> [!theoremb] Gagliardo-Nirenberg-Sobolev Inequality > > Let $U \subset \real^d$ be a bounded and open such that $\partial U$ is of class $C^1$[^1], $1 \le p < d$, $p^*$ be its [[Sobolev Conjugate|Sobolev conjugate]], and $u \in W^{1, p}(U)$. Then, there exists $C_{p, d, U} \ge 0$ such that > $ > \norm{u}_{p^*} \le C_{p, d, U}\norm{u}_{W^{1, p}(U)} > $ > the $p^*$ norm is bounded by the [[Sobolev Space|Sobolev norm]], so $u \in L^{p^*}$. # Proofs > [!theorem] Gagliardo-Nirinberg-Sobolev, p = 1 > > Let $p = 1$ and $p^* = d/(d - 1)$ be its [[Sobolev Conjugate|Sobolev conjugate]], then there exists $C \ge 0$ depending only on $p$ and $d$, such that > $ > \norm{u}_{L^{p^*}(\real^d)} \le C\norm{Du}_{L^p(\real^d)} > $ > for all $u \in C_c^1(\real^d)$, [$C^1$](Space%20of%20Continuously%20Differentiable%20Functions) with [[Compactly Supported|compact support]]. > > *Proof*. Since $u$ is compactly supported, for any $x \in \real^d$ and $j \in [d]$, > $ > \begin{align*} > u(x) &= \int_{(-\infty, x_i]}\partial_i u dy_i \\ > \abs{u(x)} &\le \int \norm{Du} dx_j > \end{align*} > $ > If $d = 1$, then we are done. Otherwise > $ > \abs{u(x)}^{d/(d-1)} \le \prod_{j \in [d]}\int \norm{Du}dx_j > $ > This satisfies the condition of the [[Iterated Hölder's Inequality]]. Therefore > $ > \norm{u}_{d/(d - 1)} \le \norm{Du}_1 > $ > [!theorem] Gagliardo-Nirenberg-Sobolev Inequality > > Let $p \in (1, n)$ and $p^* = pd/(d - p)$ be its [[Sobolev Conjugate|Sobolev conjugate]], then there exists $C \ge 0$ depending only on $p$ and $d$, such that > $ > \norm{u}_{L^{p^*}(\real^d)} \le C\norm{Du}_{L^p(\real^d)} > $ > for all $u \in C_c^1(\real^d)$, [$C^1$](Space%20of%20Continuously%20Differentiable%20Functions) with [[Compactly Supported|compact support]]. > > *Proof*. Let $\gamma > 1$, then the mapping $x \mapsto \abs{x}^\gamma$ is differentiable with derivative $x^{\gamma - 1}$. Denote > $ > L_p = P_p^{-1} \circ \mu \circ P_{p} > $ > then for any $q \ge 1$, > $ > L_q = P_{q/p}^{-1} \circ \underbrace{P_p^{-1} \circ \mu \circ P_p}_{L^p} \circ P_{q/p} > $ > > The chain rule allows applying the $p = 1$ case. > $ > \begin{align*} > L_{d/(d - 1)} \cdot (P_\gamma \cdot \abs{u}) &\le L_1 \cdot \norm{D\abs{u}^\gamma} = \gamma \angles{\abs{u}^{\gamma - 1}, \norm{Du}} > \end{align*} > $ > Let $q = p/(p - 1)$ be the Hölder conjugate of $p$, then by [[Hölder's Inequality]], > $ > \angles{\abs{u}^{\gamma - 1}, \norm{Du}} \le (L_q \circ P_{\gamma - 1} \cdot \abs{u}) \cdot L_p(\norm{Du}) > $ > Let $\gamma$ such that $\gamma d/(d - 1) = (\gamma - 1)q$, then there exists $\gamma > 1$ where > $ > \gamma = \frac{p(d - 1)}{d - p} \quad \frac{\gamma d}{d - 1} = \frac{dp}{d - p} = p^* > $ > is the Sobolev conjugate of $p$. Since > $ > \begin{align*} > L_{d/(d - 1)} \circ P_\gamma &= P_{d/(d-1)}^{-1} \circ \mu \circ P_{d/(d - 1)} \circ P_{\gamma} \\ > &= P_\gamma \circ L_{\gamma d/(d - 1)} = P_\gamma \circ L_{p^*} > \end{align*} > $ > On the other hand, > $ > \begin{align*} > L_q \circ P_{\gamma - 1} &= L_{q}^{-1} \circ \mu \circ L_q \circ P_{\gamma - 1} \\ > &= P_{\gamma - 1} \circ L_{q(\gamma - 1)} = P_{\gamma - 1} \circ L_{p^*} > \end{align*} > $ > This yields > $ > \begin{align*} > (P_{\gamma} \circ L_{p^*}) \cdot \abs{u} &\le \gamma \braks{(P_{\gamma - 1} \circ L_{p^*}) \cdot \abs{u}} \cdot L_p(\norm{Du}) \\ > \norm{u}_{p^*} &\le C\norm{Du}_{p} > \end{align*} > $ > [!theorem] > > Let $U \subset \real^d$ be a bounded and open such that $\partial U$ is of class $C^1$[^1], $1 \le p < n$, $p^*$ be its [[Sobolev Conjugate|Sobolev conjugate]], and $u \in W^{1, p}(U)$. Then, there exists $C \ge 0$, depending only on $p, n$ and $U$, such that > $ > \norm{u}_{p^*} \le C\norm{u}_{W^{1, p}(U)} > $ > therefore $u \in L^{p^*}$. > > *Proof*. Since $\partial U \in C^1$, there exists a $C \ge 0$ that only depends on $p$ and $U$, an extension operator, > $ > E: W^{1, p}(U) \to W^{1, p}(\real^d) \quad u \mapsto Eu = \ol{u} > $ > such $\norm{E} \le C$ that $\ol{u} \in W^{1, p}(\real^d)$ of [[Compactly Supported|compact support]] and $\norm{E} \le C$. Let $\seq{u_m} \subset C_c^\infty(\real^n)$ such that $u_m \to \ol{u}$ in $W^{1, p}(\real^d)$, then by the Gagliardo-Nirenberg Sobolev inequality and passing over equivalent norms, > $ > \norm{u_m - u_n}_{p^*} \le C \norm{Du_m - Du_n}_{L^p(\real^d)} \le C'\norm{u_m - u_n}_{W^{1, p}(U)} > $ > As $\norm{u_m - u_n}_{W^{1, p}} \to 0$ as $m, n \to \infty$, $\seq{u_m}$ is a [[Cauchy Sequence|Cauchy sequence]] in $L^{p^*}$. This means that $\seq{u_m}$ is a Cauchy sequence in $L^p \cap L^{p^*}$ (with respect to the norm $L_p + L_{p^*}$). By completeness of $L^{p} \cap L^{p^*}$, there exists $v \in L^p \cap L^{p^*}$ such that $u_m \to v$ in $L^p \cap L^{p^*}$. As the inclusion is continuous, $u_m \to v$ in $L^p$ and in $L^{p^*}$, thus $v = \ol{u}$, and $u_m \to \ol{u}$ in $L^{p^*}$. > > By continuity of norms, > $ > \norm{u}_{p^*} \le C' \norm{u}_{W^{1, p}} > $ [^1]: See [[Differentiable Boundary]].