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

On the finite-measure ball, the Holder inequality givesSquaring and using $|B_r|={\omega }_3{r}^3$ yieldsSince $2=1+2\alpha$ for $\alpha =1/2$, this is (6), with $\boxed{C_3={\omega }_3^{2/3}}$.