> [!definition] > > Let $W^{k, p}(\real^d)$ be the [[Sobolev Space|Sobolev space]] on $\real^d$ with $p \in [1, n)$ and suppose that an inequality of the following form holds > $ > \norm{u}_{L^q(\real^d)} \le C\norm{Du}_{L^p(\real^d)} > $ > for some $C > 0$, $q \in [1, \infty)$, and all $u \in C_c^\infty(\real^d)$, where $C$ and $q$ do not depend on $u$. Then, > $ > p^* = q = \frac{dp}{d - p} \quad \frac{1}{p^*} = \frac{1}{q} = \frac{1}{p} - \frac{1}{n} > $ > is the **Sobolev conjugate** of $p$. > > *Proof*. Let $u_t(x) = u(tx)$, then > $ > \norm{u_t}_q^q = \frac{1}{t^d}\norm{u}_q^q \quad \norm{Du_t}_p^p = t^p\norm{(Du)_t}_p^p = \frac{t^p}{t^d}\norm{Du}_p^p > $ > Plugging this into the inequality yields > $ > \frac{1}{t^{d/q}}\norm{u}_q \le C \frac{t}{t^{d/p}}\norm{Du}_p \quad \norm{u}_q \le Ct^{1 - d/p + d/q}\norm{Du}_p > $ > If $1 - d/p + d/q \ne 0$, then sending $t$ to $0$ or $\infty$ implies that $\norm{Du}_p = \infty$ or $\norm{u}_q = 0$. Thus if this equality holds, then $1 - d/p + d/q = 0$.