---- > [!definition] Definition. ([[Sobolev conjugate]]) > Assume $1 \leq p <n$. The **Sobolev conjugate** of $p$ is the number $p^{*}$, such that $\frac{n}{p}-\frac{n}{p^{*}}=1,$ > namely, $p^{*}=\frac{np}{n-p}$. ![[CleanShot 2025-11-16 at [email protected]]] Suppose that $\|u\|_{L^{q}(\Omega)} \leq C \|u\|_{W^{1,p}(\Omega)} \text{ for all }u \in W^{1,p}(\Omega).$ Then let us in particular consider, for $u \in C_{c}^{\infty}(\Omega)$, the modulations $u_{\lambda}=u(\lambda x)$, $\lambda \geq 1$, which are manifestly also in $C_{c}^{\infty}(\Omega)$. Extend $u$ by zero to all of $\mathbb{R}^{n}$. Then applying [[change of variables under a linear transformation]] with $h=|u|^{q}:\mathbb{R}^{n} \to \mathbb{R}$, we obtain $\begin{align} \|u_{\lambda}\|^{q}_{q}&= \int _{\mathbb{R}^{n}} |u(\lambda x)|^{q} \, dx \\ &= \frac{1}{\lambda^{n}} \int _{\mathbb{R}^{n}} |u(y)|^{q} \, dy \\ &= \frac{1}{\lambda^{n}} \int _{\Omega} |u(y)|^{q} \, dy \\ &=\lambda^{-n} \|u\|_{q}^{q}, \end{align}$ hence $\|u_{\lambda}\|_{q} = \lambda^{-n / q} \|u\|_{q}^{}$. Similarly, [[chain rule|we have]] $\|\nabla u_{\lambda}\|_{p}^{p}=\int _{\mathbb{R}^{n}} | \lambda (\nabla u) (\lambda x) |^{p} \, dx = \frac{\lambda^{p}}{\lambda^{n}} \int _{\mathbb{R}^{n} } |\nabla u(x)|^{p} \, dx = \lambda^{p-n} \|\nabla u\|_{p}^{p},$ hence $\|\nabla u _{\lambda}\|_{p} = \lambda^{1-n/p} \|\nabla u\|_{p}^{p}$. Thus $\begin{align} \|u_{\lambda}\|_{W^{1,p}(\Omega)}^{p} &= \|u_{\lambda}\|_{p}^{p} + \|\nabla u_{\lambda}\|_{p}^{p} \\ &= \lambda^{-n} \|u\|_{p}^{p} + \lambda^{p-n} \|\nabla u\|_{p}^{p} \\ &= \lambda^{p-n}( \lambda^{-p} \|u\|_{p}^{p} + \| \nabla u\|_{p}^{p} ). \end{align}$ Plugging in to the Sobolev-type inequality gives $\lambda^{-n/q}\|u\|_{q} \leq C (\lambda^{-n} \|u\|_{p}^{p} + \lambda^{p-n} \|\nabla u\|_{p}^{p})^{1/p} \text{ for all }u \in W^{1,p}(\Omega)$ Note that $p-n<0$ since $1 \leq p < n$. Since we assumed $\lambda \geq 1$, have $\lambda^{p-n}\geq \lambda^{-n}$, so $\|u_{\lambda}^{}\|_{1,p}^{p} \leq \lambda^{p-n}(\|u\|_{p}^{p} + \|\nabla u\|_{p}^{p}).$ Hence we further have $\lambda^{-n/q} \|u\|_{q} \leq \lambda^{(p-n)/ p} C(\|u\|_{p}^{p} + \|\nabla u\|_{p}^{p})^{1/p}$ that is $\|u\|_{q} \leq C(\|u\|_{p}^{p} + \|\nabla u\|_{p}^{p})^{1/p} \lambda^{\frac{p-n}{p}+\frac{n}{q}}.$ Writing $\alpha=\frac{p-n}{p}+\frac{n}{q}$, this inequality says $\|u\|_{q} \leq C(\|u\|_{p}^{p} + \|\nabla u\|_{p}^{p})^{1/p} \lambda^{\alpha} .$ If $\alpha<0$ then taking $\lambda \in \infty$ gives $\|u\|_{q}=0$, a contradiction (just as in lecture). So $\alpha \geq 0$, i.e., $\frac{n}{q} \geq \frac{n-p}p .$ Dividing both sides by $q$ yields the desired result. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```