Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to follow an argument in Strichartz's "A Guide to Distribution Theory and Fourier Transforms"

We consider $\langle \Delta u, \rho \rangle$ where $\Delta u$ is the two dimensional Laplacian and $\rho$ is a compactly supported smooth test-function. Swiching into polar coordinates we obtain

$$ \langle \Delta u, \rho \rangle = \lim_{\epsilon \to 0 } \left( 2 \int_0^{2\pi} \rho(\epsilon, \theta) \; d\theta - \int_0^{2\pi} \epsilon \; \log(\epsilon^2) \frac{\partial \rho}{\partial r} (\epsilon,\theta) \; d\theta \right) $$ Now looking at the first term on the RHS the argument goes that since $\rho$ is continuous and so approaches the value at the origin as $\epsilon \to 0$ we have that $$ \lim_{\epsilon \to 0 } 2 \int_0^{2\pi} \rho(\epsilon, \theta) \; d\theta \to 4\pi \langle\delta, \rho) $$ where $\delta$ is the dirac delta function. (supposedly we move the limit in using dominated convergence ?)

I think I m missing a crucial point in the argument, because I don't see how $\rho$ approaches the value at the origin in the $\theta$ coordinate. I would be very grateful for a hint that could help me clarify this misunderstanding.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Let $\phi$ the function $\rho$ written with the usual coordinates. Then $\rho(\varepsilon,\theta)=\phi(\varepsilon\cos\theta,\varepsilon\sin\theta)$ and we can use continuity at $0$ to get the wanted result.

share|improve this answer
    
thanks a lot Davide ! Just to make sure I really get it, did you mean to write $\phi(\epsilon,\theta) = \rho(\epsilon cos \theta, \epsilon sin \theta ) $ ? –  Beltrame Sep 29 '12 at 13:21
1  
Yes, I have edited. –  Davide Giraudo Sep 29 '12 at 13:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.