# Laplacian in $\Bbb R^2$ acting on compact test-function

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.

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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.
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