Split the integral at radius $R$. The <Holder inequality> on the ball and the spatial moment outside it give
$$
\|v\|_2^2
\leq |B_R|^{1/2}\|v\|_4^2
+R^{-2}\||x|v\|_2^2
\leq\sqrt\pi R\|v\|_4^2+R^{-2}\|v\|_\Sigma^2.
$$
Taking $R=\varepsilon^{-1/2}$ proves the stated estimate with $C_\varepsilon=\sqrt\pi\varepsilon^{-1/2}$. Choose $\varepsilon<1/\omega$. If $X=\|v\|_4^2$, then
$$
J(v)\geq\frac{1-\omega\varepsilon}{2}\|v\|_\Sigma^2
+\frac14X^2-\frac{\omega C_\varepsilon}{2}X.
$$
The final two terms form a quadratic polynomial bounded below, so $\inf_\Sigma J> -\infty$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 1 / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-154/1/2/solution