In some old lecture notes on the Restriction and Kakeya conjectures (Notes 5, specifically), Terence Tao gives a proof of the restriction conjecture (for the sphere) in two dimensions via a bilinear argument. Specifically, he proves the weak-type estimate $$\|\widehat{\chi_{\Omega}d\sigma}\|_{L^{q}(\mathbb{R}^{2})}\lesssim |\Omega|^{1/p},$$ where $\Omega\subset S^{1}$ and $q>4$, $q=3p'$.
Briefly, the idea is to square the above estimate, and instead consider estimates for the bilinear quantities $\|\widehat{fd\sigma}\widehat{gd\sigma}\|_{L^{\infty}(\mathbb{R}^{2})}$ and $\|\widehat{fd\sigma}\widehat{gd\sigma}\|_{L^{2}(\mathbb{R}^{2})}$, where $f,g$ are functions on $S^{1}$. The first is trivial. The second is proven by first showing that if $f,g$ are supported on distinct $\theta$-arcs with separation $\sim\theta$, then $$\|\widehat{fd\sigma}\widehat{gd\sigma}\|_{L^{2}(\mathbb{R}^{2})}\lesssim\theta^{-1/2}\|f\|_{L^{2}(S^{1})}\|g\|_{L^{2}(S^{1})}$$ To exploit this estimate, he then introduces a Whitney decomposition of $S^{1}$ into dyadic arcs, defining distinct arcs $I\simeq J$ if $I$ and $J$ belong to the same generation $A_{n}$, are not adjacent, but their parents are. From the observation $$\widehat{\chi_{\Omega}d\sigma}\widehat{\chi_{\Omega}d\sigma}=\sum_{n>1}\sum_{I,J\in A_{n}: I\sim J}\widehat{\chi_{\Omega}d\sigma_{I}}\widehat{\chi_{\Omega}d\sigma_{J}}$$ one obtains via almost orthogonality that $$\|\sum_{I,J\in A_{n} : I\sim J}\widehat{\chi_{\Omega}d\sigma_{I}}\widehat{\chi_{\Omega}d\sigma_{J}}\|_{L^{2}(\mathbb{R}^{2})}\lesssim\left(\sum_{I,J\in A_{n} : I\sim J}\|\widehat{\chi_{\Omega}d\sigma}\widehat{\chi_{\Omega}d\sigma_{J}}\|_{L^{2}(\mathbb{R}^{2})}^{2}\right)^{1/2}\lesssim 2^{n/2}\left(\sum_{I,J\in A_{n} : I\sim J}|\Omega\cap I||\Omega\cap J|\right)^{1/2}$$ One concludes by Holder's inequality and summing over $n$.
At the end of this proof, Tao briefly mentions that one can prove the Bochner-Riesz conjecture in two dimensions by a similar argument. Can anyone provide me with a reference for where this is sketched out?