integral $I(x,y,z)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{z\over 2\pi [(\alpha-x)^2+(\beta-y)^2+z^2]^{3\over2}}f(\alpha,\beta)\;d\alpha\; d\beta$

where $f$ is an arbitrary continuous function and $I$ is the solution of Laplace's equation $\nabla^2 I=0$, has the following properties:

1) Tends to $0$ as $x^2+y^2\to\infty$. Argument: As $x^2+y^2\to\infty$, the integrand tends to $0$, hence the integral tends to $0$. Concern: Does the arbitrariness of $f$ screw things up?

2) That $I(x,y,0)=f(x,y)$. Argument: For $\alpha,\beta\neq x,y$ respectively, the integrand is $0$ when $z=0$, but when $\alpha,\beta= x,y$ respectively, we have a singularity. This resembles the sampling property of the delta function, but how would I know that I can use that argument here, since we don't really have a delta function. Please help!

Thank you.

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    @DavideGiraudo:$I$am only given that $f$ is continuous. As for $I$, it is the solution to Laplace's equation s.t. $\nabla^2 I=0$.2011-12-24

1 Answers 1

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For the first question, I will assume that $f(x,y)\to 0$ when $x^2+y^2\to\infty$. Since $I(x,y,z)=\iint_{\mathbb R^2}\frac z{2\pi(t_1^2+t_2^2+z^2)^{\frac 32}}f(t_1+x,t_2+y)dt_1dt_2,$ and $f$ is bounded, for a fixed $\varepsilon>0$, we can find $R>0$ such that $\left|\iint_{t_1^2+t_2^2\geq R^2}\frac z{2\pi(t_1^2+t_2^2+z^2)^{\frac 32}}f(t_1+x,t_2+y)dt_1dt_2\right|\leq\varepsilon.$ Since for $t_1^2+t_2^2\leq R$ we have $||(t_1+x,t_2+y)|| \geq \sqrt{x^2+y^2}-\sqrt{t_1^2+t_2^2}\geq \sqrt{x^2+y^2}-R,$ and we can find R' such that if a^2+b^2\geq R' then $|f(a,b)|\leq\varepsilon$, if x^2+y^2\geq R+R' we have $|I(x,y,z)|\leq\varepsilon$.

For the second question, fix $\varepsilon>0$. Take $R>0$ such that $\left|\iint_{t_1^2+t_2^2\geq R^2}\frac z{2\pi(t_1^2+t_2^2+z^2)^{\frac 32}}dt_1dt_2\right|\leq\varepsilon.$ (note that such a $R$ doesn't depend on $z$ since it's the case for $\int_{\mathbb R^2}\frac{z}{(u^2+v^2+z^2)}dudv$) Then we use the uniform continuity of $f$ on $\overline{C(0,R)}$, we take $\delta>0$ such that if $x_1^2+x_2^2\leq R^2$, x_1'^2+x_2'^2\leq R^2 and (x_1-x'_1)^2+(x_2-x'_2)^2\leq\delta^2 then |f(x_1,x_2)-f(x'_1,x'_2)|\leq\varepsilon. Then \begin{align*} |I(x,y,z)-f(x,y)|&\leq\frac 1{2\pi}\iint_{\mathbb R^2}\frac z{(t_1^2+t_2^2+z^2)^{3/2}}|f(t_1+x,t_2+y)-f(x,y)|dt_1dt_2\\ &\leq 2\sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|\varepsilon+\frac 1{2\pi}\iint_{t_1^2+t_2^2\leq R^2}\frac{|z||f(t_1+x,t_2+y)-f(x,y)|}{(t_1^2+t_2^2+z^2)^{3/2}}dt_1dt_2\\ &=2\sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|\varepsilon+ \frac 1{2\pi}\iint_{t_1^2+t_2^2\leq \delta^2}\frac{|z||f(t_1+x,t_2+y)-f(x,y)|}{(t_1^2+t_2^2+z^2)^{3/2}}dt_1dt_2\\ &+\frac 1{2\pi}\iint_{\delta^2\le t_1^2+t_2^2\leq R^2}\frac{|z||f(t_1+x,t_2+y)-f(x,y)|}{(t_1^2+t_2^2+z^2)^{3/2}}dt_1dt_2\\ &\leq\varepsilon(2\sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|+|z|)\\ &+|z|\sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|\iint_{R^2\geq t_1^2+t_2^2\geq \delta}\frac{dt_1dt_2}{(\delta^2+z^2)^{3/2}}, \end{align*} so, for all $\varepsilon>0$ $\limsup_{z\to 0}|I(x,y,z)-f(x,y)|\leq 2\varepsilon \sup_{(s_1,s_2)\in\mathbb R^2}|f(s_1,s_2)|,$ which gives the result.