The Sobolev conjugate exponent and the scaling identity

Let $n\ge 2$ and $1\le p<n$. The Sobolev conjugate of $p$ is $p^{\ast }=\frac{np}{n-p}\mathrm{.}$ The denominator $n-p$ is positive, so $p^{\ast }$ is a finite real number greater than $1$. Conjugate exponents in the sense of Conjugate exponents, including the endpoint conventions and real powers in the sense of Real powers for positive bases, with the zero-base positive-exponent convention are used below.

The following elementary identities are part of the definition and record how $p^{\ast }$ is used. $\frac{1}{p^{\ast }}=\frac{n-p}{np}=\frac{1}{p}-\frac{1}{n},p^{\ast }-p=\frac{np-p(n-p)}{n-p}=\frac{p^2}{n-p}>0,$ so $p^{\ast }>p$ and $1\mathrm{/}{p}^{\ast }+1\mathrm{/}n=1\mathrm{/}p$; in particular $p^{\ast }=n\mathrm{/}(n-1)$ when $p=1$. Equivalently $p^{\ast }=n/(\frac{n}{p}-1)$.

For $k\ge 1$ with $kp<n$ the $k$-fold Sobolev conjugate $p_k^{\ast }$ is defined recursively by $\frac{1}{p_1^{\ast }}=\frac{1}{p}-\frac{1}{n},\frac{1}{p_{j+1}^{\ast }}=\frac{1}{p_j^{\ast }}-\frac{1}{n}(1\le j<k)\mathrm{.}$ Induction on $j$ gives $\frac{1}{p_j^{\ast }}=\frac{1}{p}-\frac{j}{n}$ and hence $p_k^{\ast }=\frac{np}{n-kp};$ the recursion is well posed because $jp<n$ for $j\le k$ keeps every $p_j^{\ast }$ finite and greater than $1$.

Finally, the map $p\mapsto p^{\ast }=n\mathrm{/}(\frac{n}{p}-1)$ is continuous on $(1,n)$, since $n\mathrm{/}p-1>0$ there and the reciprocal and affine maps are continuous; it is strictly increasing on $(1,n)$, because $p\mapsto n\mathrm{/}p$ is strictly decreasing and $t\mapsto n\mathrm{/}(t-1)$ is strictly decreasing on $(1,\mathrm{\infty })$; and $p^{\ast }\to \mathrm{\infty }$ as $p\to n^{-}$, because $n\mathrm{/}p-1\to 0^{+}$. No finite Sobolev conjugate is attached to $p\ge n$: for $p=n$ the formula has denominator $0$, and for $p>n$ the expression $np\mathrm{/}(n-p)$ is negative.