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Let $u=(-R_2\theta,R_1\theta)$ where $R_1,R_2$ are the usual Riesz Transforms in $\mathbb{R}^2$, $\mathbb{T}^2$ denotes the torus, $\theta\in C^{\infty}(\mathbb{T}^2)$ and $-\theta=(-\nabla)^{\frac{1}{2}}\psi=\Lambda\psi$. How do I show that $$-\int_{\mathbb{T}^2}\Lambda\psi(\nabla^{\perp}\psi\nabla\phi)dx=\int_{\mathbb{T}^2}\nabla^{\perp}\psi\Lambda(\psi\nabla\phi)dx.$$

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Let's write the integrand in components: $$ \Lambda\psi(-\partial_2\psi\partial_1\theta+\partial_1\psi\partial_2\theta). $$ Now we integrate by parts (there are no boundary terms because of the periodicity) $$ \int \Lambda\psi(-\partial_2\psi\partial_1\theta+\partial_1\psi\partial_2\theta)=\int \partial_2(\Lambda\psi\partial_1\theta)\psi-\psi\partial_1(\Lambda\psi\partial_2\theta) $$ Developing the terms with parenthesis, we get $$ \int (\partial_2\Lambda\psi\partial_1\theta+\Lambda\psi\partial_2\partial_1\theta)\psi-\psi(\partial_1\Lambda\psi\partial_2\theta+\Lambda\psi\partial_1\partial_2\theta)=\int \psi\partial_2\Lambda\psi\partial_1\theta-\psi\partial_1\Lambda\psi\partial_2\theta $$ Writing in vectorial form: $$ -\int\Lambda\nabla^\perp\psi\nabla\theta\psi. $$ To conclude we use that $\Lambda$ is self-adjoint (to see this, you may use that the multiplier is a real valude function or equivalently, you may symmetrize the kernel expression)

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