Let $u_0: \Omega \subset \mathbb{R}^2 \to \mathbb{R}^2, n \geq 1 $, be a continuous function and the functional $E$ given by $$ E(u) = \int_\Omega \dfrac{1}{2} \| u(x) - u_0(x) \|_2^2 dx - \dfrac{\gamma}{4} \int_\Omega \int_\Omega w(x-y) \| u(x) - u(y) \|_2^2 dx dy $$ with $w(\cdot) $ is a normalized Gaussian kernel of standard deviation $\sigma$. Determinate an expression of the minimum of the functional $E$ using Fourier.
Ideas
I update the minimum using gradient descent finding the expression for $\nabla E$ but there is a direct solution using Fourier. The first part of the expression using Fourier is
$$ \dfrac{1}{2} \| F(u) - F(u_0) \|^2 $$
but I'm stuck with the second part.
Perhaps just a partial answer as I will consider $\Omega = \Bbb R^d$. Let $\mathcal F(u) = \widehat{u}$ denote the Fourier transform and $|x| = \sqrt{x_1^2+x_2^2}$ denote the Euclidean norm in $\Bbb R^2$. Then since $$ \iint w(x-y)\, |u(x)-u(y)|^2\,\mathrm d x\,\mathrm d y = 2\int w\int |u|^2 - 2\int (w*u) \cdot u $$ so if $v=\widehat u$, one gets $$ E(u) = \frac{1}{2}\int|v-v_0|^2 - \gamma\left(\widehat{w}(0)-\widehat{w}\right)\,|v|^2. $$ The associated Euler Lagrange problem gives $$ \left(1 - \gamma \left(\widehat{w}(0)-\widehat{w}\right)\right)v = v_0 $$ that is $$ u = \mathcal{F}\left(\frac{1}{1 - \gamma \left(\widehat{w}(0)-\widehat{w}\right)}\right) * u_0. $$