[proofplan] We prove the Gagliardo-Nirenberg-Sobolev inequality in three stages. First, we establish the critical case $p = 1$ for compactly supported smooth functions by integrating partial derivatives along each coordinate axis via the Fundamental Theorem of Calculus, then combining the resulting one-dimensional bounds with the Generalised Holder Inequality to produce the $L^{n/(n-1)}$ estimate. Second, we reduce the general case $1 < p < n$ to the $p = 1$ result by applying the established inequality to the auxiliary function $|u|^\gamma$ for a carefully chosen exponent $\gamma = p(n-1)/(n-p)$, then separating the resulting integral via Holder's inequality with conjugate pair $(p, p/(p-1))$. Finally, a density argument extends the inequality from $C_c^\infty(\mathbb{R}^n)$ to the full Sobolev space $W^{1,p}(\mathbb{R}^n)$. [/proofplan] [step:Establish the $p = 1$ inequality for compactly supported smooth functions] Let $u \in C_c^1(\mathbb{R}^n)$. Since $u$ has compact support, $u(x) \to 0$ as $|x| \to \infty$ along every coordinate direction. For each fixed $x \in \mathbb{R}^n$ and each index $i \in \{1, \dots, n\}$, the [Fundamental Theorem of Calculus](/theorems/632) applied along the $i$-th coordinate axis yields \begin{align*} u(x) = \int_{-\infty}^{x_i} \partial_i u(x_1, \dots, x_{i-1}, t, x_{i+1}, \dots, x_n) \, d\mathcal{L}^1(t). \end{align*} Taking absolute values and enlarging the integration domain from $(-\infty, x_i]$ to $\mathbb{R}$: \begin{align*} |u(x)| \le \int_{-\infty}^{\infty} |\partial_i u(x_1, \dots, x_{i-1}, y_i, x_{i+1}, \dots, x_n)| \, d\mathcal{L}^1(y_i). \end{align*} This holds for each $i = 1, \dots, n$. Raising each bound to the power $1/(n-1)$ and multiplying over all $i$: \begin{align*} |u(x)|^{n/(n-1)} \le \prod_{i=1}^n \left( \int_{-\infty}^{\infty} |\partial_i u(x_1, \dots, x_{i-1}, y_i, x_{i+1}, \dots, x_n)| \, d\mathcal{L}^1(y_i) \right)^{1/(n-1)}. \end{align*} We integrate over $x_1 \in \mathbb{R}$ with respect to $\mathcal{L}^1$. The factor corresponding to $i = 1$ does not depend on $x_1$ (it integrates over $y_1$), so it factors out. The remaining $n - 1$ factors are functions of $x_1$ raised to the power $1/(n-1)$. We apply the Generalised Holder Inequality with $n - 1$ functions each in $L^{n-1}$, since $\sum_{i=2}^n 1/(n-1) = 1$: \begin{align*} \int_{-\infty}^{\infty} |u(x)|^{n/(n-1)} \, d\mathcal{L}^1(x_1) \le \left( \int_{-\infty}^{\infty} |\partial_1 u| \, d\mathcal{L}^1(y_1) \right)^{1/(n-1)} \prod_{i=2}^n \left( \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |\partial_i u| \, d\mathcal{L}^1(y_i) \, d\mathcal{L}^1(x_1) \right)^{1/(n-1)}. \end{align*} Repeating this integration-and-factoring procedure successively over $x_2, x_3, \dots, x_n$, applying the Generalised Holder Inequality at each stage, we obtain after integrating over all $n$ variables: \begin{align*} \int_{\mathbb{R}^n} |u(x)|^{n/(n-1)} \, d\mathcal{L}^n(x) \le \prod_{i=1}^n \left( \int_{\mathbb{R}^n} |\partial_i u(x)| \, d\mathcal{L}^n(x) \right)^{1/(n-1)}. \end{align*} Taking the $(n-1)/n$-th power of both sides: \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \left( \prod_{i=1}^n \|\partial_i u\|_{L^1(\mathbb{R}^n)} \right)^{1/n}. \end{align*} The arithmetic-geometric mean inequality gives $\left( \prod_{i=1}^n a_i \right)^{1/n} \le \frac{1}{n} \sum_{i=1}^n a_i$ for $a_i \ge 0$. The norm equivalence $\sum_{i=1}^n \|\partial_i u\|_{L^1} = \|\nabla u\|_{L^1(\mathbb{R}^n)}$ (where the $L^1$ norm of the gradient is defined as $\int_{\mathbb{R}^n} |\nabla u| \, d\mathcal{L}^n$) yields \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \frac{1}{n} \|\nabla u\|_{L^1(\mathbb{R}^n)}. \end{align*} This establishes the theorem for $p = 1$ with Sobolev conjugate $1^* = n/(n-1)$. [guided] The goal of this step is to prove the GNS inequality in the special case $p = 1$, where the Sobolev conjugate exponent is $1^* = n/(n-1)$. The strategy is purely one-dimensional: we express $u(x)$ as an integral of its partial derivative along each coordinate axis, then combine these $n$ one-dimensional bounds into a single $n$-dimensional estimate. Since $u \in C_c^1(\mathbb{R}^n)$ has compact support, $u$ vanishes outside a large ball, so for each fixed $x \in \mathbb{R}^n$ and each coordinate index $i \in \{1, \dots, n\}$, the [Fundamental Theorem of Calculus](/theorems/632) gives \begin{align*} u(x) = \int_{-\infty}^{x_i} \partial_i u(x_1, \dots, x_{i-1}, t, x_{i+1}, \dots, x_n) \, d\mathcal{L}^1(t). \end{align*} Taking absolute values and extending the upper limit from $x_i$ to $+\infty$ (which only enlarges the integral since the integrand is non-negative): \begin{align*} |u(x)| \le \int_{-\infty}^{\infty} |\partial_i u(x_1, \dots, x_{i-1}, y_i, x_{i+1}, \dots, x_n)| \, d\mathcal{L}^1(y_i). \end{align*} Denote the right-hand side by $I_i(x)$. This bound holds for every $i = 1, \dots, n$, so we have $n$ independent upper bounds on $|u(x)|$. The key idea is to combine them multiplicatively. Raising each to the power $1/(n-1)$ and multiplying: \begin{align*} |u(x)|^{n/(n-1)} \le \prod_{i=1}^n I_i(x)^{1/(n-1)}. \end{align*} Why exponent $1/(n-1)$? Because we have $n$ factors, each raised to $1/(n-1)$, and $n \cdot \frac{1}{n-1} = \frac{n}{n-1} = 1^*$, which is exactly the Sobolev conjugate for $p = 1$. Now we integrate over $\mathbb{R}^n$, one variable at a time. Integrating first over $x_1$: the factor $I_1(x)^{1/(n-1)}$ depends only on $(x_2, \dots, x_n)$ (it integrates over $y_1$, not $x_1$), so it pulls outside the $x_1$-integral. The remaining $n - 1$ factors $I_2^{1/(n-1)}, \dots, I_n^{1/(n-1)}$ each depend on $x_1$, and each is raised to the power $1/(n-1)$. Since $\sum_{i=2}^n 1/(n-1) = 1$, the Generalised Holder Inequality applies with $n - 1$ functions each in $L^{n-1}$: \begin{align*} \int_{\mathbb{R}} \prod_{i=2}^n I_i(x)^{1/(n-1)} \, d\mathcal{L}^1(x_1) \le \prod_{i=2}^n \left( \int_{\mathbb{R}} I_i(x) \, d\mathcal{L}^1(x_1) \right)^{1/(n-1)}. \end{align*} Each inner integral $\int_{\mathbb{R}} I_i(x) \, d\mathcal{L}^1(x_1) = \int_{\mathbb{R}} \int_{\mathbb{R}} |\partial_i u| \, d\mathcal{L}^1(y_i) \, d\mathcal{L}^1(x_1)$ is a double integral over two of the $n$ variables. We repeat this procedure for $x_2, x_3, \dots, x_n$, each time pulling out one factor and applying Holder to the rest. After all $n$ integrations, each factor has been integrated over all $n$ variables: \begin{align*} \int_{\mathbb{R}^n} |u(x)|^{n/(n-1)} \, d\mathcal{L}^n(x) \le \prod_{i=1}^n \left( \int_{\mathbb{R}^n} |\partial_i u(x)| \, d\mathcal{L}^n(x) \right)^{1/(n-1)}. \end{align*} Taking the $(n-1)/n$-th power gives $\|u\|_{L^{n/(n-1)}} \le \left( \prod_{i=1}^n \|\partial_i u\|_{L^1} \right)^{1/n}$. Applying the arithmetic-geometric mean inequality $(\prod a_i)^{1/n} \le \frac{1}{n} \sum a_i$ and noting $\sum_{i=1}^n \|\partial_i u\|_{L^1} = \|\nabla u\|_{L^1}$: \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \frac{1}{n} \|\nabla u\|_{L^1(\mathbb{R}^n)}. \end{align*} [/guided] [/step] [step:Reduce the general case $1 < p < n$ to the $p = 1$ result via a power substitution] Let $u \in C_c^1(\mathbb{R}^n)$ and define the exponent $\gamma = p(n-1)/(n-p)$. Since $1 < p < n$, we verify $\gamma > 1$: indeed $\gamma = p(n-1)/(n-p) > p \cdot 1 / (n-p) \cdot (n-1) > 1$ since $p(n-1) > n - p$ follows from $pn - p > n - p$, i.e., $pn > n$, which holds for $p > 1$. Define the auxiliary function \begin{align*} v : \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto |u(x)|^\gamma. \end{align*} Since $u \in C_c^1(\mathbb{R}^n)$ and $\gamma > 1$, the function $v$ lies in $C_c^1(\mathbb{R}^n)$ with gradient \begin{align*} |\nabla v(x)| = \gamma |u(x)|^{\gamma - 1} |\nabla u(x)|. \end{align*} Applying the $p = 1$ inequality established above to $v$: \begin{align*} \left( \int_{\mathbb{R}^n} |u|^{\gamma n/(n-1)} \, d\mathcal{L}^n \right)^{(n-1)/n} \le \frac{\gamma}{n} \int_{\mathbb{R}^n} |u|^{\gamma - 1} |\nabla u| \, d\mathcal{L}^n. \end{align*} The exponent on the left-hand side satisfies $\gamma n/(n-1) = p(n-1)/(n-p) \cdot n/(n-1) = np/(n-p) = p^*$. We apply Holder's inequality to the right-hand side with the conjugate pair $(p, p' = p/(p-1))$, pairing $|\nabla u|$ with $|u|^{\gamma - 1}$: \begin{align*} \int_{\mathbb{R}^n} |u|^{\gamma - 1} |\nabla u| \, d\mathcal{L}^n \le \left( \int_{\mathbb{R}^n} |\nabla u|^p \, d\mathcal{L}^n \right)^{1/p} \left( \int_{\mathbb{R}^n} |u|^{(\gamma - 1)p/(p-1)} \, d\mathcal{L}^n \right)^{(p-1)/p}. \end{align*} We verify the exponent: $(\gamma - 1) \cdot p/(p-1) = \left( \frac{p(n-1)}{n-p} - 1 \right) \frac{p}{p-1} = \frac{pn - p - n + p}{n - p} \cdot \frac{p}{p-1} = \frac{n(p-1)}{n-p} \cdot \frac{p}{p-1} = \frac{np}{n-p} = p^*$. Therefore the inequality reads \begin{align*} \|u\|_{L^{p^*}}^{\gamma} \le \frac{\gamma}{n} \|u\|_{L^{p^*}}^{\gamma - 1} \|\nabla u\|_{L^p}. \end{align*} Since $u$ has compact support, $\|u\|_{L^{p^*}} < \infty$. Dividing both sides by $\|u\|_{L^{p^*}}^{\gamma - 1}$ (when $u \equiv 0$, both sides vanish and the inequality holds): \begin{align*} \|u\|_{L^{p^*}(\mathbb{R}^n)} \le \frac{\gamma}{n} \|\nabla u\|_{L^p(\mathbb{R}^n)}. \end{align*} Setting $C = \gamma/n = p(n-1)/(n(n-p))$ completes the inequality for $C_c^1$ functions. [guided] The idea is to bootstrap from the $p = 1$ case to general $p$ by applying the already-proven inequality to a power of $|u|$. The exponent $\gamma$ is chosen so that the $L^{1^*}$ norm of $|u|^\gamma$ becomes the $L^{p^*}$ norm of $u$. Define $\gamma = p(n-1)/(n-p)$ and $v(x) = |u(x)|^\gamma$. Why this choice? We need $\gamma \cdot 1^* = \gamma \cdot n/(n-1) = p^*$. Solving: $\gamma = p^* \cdot (n-1)/n = \frac{np}{n-p} \cdot \frac{n-1}{n} = \frac{p(n-1)}{n-p}$. Since $u \in C_c^1(\mathbb{R}^n)$ and $\gamma > 1$, the chain rule gives $\nabla v = \gamma |u|^{\gamma-1} (\operatorname{sgn} u) \nabla u$, so $|\nabla v| = \gamma |u|^{\gamma-1} |\nabla u|$. The function $v$ lies in $C_c^1(\mathbb{R}^n)$ because $\gamma > 1$ ensures differentiability at zeros of $u$. Applying the $p = 1$ GNS inequality: \begin{align*} \|v\|_{L^{n/(n-1)}} \le \frac{1}{n} \|\nabla v\|_{L^1}. \end{align*} The left-hand side equals $\left( \int_{\mathbb{R}^n} |u|^{\gamma n/(n-1)} \, d\mathcal{L}^n \right)^{(n-1)/n} = \|u\|_{L^{p^*}}^{\gamma}$ since $\gamma n/(n-1) = p^*$. The right-hand side equals $\frac{\gamma}{n} \int_{\mathbb{R}^n} |u|^{\gamma-1} |\nabla u| \, d\mathcal{L}^n$. Now we need to separate $|u|^{\gamma-1}$ from $|\nabla u|$ using Holder's inequality. The natural pairing is to put $|\nabla u|$ in $L^p$ and $|u|^{\gamma-1}$ in $L^{p'}$ where $p' = p/(p-1)$ is the Holder conjugate. This requires computing $(\gamma-1) \cdot p' = (\gamma - 1) \cdot p/(p-1)$. Substituting $\gamma = p(n-1)/(n-p)$: \begin{align*} (\gamma - 1) \frac{p}{p-1} = \frac{p(n-1) - (n-p)}{n-p} \cdot \frac{p}{p-1} = \frac{pn - n}{n-p} \cdot \frac{p}{p-1} = \frac{n(p-1)}{n-p} \cdot \frac{p}{p-1} = \frac{np}{n-p} = p^*. \end{align*} This is exactly $p^*$, so the $L^{p'}$ factor becomes $\|u\|_{L^{p^*}}^{\gamma - 1}$. After applying Holder: \begin{align*} \|u\|_{L^{p^*}}^{\gamma} \le \frac{\gamma}{n} \|u\|_{L^{p^*}}^{\gamma - 1} \|\nabla u\|_{L^p}. \end{align*} Dividing by $\|u\|_{L^{p^*}}^{\gamma - 1}$ (justified because $u$ has compact support, so the $L^{p^*}$ norm is finite; if $\|u\|_{L^{p^*}} = 0$ then $u = 0$ a.e. and the inequality holds with both sides equal to zero): \begin{align*} \|u\|_{L^{p^*}(\mathbb{R}^n)} \le \frac{p(n-1)}{n(n-p)} \|\nabla u\|_{L^p(\mathbb{R}^n)}. \end{align*} [/guided] [/step] [step:Extend to all of $W^{1,p}(\mathbb{R}^n)$ by density] Since $C_c^\infty(\mathbb{R}^n)$ is dense in $W^{1,p}(\mathbb{R}^n)$, for any $u \in W^{1,p}(\mathbb{R}^n)$ there exists a sequence $(u_m)_{m=1}^\infty \subset C_c^\infty(\mathbb{R}^n)$ with $\|u_m - u\|_{W^{1,p}(\mathbb{R}^n)} \to 0$. The inequality holds for each $u_m$: \begin{align*} \|u_m\|_{L^{p^*}(\mathbb{R}^n)} \le C \|\nabla u_m\|_{L^p(\mathbb{R}^n)}. \end{align*} The sequence $(u_m)$ is Cauchy in $L^{p^*}(\mathbb{R}^n)$: for any $m, l \in \mathbb{N}$, \begin{align*} \|u_m - u_l\|_{L^{p^*}} \le C \|\nabla(u_m - u_l)\|_{L^p} \le C \|u_m - u_l\|_{W^{1,p}} \to 0. \end{align*} Since $L^{p^*}(\mathbb{R}^n)$ is complete, $u_m \to \tilde{u}$ in $L^{p^*}$. Because $u_m \to u$ in $L^p$ and both limits must agree a.e. (by extracting a common subsequence converging a.e.), we have $\tilde{u} = u$ a.e. Passing to the limit in the inequality: \begin{align*} \|u\|_{L^{p^*}(\mathbb{R}^n)} = \lim_{m \to \infty} \|u_m\|_{L^{p^*}(\mathbb{R}^n)} \le C \lim_{m \to \infty} \|\nabla u_m\|_{L^p(\mathbb{R}^n)} = C \|\nabla u\|_{L^p(\mathbb{R}^n)}. \end{align*} [/step]