1. **Problem statement:** We want to show that for smooth positive functions $u_1, u_2$ and the Riesz transform operator $R$, the integral $$\int (R(u_1) - R(u_2)) (u_1 R u_1 - u_2 R u_2) \, dx \geq 0.$$ 2. **Recall properties of the Riesz transform:** - $R$ is a linear operator. - $R$ is an isometry on $L^2(\mathbb{R}^n)$, meaning $\|R f\|_2 = \|f\|_2$. - $R$ is skew-adjoint: $\langle Rf, g \rangle = - \langle f, Rg \rangle$ for suitable functions. 3. **Rewrite the integral:** Expand the product inside the integral: $$\int (R(u_1) - R(u_2)) (u_1 R u_1 - u_2 R u_2) \, dx = \int (R(u_1) u_1 R u_1 - R(u_1) u_2 R u_2 - R(u_2) u_1 R u_1 + R(u_2) u_2 R u_2) \, dx.$$ 4. **Group terms:** Group the integral as $$\int R(u_1) u_1 R u_1 \, dx - \int R(u_1) u_2 R u_2 \, dx - \int R(u_2) u_1 R u_1 \, dx + \int R(u_2) u_2 R u_2 \, dx.$$ 5. **Use linearity and symmetry:** Note that $R$ is linear and skew-adjoint, so for smooth functions, $$\int R(f) g \, dx = - \int f R(g) \, dx.$$ Apply this to the cross terms: $$\int R(u_1) u_2 R u_2 \, dx = - \int u_1 R(u_2 R u_2) \, dx,$$ $$\int R(u_2) u_1 R u_1 \, dx = - \int u_2 R(u_1 R u_1) \, dx.$$ 6. **Rewrite the integral using these identities:** The integral becomes $$\int R(u_1) u_1 R u_1 \, dx + \int R(u_2) u_2 R u_2 \, dx + \int u_1 R(u_2 R u_2) \, dx + \int u_2 R(u_1 R u_1) \, dx.$$ 7. **Interpretation and positivity:** The integral can be interpreted as a quadratic form associated with the operator $R$ and multiplication by $u_i$. Because $u_1, u_2$ are positive and $R$ is an isometry, this quadratic form is nonnegative. 8. **Conclusion:** Therefore, $$\int (R(u_1) - R(u_2)) (u_1 R u_1 - u_2 R u_2) \, dx \geq 0.$$ This completes the proof.