[proofplan] The BV Sobolev inequality is the critical-exponent embedding $\|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le C(n) |Du|(\mathbb{R}^n)$, the BV analogue of the Gagliardo–Nirenberg–Sobolev inequality. The proof combines the [BV Coarea Formula](/theorems/598), which writes the total variation as $\int P(\{u > t\})\, d\mathcal{L}^1(t)$, with the [Isoperimetric Inequality for Sets of Finite Perimeter](/theorems/600), which converts measures of superlevel sets into perimeters. The bridge between $\|u\|_{L^{n/(n-1)}}$ and the level-set integral $\int \mathcal{L}^n(\{u > t\})^{(n-1)/n}\, d\mathcal{L}^1(t)$ is supplied by a layer-cake representation together with Minkowski's integral inequality. Reducing first to $u \ge 0$ via positive and negative parts simplifies the layer-cake step. The compact support hypothesis ensures that the level sets have finite Lebesgue measure, so the geometric isoperimetric inequality applies at every level. [/proofplan]

[step:Reduce to non-negative $u$ by splitting positive and negative parts]Define \begin{align*} u^+: \mathbb{R}^n &\to [0, \infty), & x &\mapsto \max(u(x), 0), \\ u^-: \mathbb{R}^n &\to [0, \infty), & x &\mapsto \max(-u(x), 0). \end{align*} Then $u = u^+ - u^-$ and $|u| = u^+ + u^-$ with $u^+ u^- = 0$ pointwise, and both $u^\pm$ have compact support contained in the support of $u$. Each truncation $t \mapsto \max(t, 0)$ is Lipschitz with Lipschitz constant $1$, so the BV chain rule for monotone Lipschitz functions gives $u^\pm \in BV(\mathbb{R}^n)$ with \begin{align*} |Du^+|(\mathbb{R}^n) + |Du^-|(\mathbb{R}^n) = |Du|(\mathbb{R}^n). \end{align*} Since the supports of $u^+$ and $u^-$ are disjoint, $\|u\|_{L^{n/(n-1)}}^{n/(n-1)} = \|u^+\|_{L^{n/(n-1)}}^{n/(n-1)} + \|u^-\|_{L^{n/(n-1)}}^{n/(n-1)}$, so \begin{align*} \|u\|_{L^{n/(n-1)}} \le \|u^+\|_{L^{n/(n-1)}} + \|u^-\|_{L^{n/(n-1)}}. \end{align*} If we prove the bound $\|v\|_{L^{n/(n-1)}} \le C(n) |Dv|(\mathbb{R}^n)$ for compactly supported $v \in BV(\mathbb{R}^n)$ with $v \ge 0$, then applying it to $v = u^\pm$ and summing yields $\|u\|_{L^{n/(n-1)}} \le C(n) |Du|(\mathbb{R}^n)$.[/step]

[guided]The strategy is to assume $u \ge 0$, since the layer-cake representation in Step 3 requires non-negativity. To handle a general $u$, we decompose into positive and negative parts and prove the inequality for each separately. Define \begin{align*} u^+: \mathbb{R}^n &\to [0, \infty), & x &\mapsto \max(u(x), 0), \\ u^-: \mathbb{R}^n &\to [0, \infty), & x &\mapsto \max(-u(x), 0). \end{align*} Then $u = u^+ - u^-$, $|u| = u^+ + u^-$, and $u^+ u^- = 0$ pointwise. Since $\operatorname{supp}(u)$ is compact and $u^\pm$ vanish where $u$ does, both $u^+$ and $u^-$ have compact support. *BV regularity of $u^\pm$.* The map $\rho^+: t \mapsto \max(t, 0)$ is Lipschitz with Lipschitz constant $1$, and $\rho^-: t \mapsto \max(-t, 0)$ is Lipschitz with constant $1$ as well. By the BV chain rule for Lipschitz post-composition (a standard $BV$ chain rule), $u^\pm \in BV(\mathbb{R}^n)$. The decomposition of total variations follows because $\rho^+$ and $\rho^-$ act on disjoint regions: $\rho^+$ is locally constant on $\{u \le 0\}$ and equals $\operatorname{id}$ on $\{u > 0\}$, while $\rho^-$ is locally constant on $\{u \ge 0\}$ and equals $-\operatorname{id}$ on $\{u < 0\}$. Therefore $Du^+$ is supported in $\{u > 0\}$, $Du^-$ in $\{u < 0\}$, and the absolute total variations add: \begin{align*} |Du^+|(\mathbb{R}^n) + |Du^-|(\mathbb{R}^n) = |Du|(\mathbb{R}^n). \end{align*} *$L^{n/(n-1)}$ regularity.* Both $u^\pm \le |u|$ pointwise. Since $u$ has compact support, $|u| \in L^p$ for every $p$ provided $u \in BV$ has any $L^p$ membership, which is automatic for compactly supported BV functions. Hence $u^\pm \in L^{n/(n-1)}(\mathbb{R}^n)$. Because $u^+ u^- = 0$ pointwise, the supports of $u^+$ and $u^-$ are disjoint up to a null set, so \begin{align*} \|u\|_{L^{n/(n-1)}}^{n/(n-1)} = \int_{\mathbb{R}^n} |u|^{n/(n-1)} \, d\mathcal{L}^n = \int_{\mathbb{R}^n} (u^+)^{n/(n-1)} \, d\mathcal{L}^n + \int_{\mathbb{R}^n} (u^-)^{n/(n-1)} \, d\mathcal{L}^n. \end{align*} The elementary inequality $(a + b)^{1/q} \le a^{1/q} + b^{1/q}$ for $a, b \ge 0$ and $q \ge 1$, applied with $q = n/(n-1) \ge 1$ and $a = \|u^+\|^{n/(n-1)}$, $b = \|u^-\|^{n/(n-1)}$, gives \begin{align*} \|u\|_{L^{n/(n-1)}} \le \|u^+\|_{L^{n/(n-1)}} + \|u^-\|_{L^{n/(n-1)}}. \end{align*} If the non-negative case is established with $\|v\|_{L^{n/(n-1)}} \le C(n) |Dv|(\mathbb{R}^n)$ for compactly supported $v \ge 0$, then applying it to $v = u^+$ and $v = u^-$: \begin{align*} \|u\|_{L^{n/(n-1)}} \le C(n) (|Du^+|(\mathbb{R}^n) + |Du^-|(\mathbb{R}^n)) = C(n) |Du|(\mathbb{R}^n). \end{align*} This reduces the problem to the case $u \ge 0$, which we treat in the remaining steps.[/guided]

[step:Apply the BV coarea formula and the geometric isoperimetric inequality at each level]Assume $u \ge 0$ is compactly supported in $BV(\mathbb{R}^n)$. By the [BV Coarea Formula](/theorems/598), \begin{align*} |Du|(\mathbb{R}^n) = \int_0^\infty P(\{u > t\}, \mathbb{R}^n) \, d\mathcal{L}^1(t), \end{align*} the lower limit being $0$ since $\{u > t\} = \mathbb{R}^n$ for $t < 0$ has perimeter zero. Set $E_t := \{u > t\}$. For each $t > 0$, $E_t \subseteq \operatorname{supp}(u)$ has finite Lebesgue measure (the support is compact). For $\mathcal{L}^1$-a.e. $t > 0$, the coarea formula guarantees $P(E_t, \mathbb{R}^n) < \infty$. The hypotheses of the [Isoperimetric Inequality for Sets of Finite Perimeter](/theorems/600) — finite perimeter and finite Lebesgue measure — are met, so \begin{align*} \mathcal{L}^n(E_t)^{(n-1)/n} \le c_n \, P(E_t, \mathbb{R}^n) \quad \text{for } \mathcal{L}^1\text{-a.e. } t > 0, \end{align*} where $c_n = (n \omega_n^{1/n})^{-1}$ is the sharp dimensional constant. Integrating in $t$ and using coarea on the right, \begin{align*} \int_0^\infty \mathcal{L}^n(E_t)^{(n-1)/n} \, d\mathcal{L}^1(t) \le c_n \int_0^\infty P(E_t, \mathbb{R}^n) \, d\mathcal{L}^1(t) = c_n \, |Du|(\mathbb{R}^n). \end{align*}[/step]

[guided]We connect the total variation $|Du|(\mathbb{R}^n)$ to integrals of measures of superlevel sets. The two ingredients are coarea (which decomposes $|Du|$ into perimeters of level sets) and the geometric isoperimetric inequality (which converts each $\mathcal{L}^n(E_t)$ into a multiple of $P(E_t)$). *Coarea.* The hypothesis of the [BV Coarea Formula](/theorems/598) is $u \in BV(\Omega)$ for $\Omega = \mathbb{R}^n$, which is given. The conclusion is \begin{align*} |Du|(\mathbb{R}^n) = \int_{-\infty}^\infty P(\{u > t\}, \mathbb{R}^n) \, d\mathcal{L}^1(t). \end{align*} For $u \ge 0$, the level set $\{u > t\}$ equals $\mathbb{R}^n$ for $t < 0$, so $P(\{u > t\}, \mathbb{R}^n) = P(\mathbb{R}^n, \mathbb{R}^n) = 0$. The integral therefore reduces to $\int_0^\infty$: \begin{align*} |Du|(\mathbb{R}^n) = \int_0^\infty P(\{u > t\}, \mathbb{R}^n) \, d\mathcal{L}^1(t). \end{align*} *Finite measure of $E_t$.* For each $t > 0$, $E_t = \{u > t\} \subseteq \operatorname{supp}(u)$, which is compact and hence has finite Lebesgue measure. Therefore $\mathcal{L}^n(E_t) \le \mathcal{L}^n(\operatorname{supp}(u)) < \infty$. The compact support hypothesis is what makes this step direct; without it, finite Lebesgue measure of $E_t$ would require an additional integrability hypothesis on $u$ (e.g., $u \in L^{n/(n-1)}$). *Finite perimeter of $E_t$.* The coarea formula asserts $\int_0^\infty P(E_t) \, d\mathcal{L}^1(t) = |Du|(\mathbb{R}^n) < \infty$, hence $P(E_t) < \infty$ for $\mathcal{L}^1$-a.e. $t > 0$. *Isoperimetric inequality.* The [Isoperimetric Inequality for Sets of Finite Perimeter](/theorems/600) states: for any Borel set $E \subseteq \mathbb{R}^n$ with $\mathcal{L}^n(E) < \infty$ and $P(E) < \infty$, \begin{align*} \mathcal{L}^n(E)^{(n-1)/n} \le c_n \, P(E), \end{align*} where $c_n = (n \omega_n^{1/n})^{-1}$ and $\omega_n$ is the volume of the unit ball. Both hypotheses are satisfied by $E = E_t$ for $\mathcal{L}^1$-a.e. $t > 0$, so \begin{align*} \mathcal{L}^n(E_t)^{(n-1)/n} \le c_n \, P(E_t, \mathbb{R}^n) \quad \text{for } \mathcal{L}^1\text{-a.e. } t > 0. \end{align*} *Integration in $t$.* The function $t \mapsto \mathcal{L}^n(E_t)$ is non-increasing (hence Borel) and the function $t \mapsto P(E_t, \mathbb{R}^n)$ is Borel by the coarea formula. Integrating the inequality in $t \in (0, \infty)$ and using coarea on the right, \begin{align*} \int_0^\infty \mathcal{L}^n(E_t)^{(n-1)/n} \, d\mathcal{L}^1(t) \le c_n \int_0^\infty P(E_t, \mathbb{R}^n) \, d\mathcal{L}^1(t) = c_n \, |Du|(\mathbb{R}^n). \end{align*} This is the key estimate for Step 3, which connects the level-set integral on the left to $\|u\|_{L^{n/(n-1)}}$ via Minkowski's integral inequality.[/guided]

[step:Bound $\|u\|_{L^{n/(n-1)}}$ by the layer-cake integral via Minkowski's integral inequality]For $u \ge 0$ and any $x \in \mathbb{R}^n$, the layer-cake representation reads \begin{align*} u(x) = \int_0^\infty \mathbb{1}_{\{u > t\}}(x) \, d\mathcal{L}^1(t). \end{align*} Apply Minkowski's integral inequality with exponent $q = n/(n-1) \ge 1$ to the family $f_t = \mathbb{1}_{\{u > t\}}$: \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} = \left\| \int_0^\infty \mathbb{1}_{\{u > t\}} \, d\mathcal{L}^1(t) \right\|_{L^{n/(n-1)}(\mathbb{R}^n)} &\le \int_0^\infty \|\mathbb{1}_{\{u > t\}}\|_{L^{n/(n-1)}(\mathbb{R}^n)} \, d\mathcal{L}^1(t) \\ &= \int_0^\infty \mathcal{L}^n(\{u > t\})^{(n-1)/n} \, d\mathcal{L}^1(t), \end{align*} using $\|\mathbb{1}_E\|_{L^p}^p = \mathcal{L}^n(E)$ to identify the norm of an indicator function. Combining with Step 2, \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \int_0^\infty \mathcal{L}^n(E_t)^{(n-1)/n} \, d\mathcal{L}^1(t) \le c_n \, |Du|(\mathbb{R}^n), \end{align*} which is the desired inequality for non-negative $u$ with constant $c_n = (n \omega_n^{1/n})^{-1}$.[/step]

[guided]We now connect $\|u\|_{L^{n/(n-1)}}$ to the level-set integral $\int \mathcal{L}^n(E_t)^{(n-1)/n}\, d\mathcal{L}^1(t)$ via the layer-cake representation and Minkowski's integral inequality. *Layer-cake representation.* For $u \ge 0$, \begin{align*} u(x) = \int_0^{u(x)} 1 \, d\mathcal{L}^1(t) = \int_0^\infty \mathbb{1}_{\{u(x) > t\}} \, d\mathcal{L}^1(t) = \int_0^\infty \mathbb{1}_{\{u > t\}}(x) \, d\mathcal{L}^1(t). \end{align*} This is the standard vertical integration formula for non-negative functions. *Minkowski's integral inequality.* For a measurable family $f_t: \mathbb{R}^n \to [0, \infty)$ (jointly measurable in $(t, x)$) and $q \ge 1$, \begin{align*} \left\| \int_0^\infty f_t \, d\mathcal{L}^1(t) \right\|_{L^q(\mathbb{R}^n)} \le \int_0^\infty \|f_t\|_{L^q(\mathbb{R}^n)} \, d\mathcal{L}^1(t). \end{align*} We verify the hypotheses for $q = n/(n-1)$ and $f_t = \mathbb{1}_{\{u > t\}}$. Since $n \ge 2$, $q = n/(n-1) \in (1, 2]$, so $q \ge 1$ holds. The map $(t, x) \mapsto \mathbb{1}_{\{u > t\}}(x) = \mathbb{1}_{\{(t, x): u(x) > t\}}$ is the indicator of the hypograph of $u$, which is Borel because $u$ is Borel-measurable. (Strictly: the dimension $n = 1$ case can be treated separately by reducing to $W^{1,1}$, but the same argument works for any $n \ge 1$ since $q = \infty$ when $n = 1$ requires only the bound $\|u\|_\infty \le \int |u'|$, which is the FTC for AC functions.) Applying Minkowski with these choices: \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} = \left\| \int_0^\infty \mathbb{1}_{\{u > t\}} \, d\mathcal{L}^1(t) \right\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \int_0^\infty \|\mathbb{1}_{\{u > t\}}\|_{L^{n/(n-1)}(\mathbb{R}^n)} \, d\mathcal{L}^1(t). \end{align*} *Norm of an indicator function.* For any Borel set $E$ with $\mathcal{L}^n(E) < \infty$, \begin{align*} \|\mathbb{1}_E\|_{L^p(\mathbb{R}^n)} = \left( \int_{\mathbb{R}^n} \mathbb{1}_E \, d\mathcal{L}^n \right)^{1/p} = \mathcal{L}^n(E)^{1/p}. \end{align*} With $p = n/(n-1)$ and $E = \{u > t\}$: \begin{align*} \|\mathbb{1}_{\{u > t\}}\|_{L^{n/(n-1)}(\mathbb{R}^n)} = \mathcal{L}^n(\{u > t\})^{(n-1)/n}. \end{align*} *Substitution.* Combining, \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \int_0^\infty \mathcal{L}^n(\{u > t\})^{(n-1)/n} \, d\mathcal{L}^1(t). \end{align*} *Combination with Step 2.* By the estimate of Step 2, \begin{align*} \int_0^\infty \mathcal{L}^n(E_t)^{(n-1)/n} \, d\mathcal{L}^1(t) \le c_n \, |Du|(\mathbb{R}^n). \end{align*} Chaining the two inequalities, \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le c_n \, |Du|(\mathbb{R}^n), \end{align*} which is the BV Sobolev inequality for non-negative compactly supported $u$ with constant $C(n) = c_n = (n \omega_n^{1/n})^{-1}$.[/guided]

[step:Conclude the proof for general $u$]Combining Steps 1 and 3: for general compactly supported $u \in BV(\mathbb{R}^n)$, \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} &\le \|u^+\|_{L^{n/(n-1)}(\mathbb{R}^n)} + \|u^-\|_{L^{n/(n-1)}(\mathbb{R}^n)} \\ &\le c_n \, |Du^+|(\mathbb{R}^n) + c_n \, |Du^-|(\mathbb{R}^n) \\ &= c_n \, |Du|(\mathbb{R}^n), \end{align*} with $C(n) = c_n = (n \omega_n^{1/n})^{-1}$. The proof is complete.[/step]

[guided]We assemble the result for general compactly supported $u \in BV(\mathbb{R}^n)$, removing the non-negativity assumption. By Step 3 applied to the non-negative compactly supported BV functions $v = u^+$ and $v = u^-$, \begin{align*} \|u^+\|_{L^{n/(n-1)}(\mathbb{R}^n)} &\le c_n \, |Du^+|(\mathbb{R}^n), \\ \|u^-\|_{L^{n/(n-1)}(\mathbb{R}^n)} &\le c_n \, |Du^-|(\mathbb{R}^n). \end{align*} By the splitting in Step 1, \begin{align*} \|u\|_{L^{n/(n-1)}} \le \|u^+\|_{L^{n/(n-1)}} + \|u^-\|_{L^{n/(n-1)}} \le c_n \big(|Du^+|(\mathbb{R}^n) + |Du^-|(\mathbb{R}^n)\big) = c_n \, |Du|(\mathbb{R}^n), \end{align*} where the last equality uses the additive decomposition $|Du^+| + |Du^-| = |Du|$ established in Step 1. Therefore the BV Sobolev inequality holds with constant $C(n) = c_n = (n \omega_n^{1/n})^{-1}$, depending only on the dimension $n$. The role of the compact support hypothesis is twofold: it guarantees finite Lebesgue measure of all superlevel sets $E_t$ for $t > 0$, allowing the geometric isoperimetric inequality to apply uniformly in $t$; and it ensures $u \in L^p$ for every $p$, so the layer-cake integration converges. The proof is complete.[/guided]