On the finite-measure ball, the <Holder inequality> gives
$$
\|f\|_{L^{6/5}(B_r)}
\leq |B_r|^{1/3}\|f\|_{L^2(B_r)}.
$$
Squaring and using $|B_r|=\omega_3r^3$ yields
$$
\left(\int_{B_r}|f|^{6/5}\right)^{5/3}
\leq\omega_3^{2/3}r^2\int_{B_r}|f|^2.
$$
Since $2=1+2\alpha$ for $\alpha=1/2$, this is (6), with $\boxed{C_3=\omega_3^{2/3}}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 107 / 3 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-107/3/ii/solution