For $1\leq p<n$, define the <Sobolev conjugate exponent>
$$
p^*=\frac{np}{n-p},
\qquad
\frac1{p^*}=\frac1p-\frac1n.
$$
The <Sobolev inequality>, also called the Gagliardo--Nirenberg--Sobolev inequality, states that there is a constant $C=C(n,p)$ such that
$$
\boxed{\|u\|_{L^{p^*}(\mathbb R^n)}\leq C\|Du\|_{L^p(\mathbb R^n)}}
$$
for every $u\in C_c^\infty(\mathbb R^n)$, and hence by completion for every $u\in W^{1,p}(\mathbb R^n)$ for which the right formulation applies.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 1 / c / i / Solution

{title2=$p^*=np/(n-p)$}

For $1\leq p<n$, the Sobolev conjugate exponent $p^*$ is defined by
$$
\frac1{p^*}=\frac1p-\frac1n,
\qquad
p^*=\frac{np}{n-p}.
$$
The first-order <Sobolev inequality> maps one derivative in $L^p(\mathbb R^n)$ to the function in $L^{p^*}(\mathbb R^n)$.

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