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I am trying to make sense of the following operator, acting un the ser of continuos functions from $\mathbb{R}^2$ to $\mathbb{R}$: $$\mathbf{L}[u]=lim_{r\to0}\frac{1}{r^2}(\frac{1}{r^2}\oint_{\left \| y \right \|<r}^{ } (u(x_1+y_1,x_2+y_2)-u(x_1,x_2))dy_1dy_2)$$

Note that the integral is the mean of the values taken in a ball of radius r minus the value at the centre, so by the property of the mean value of armonic functions L[u] vanishes if u is armonic.

Because of this, I am thinking this operator might be some convoluted way to write the Laplacian, and if not, I'm interested on its relation to the Laplacian anyway.

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Suppose that $u$ is twice-differentiable (else it's rather hard to make sense of what its Laplacian would be). WLOG we can assume $(x_1,x_2)=(0,0)$. Consider the integral over the circle of radius $\rho$ (we can then calculate the radial integral later): put $y_1=\rho\cos{\theta}$, $y_2 = \rho\sin{\theta}$, so $dy_1 \, dy_2 = \rho \, d\rho \, d\theta $, so the angular integral becomes $$ \int_0^{2\pi} (u(\rho\cos{\theta},\rho\sin{\theta})-u(0,0))\rho \, d\theta. $$ We now expand $u$ to second order in $\rho$: $$ u(\rho\cos{\theta},\rho\sin{\theta}) = u(0,0) + \rho(u_1\cos{\theta}+u_2\sin{\theta}) + \frac{1}{2}\rho^2( u_{11}\cos^2{\theta} + 2u_{12}\cos{\theta}\sin{\theta} + u_{22}\sin^2{\theta} ) + o(\rho^2) $$ where $u_1=\frac{\partial u}{\partial y_1}(0,0)$ and so on. Now we can plug this into the integral and integrate term-by-term: $$ \int_0^{2\pi} (u(\rho\cos{\theta},\rho\sin{\theta})-u(0,0))\rho \, d\theta \\ = \int_0^{2\pi} \left( \rho(u_1\cos{\theta}+u_2\sin{\theta}) + \frac{1}{2}\rho^2( u_{11}\cos^2{\theta} + 2u_{12}\cos{\theta}\sin{\theta} + u_{22}\sin{\theta} ) + o(\rho^2) \right) \rho \, d\theta \\ = 0\rho^2 + \frac{\pi}{2} (u_{11}+u_{22})\rho^3 + o(\rho^3) = \pi \rho^3 \Delta u(0,0) + o(\rho^3), $$ where $\Delta u$ is the Laplacian, since $$\int_0^{2\pi} \cos{\theta} \, d\theta = \int_0^{2\pi} \sin{\theta} \, d\theta = \int_0^{2\pi} \sin{\theta}\cos{\theta} \, d\theta = 0 \\ \int_0^{2\pi} \cos^2{\theta} \, d\theta = \int_0^{2\pi} \sin^2{\theta} \, d\theta = \pi. $$ So we have $$\lim_{r \to 0} \frac{1}{r^3} \int_{\lVert \mathbf{y} \rVert = r} (u(\mathbf{x}+\mathbf{y})-u(\mathbf{x})) \, dy = \frac{\pi}{2} \Delta u(\mathbf{x}), $$ while integrating from $0$ to $r$ gives $$ \int_0^r\int_0^{2\pi} (u(\rho\cos{\theta},\rho\sin{\theta})-u(0,0))\rho \, d\theta \, d\rho \\ = \pi \frac{r^4}{8} \Delta u(0,0) + o(r^4), $$ which gives $$ \lim_{r \to 0} \frac{1}{r^4} \int_{\lVert \mathbf{y} \rVert \leq r} (u(\mathbf{x}+\mathbf{y})-u(\mathbf{x})) \, dy = \frac{\pi}{8} \Delta u(\mathbf{x}). $$

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