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.
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.