[proofplan] We prove the critical embedding $W^{1,n}(U) \subset L^q(U)$ for every $q \in [n, \infty)$ by reducing to the subcritical [Gagliardo-Nirenberg-Sobolev inequality](/theorems/61). For $q = n$ the embedding holds by definition. For $q > n$, we choose $p = qn/(n + q) \in [1, n)$ so that the Sobolev conjugate $p^* = np/(n-p)$ equals $q$. Since $U$ is bounded, Hölder's inequality gives the continuous inclusion $L^n(U) \hookrightarrow L^p(U)$, which yields $W^{1,n}(U) \hookrightarrow W^{1,p}(U)$. The GNS inequality on bounded domains with $C^1$ boundary (via the Extension Theorem) then provides $W^{1,p}(U) \hookrightarrow L^{p^*}(U) = L^q(U)$. [/proofplan] [step:Handle the trivial case $q = n$] When $q = n$, the inclusion $W^{1,n}(U) \subset L^n(U)$ holds by definition of the [Sobolev space](/page/Sobolev%20Space) $W^{1,n}(U)$, since membership in $W^{1,n}(U)$ requires $u \in L^n(U)$. The embedding constant is $C_n = 1$. [/step] [step:Select $p \in [1, n)$ so that the Sobolev conjugate $p^*$ equals the target exponent $q$] Let $q > n$ be fixed. We solve $p^* = np/(n - p) = q$ for $p$: \begin{align*} np = q(n - p) \implies np + qp = qn \implies p = \frac{qn}{n + q}. \end{align*} We verify $1 \le p < n$. For the upper bound: $p = qn/(n + q) < n$ because $qn < n(n + q)$ reduces to $q < n + q$, which holds. For the lower bound (assuming $n \ge 2$): $p \ge 1$ requires $qn \ge n + q$, equivalently $q(n - 1) \ge n$, which holds since $q > n \ge 2$ gives $q(n-1) \ge n(n-1) \ge n$. [guided] The goal is to find $p \in [1, n)$ so that the [Gagliardo-Nirenberg-Sobolev inequality](/theorems/61), which embeds $W^{1,p}(\mathbb{R}^n)$ into $L^{p^*}(\mathbb{R}^n)$ with $p^* = np/(n-p)$, delivers exactly the target exponent $q$. We need $p^* = q$, which means solving $np/(n-p) = q$ for $p$. Cross-multiplying: $np = q(n - p) = qn - qp$, so $np + qp = qn$, giving $p = qn/(n + q)$. This is the unique solution because $p^* = np/(n-p)$ is a strictly increasing function of $p$ on $[1, n)$: differentiating, $\frac{d}{dp}\left(\frac{np}{n-p}\right) = \frac{n^2}{(n-p)^2} > 0$. As $p$ ranges over $[1, n)$, the Sobolev conjugate $p^*$ increases from $1^* = n/(n-1)$ to $+\infty$ as $p \nearrow n$. Since $q > n > n/(n-1)$, the value $q$ lies in the range of $p^*$, so there is a unique $p \in (1, n)$ with $p^* = q$. We verify that $p = qn/(n+q)$ satisfies $1 \le p < n$. For the upper bound: \begin{align*} \frac{qn}{n+q} < n \iff qn < n(n+q) = n^2 + nq \iff 0 < n^2, \end{align*} which always holds. For the lower bound, assuming $n \ge 2$: \begin{align*} \frac{qn}{n+q} \ge 1 \iff qn \ge n + q \iff q(n-1) \ge n. \end{align*} Since $q > n$ and $n \ge 2$, we have $q(n-1) \ge n(n-1) = n^2 - n \ge n$ (using $n^2 - n \ge n \iff n \ge 2$). So $p \in [1, n)$ as required by the [Gagliardo-Nirenberg-Sobolev inequality](/theorems/61). [/guided] [/step] [step:Embed $W^{1,n}(U)$ into $W^{1,p}(U)$ via Hölder's inequality on the bounded domain] Since $U$ is bounded with $\mathcal{L}^n(U) < \infty$ and $1 \le p < n$, [Hölder's inequality](/theorems/???) with exponents $n/p > 1$ and $n/(n - p)$ (which are conjugate since $p/n + (n-p)/n = 1$) gives the inclusion $L^n(U) \hookrightarrow L^p(U)$: for any $f \in L^n(U)$, \begin{align*} \|f\|_{L^p(U)} \le \mathcal{L}^n(U)^{1/p - 1/n} \|f\|_{L^n(U)}. \end{align*} Applying this to both $u$ and each component $\partial_i u$ of the weak gradient: \begin{align*} \|u\|_{W^{1,p}(U)} = \|u\|_{L^p(U)} + \|\nabla u\|_{L^p(U)} \le \mathcal{L}^n(U)^{1/p - 1/n} \left( \|u\|_{L^n(U)} + \|\nabla u\|_{L^n(U)} \right) = C_{U,p,n} \|u\|_{W^{1,n}(U)} \end{align*} where $C_{U,p,n} = \mathcal{L}^n(U)^{1/p - 1/n}$. [guided] We need to embed $W^{1,n}(U)$ into $W^{1,p}(U)$ for $p < n$. Since $p < n$, the $L^n$ norm controls the $L^p$ norm on bounded domains — this is a standard consequence of [Hölder's inequality](/theorems/???). Concretely, writing $|f|^p = |f|^p \cdot 1$ and applying Hölder's inequality with exponents $n/p$ and $n/(n-p)$ (these are conjugate since $p/n + (n-p)/n = 1$): \begin{align*} \int_U |f|^p \, d\mathcal{L}^n = \int_U |f|^p \cdot 1 \, d\mathcal{L}^n \le \left(\int_U |f|^n \, d\mathcal{L}^n\right)^{p/n} \mathcal{L}^n(U)^{1 - p/n}. \end{align*} Taking $p$-th roots: $\|f\|_{L^p(U)} \le \mathcal{L}^n(U)^{1/p - 1/n} \|f\|_{L^n(U)}$. This is where the hypothesis that $U$ is bounded (hence $\mathcal{L}^n(U) < \infty$) is consumed — the estimate fails on unbounded domains. Applying this bound to both $u$ and each component $\partial_i u$ of the weak gradient: \begin{align*} \|u\|_{W^{1,p}(U)} = \|u\|_{L^p(U)} + \|\nabla u\|_{L^p(U)} \le \mathcal{L}^n(U)^{1/p - 1/n} \left( \|u\|_{L^n(U)} + \|\nabla u\|_{L^n(U)} \right) = C_{U,p,n} \|u\|_{W^{1,n}(U)} \end{align*} where $C_{U,p,n} = \mathcal{L}^n(U)^{1/p - 1/n}$. This establishes $W^{1,n}(U) \hookrightarrow W^{1,p}(U)$ with explicit embedding constant. [/guided] [/step] [step:Apply the Gagliardo-Nirenberg-Sobolev inequality on the bounded domain to reach $L^q(U)$] Since $U$ has $C^1$ boundary, the [Extension Theorem](/theorems/58) provides a bounded linear operator $E: W^{1,p}(U) \to W^{1,p}(\mathbb{R}^n)$ with $\|Eu\|_{W^{1,p}(\mathbb{R}^n)} \le C_E \|u\|_{W^{1,p}(U)}$. The [Gagliardo-Nirenberg-Sobolev inequality](/theorems/61) on $\mathbb{R}^n$ gives the continuous embedding $W^{1,p}(\mathbb{R}^n) \hookrightarrow L^{p^*}(\mathbb{R}^n)$. Restricting to $U$ and composing with the extension: \begin{align*} \|u\|_{L^{p^*}(U)} \le \|Eu\|_{L^{p^*}(\mathbb{R}^n)} \le C_{\text{GNS}} \|Eu\|_{W^{1,p}(\mathbb{R}^n)} \le C_{\text{GNS}} \, C_E \|u\|_{W^{1,p}(U)}. \end{align*} Since $p^* = q$ by construction, this reads $\|u\|_{L^q(U)} \le C_{\text{GNS}} \, C_E \|u\|_{W^{1,p}(U)}$. [guided] The [Gagliardo-Nirenberg-Sobolev inequality](/theorems/61) is stated for functions in $W^{1,p}(\mathbb{R}^n)$, not for functions on a bounded domain $U$. To bridge this gap, we use the [Extension Theorem](/theorems/58): since $\partial U$ is $C^1$, there exists a bounded linear operator $E: W^{1,p}(U) \to W^{1,p}(\mathbb{R}^n)$ satisfying $Eu|_U = u$ and $\|Eu\|_{W^{1,p}(\mathbb{R}^n)} \le C_E \|u\|_{W^{1,p}(U)}$. This is precisely where the $C^1$ boundary hypothesis is essential — without regularity of $\partial U$, the extension operator may not exist. Applying the GNS inequality to $Eu \in W^{1,p}(\mathbb{R}^n)$ (with $1 \le p < n$, which is verified): \begin{align*} \|Eu\|_{L^{p^*}(\mathbb{R}^n)} \le C_{\text{GNS}} \|Eu\|_{W^{1,p}(\mathbb{R}^n)} \le C_{\text{GNS}} \, C_E \|u\|_{W^{1,p}(U)}. \end{align*} Restricting to $U$: since $Eu|_U = u$, we have $\|u\|_{L^{p^*}(U)} \le \|Eu\|_{L^{p^*}(\mathbb{R}^n)}$. Combining gives $\|u\|_{L^{p^*}(U)} \le C_{\text{GNS}} \, C_E \|u\|_{W^{1,p}(U)}$. Since $p^* = q$ by our choice of $p$ in the previous step, we have $\|u\|_{L^q(U)} \le C_{\text{GNS}} \, C_E \|u\|_{W^{1,p}(U)}$. [/guided] [/step] [step:Combine the chain of embeddings to conclude] Chaining the estimates from the previous two steps: \begin{align*} \|u\|_{L^q(U)} \le C_{\text{GNS}} \, C_E \, C_{U,p,n} \, \|u\|_{W^{1,n}(U)}. \end{align*} Setting $C_q = C_{\text{GNS}} \, C_E \, C_{U,p,n}$, which depends on $U$, $n$, and $q$ (through $p = qn/(n+q)$), we obtain the continuous embedding $W^{1,n}(U) \hookrightarrow L^q(U)$ for every $q \in [n, \infty)$. [guided] We assemble the full chain of embeddings. For any $q > n$, the three steps give: 1. $W^{1,n}(U) \hookrightarrow W^{1,p}(U)$ with constant $C_{U,p,n} = \mathcal{L}^n(U)^{1/p - 1/n}$ (from Hölder's inequality on the bounded domain), 2. $W^{1,p}(U) \hookrightarrow L^{p^*}(U) = L^q(U)$ with constant $C_{\text{GNS}} \, C_E$ (from the Extension Theorem and GNS inequality). Composing: $\|u\|_{L^q(U)} \le C_{\text{GNS}} \, C_E \, C_{U,p,n} \, \|u\|_{W^{1,n}(U)}$. Setting $C_q = C_{\text{GNS}} \, C_E \, C_{U,p,n}$, which depends on $U$, $n$, and $q$ (through the intermediate exponent $p = qn/(n+q)$), we conclude that $W^{1,n}(U) \hookrightarrow L^q(U)$ continuously for every $q \in [n, \infty)$. Together with the trivial case $q = n$ from the first step, this completes the proof. [/guided] [/step]