# Does the laplacian of a $C^2$ function with compact support uniquely determine the second derivatives?

Let $$f\in C^2(\mathbb R^n)$$ be a function with compact support. Can we recover all of the partial derivatives $$f_{x_i x_j}$$ just from knowing the Laplacian $$\sum_{i=1}^n f_{x_i x_i}$$?

This sounds absurd, but the Riesz transforms do exactly that. I am asking for an intuitive explanation of how that can be, or a simpler proof.

• I admit that I had to look on the Wikipedia to make full sense of your last comment, here; en.wikipedia.org/wiki/… May 25 at 17:08
• There is an important point to be made, which your formulation of the problem slightly obscures. To do what you suggest, it is necessary to know the Laplacian at all points. The way you phrased it, it seems that just knowing the Laplacian at a point you can reconstruct all derivatives there, which of course is false. Still, I agree with you that it is a surprising thing. May 25 at 17:11
• Since there are no harmonic functions with compact support, you can recover $f$ from its laplacian May 25 at 17:51

Let us consider the 2d case only. This is only due to my laziness in typing and is not a fundamental obstruction. So, we have a function $$f\colon \mathbb R^2\to \mathbb C$$ which is smooth and compactly supported. We denote its Fourier transform by $$\hat{f}(\xi, \eta)=\int_{\mathbb R^2} f(x, y)e^{-ix\xi-iy\eta}\, dxdy.$$
Now, knowing the Laplacian $$L(x, y)=-\Delta f(x, y)$$ at all points $$(x, y)\in \mathbb R^2$$ amounts to knowing its Fourier transform $$\widehat{L}(\xi, \eta)=(\xi^2+\eta^2)\hat{f}(\xi, \eta)$$ at all $$(\xi, \eta)\in\mathbb R^2$$. From this, we want to recover the Hessian matrix of $$f$$, which corresponds by Fourier transform to the matrix $$\tag{1} -\begin{bmatrix} \xi^2 \hat{f}(\xi, \eta) & \xi\eta \hat{f}(\xi, \eta) \\ \xi \eta \hat{f}(\xi, \eta) & \eta^2 \hat{f}(\xi, \eta)\end{bmatrix}.$$ Now, the Riesz transform $$R=(R_x, R_y)$$ is the operator $$\widehat{Rf}(\xi, \eta)=\left( \frac{\xi \hat{f}}{\sqrt{\xi^2+\eta^2}}, \frac{\eta \hat{f}}{\sqrt{\xi^2+\eta^2}}\right).$$ So we can compute directly that $$\widehat{R_xR_x L}(\xi, \eta) = \frac{\xi^2+\eta^2}{(\sqrt{\xi^2+\eta^2})^2}\xi^2 \hat{f}(\xi, \eta)=\xi^2\hat{f}(\xi, \eta),$$ and similarly $$\widehat{R_y R_y L}(\xi, \eta)=\eta^2\hat{f}(\xi, \eta),$$ and $$\widehat{R_xR_y L}(\xi, \eta)=\xi\eta \hat{f}(\xi, \eta).$$
We have thus recovered the Hessian matrix (1) from the function $$L$$ by applying Riesz transforms. At Fourier side, this is a purely algebraic process and it looks much less mysterious.
As @SolubleFish wrote in a comment, $$\Delta f$$ determines the whole function $$f$$ because there are no harmonic functions with compact support.