Consider the Poisson equation $-\Delta u=f$ on $\mathbb R^3$.Show that a bounded solution of this equation satisfies $\int|\nabla u(x)|^2dx=\frac{1}{4\pi}\int\int\frac{f(x)f(y)}{|x-y|^\lambda}dxdy$ for some $\lambda$.
Here is what I tried. I know bounded solution of 3-d Poisson equation is of the form $$u(x)=\frac{1}{4\pi}\int \frac{1}{|x-y|}f(y)dy=\frac{1}{4\pi}\int \frac{1}{|y|}f(x-y)dy$$. $$\frac{\partial u}{\partial x_i}(x)=\frac{1}{4\pi}\int \frac{1}{|y|}\frac{\partial f(x-y)}{\partial x_i}dy$$ Hence $$(\frac{\partial u}{\partial x_i}(x))^2=(\frac{1}{4\pi}\int \frac{1}{|y|}\frac{\partial f(x-y)}{\partial x_i}dy)^2$$ But I don't know how to integrate $|\nabla u(x)|^2$. Any advice is appreciated.
To integrate $|\nabla u |^2$ use integration by parts.
First note that, $$ \nabla\cdot (u \nabla u) = \nabla u \cdot \nabla u + u \nabla^2 u $$
Then we can write,
$$ \int_R |\nabla u |^2 \ d\tau = \int_R \nabla \cdot (u \nabla u) - u \nabla^2 u \ d\tau = \int_{\partial R} u \nabla u \cdot d\sigma - \int_R u \nabla^2 u $$
The boundary integral can be eliminated using your boundary conditions (usually either $u$ is zero on $\partial R$ or its normal derivative is).
$$ \int_R |\nabla u |^2 \ d\tau = - \int_R u \nabla^2 u $$
Based on your work, you should be able to finish the problem from here.