Sharp $L^2\to L^6$ extension inequality conjecture on the circle

Let $\mathbb{S}^1$ be the unit circle with surface measure $\sigma$, let $\widehat{f\sigma}$ denote the Fourier extension of $f\in L^2(\mathbb{S}^1)$, and let $J_0$ be the Bessel function of the first kind of order zero. Define

$$C_{\mathrm{opt}}:=\sup_{f\in L^2(\mathbb{S}^1),\ f\neq 0}\frac{\|\widehat{f\sigma}\|_{L^6(\mathbb{R}^2)}^6}{\|f\|_{L^2(\mathbb{S}^1)}^6}.$$

Sharp $L^2\to L^6$ extension conjecture. For all $f\in L^2(\mathbb{S}^1)\setminus\{0\}$,

$$\frac{\|\widehat{f\sigma}\|_{L^6(\mathbb{R}^2)}^6}{\|f\|_{L^2(\mathbb{S}^1)}^6}\leq C_{\mathrm{opt}}=\frac{\|\widehat{\sigma}\|_{L^6(\mathbb{R}^2)}^6}{\|\boldsymbol{1}\|_{L^2(\mathbb{S}^1)}^6}=(2\pi)^4\left(\int_0^{\infty}J_0^6(r)r\,dr\right).$$

This asserts that constant functions attain the sharp extension constant in the remaining even-exponent case for the circle. Extremizers are known to exist and to have additional regularity and symmetry, but the stated sharp inequality was not assigned a resolution status in the source.