[proofplan] We prove a half-volume $L^1$ Sobolev inequality from the definition of the Cheeger constant by applying the coarea formula to superlevel sets. Given an arbitrary smooth nonconstant function, we subtract a median so that the positive and negative parts each live on sets of volume at most half of $M$. Applying the half-volume inequality to the squares of these two parts bounds the $L^2$ size of $f-m$ by a mixed gradient term involving $|f-m|$ and $|\nabla f|_g$. The [Cauchy-Schwarz inequality](/theorems/432) then converts this into the Rayleigh quotient estimate, and taking the infimum gives the claimed bound. [/proofplan]

[step:Choose a median so both signs occupy at most half the manifold] Let $n := \dim M$. Let $\operatorname{vol}_g$ denote the Riemannian volume measure on $M$, and let $\mathcal{H}^{n-1}_g$ denote the Riemannian $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) induced by $g$. Let $V := \operatorname{vol}_g(M)$. We use the Cheeger constant \begin{align*} h(M) := \inf_{\Omega}\frac{\mathcal{H}^{n-1}_g(\partial \Omega)}{\min\{\operatorname{vol}_g(\Omega),\operatorname{vol}_g(M\setminus \Omega)\}}, \end{align*} where the infimum is taken over smooth domains $\Omega \subset M$ with $0<\operatorname{vol}_g(\Omega)<V$. We also use the convention that $\lambda_1(M)$ is the first positive eigenvalue of the nonnegative Laplace-Beltrami operator $-\Delta_g$, equivalently \begin{align*} \lambda_1(M) = \inf_{f\in C^\infty(M),\ f\not\equiv \operatorname{constant}} \frac{\int_M |\nabla f|_g^2\, d\operatorname{vol}_g} {\inf_{a\in\mathbb{R}}\int_M |f-a|^2\, d\operatorname{vol}_g}. \end{align*} For $f \in C^\infty(M)$, define a median of $f$ to be a real number $m \in \mathbb{R}$ such that \begin{align*} \operatorname{vol}_g(\{x \in M : f(x) > m\}) \leq \frac{V}{2}, \qquad \operatorname{vol}_g(\{x \in M : f(x) < m\}) \leq \frac{V}{2}. \end{align*} Such an $m$ exists because the distribution function \begin{align*} F:\mathbb{R} &\to [0,V]\\ t &\mapsto \operatorname{vol}_g(\{x \in M : f(x) \leq t\}) \end{align*} is nondecreasing, satisfies $\lim_{t \to -\infty}F(t)=0$ and $\lim_{t \to \infty}F(t)=V$. Define \begin{align*} m := \inf\{t \in \mathbb{R} : F(t) \geq V/2\}. \end{align*} For every $s<m$, the definition of the infimum gives $F(s)<V/2$. Since $\{x \in M : f(x)<m\}=\bigcup_{k=1}^{\infty}\{x \in M : f(x)\leq m-1/k\}$ after discarding empty terms with $m-1/k$ below the range of $f$, continuity from below of the finite measure $\operatorname{vol}_g$ gives \begin{align*} \operatorname{vol}_g(\{x \in M : f(x)<m\}) \leq \frac{V}{2}. \end{align*} For every $s>m$, monotonicity and the definition of $m$ give $F(s)\geq V/2$. Since $\{x \in M : f(x)>m\}=\bigcup_{k=1}^{\infty}\{x \in M : f(x)>m+1/k\}$, continuity from below gives \begin{align*} \operatorname{vol}_g(\{x \in M : f(x)>m\}) \leq V-\frac{V}{2} =\frac{V}{2}. \end{align*} Thus $m$ has the two stated median properties. Fix such an $m$ and define \begin{align*} v: M &\to \mathbb{R}\\ x &\mapsto f(x)-m. \end{align*} Define the positive and negative parts \begin{align*} v_+: M &\to [0,\infty)\\ x &\mapsto \max\{v(x),0\}, \end{align*} and \begin{align*} v_-: M &\to [0,\infty)\\ x &\mapsto \max\{-v(x),0\}. \end{align*} Then $v=v_+-v_-$, $|v|=v_++v_-$, and \begin{align*} \operatorname{vol}_g(\{x \in M : v_+(x)>0\}) \leq \frac{V}{2}, \qquad \operatorname{vol}_g(\{x \in M : v_-(x)>0\}) \leq \frac{V}{2}. \end{align*} [/step]

[step:Derive the half-volume $L^1$ inequality from the coarea formula]Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Let $u: M \to [0,\infty)$ be a Lipschitz function such that \begin{align*} \operatorname{vol}_g(\{x \in M : u(x)>0\}) \leq \frac{V}{2}. \end{align*} We use the standard finite-perimeter extension of the Cheeger lower bound: if $E \subset M$ is a set of finite perimeter and $\operatorname{vol}_g(E)\leq V/2$, then \begin{align*} P_g(E)\geq h(M)\operatorname{vol}_g(E), \end{align*} where $P_g(E)$ denotes the Riemannian perimeter of $E$. This extension follows from the [strict approximation theorem for finite-perimeter sets](/page/Finite%20Perimeter%20Set) on compact Riemannian manifolds: since $E$ has finite perimeter in the compact manifold $M$, there are smooth domains $E_j\subset M$ such that \begin{align*} \operatorname{vol}_g(E_j)&\to \operatorname{vol}_g(E),\\ \mathcal{H}^{n-1}_g(\partial E_j)&\to P_g(E). \end{align*} The defining Cheeger inequality for each smooth domain $E_j$ gives \begin{align*} \mathcal{H}^{n-1}_g(\partial E_j) \geq h(M)\min\{\operatorname{vol}_g(E_j),\operatorname{vol}_g(M\setminus E_j)\}. \end{align*} Passing to the limit and using continuity of the minimum function gives \begin{align*} P_g(E) \geq h(M)\min\{\operatorname{vol}_g(E),\operatorname{vol}_g(M\setminus E)\}. \end{align*} Since $\operatorname{vol}_g(E)\leq V/2$, the minimum equals $\operatorname{vol}_g(E)$. We claim that \begin{align*} \int_M |\nabla u|_g\, d\operatorname{vol}_g \geq h(M)\int_M u\, d\operatorname{vol}_g. \end{align*} Indeed, for $t>0$ define the superlevel set \begin{align*} E_t := \{x \in M : u(x)>t\}. \end{align*} Then $E_t \subset \{u>0\}$, so \begin{align*} \operatorname{vol}_g(E_t) \leq \frac{V}{2}. \end{align*} By the [coarea formula](/page/Coarea%20Formula) for Lipschitz functions on Riemannian manifolds, applied to the map $u:M\to[0,\infty)$, \begin{align*} \int_M |\nabla u|_g\, d\operatorname{vol}_g = \int_0^\infty P_g(E_t)\, d\mathcal{L}^1(t). \end{align*} For $\mathcal{L}^1$-almost every $t>0$, the superlevel set $E_t$ has finite perimeter; this finite-perimeter property is one of the conclusions of the coarea formula for Lipschitz functions. Since $\operatorname{vol}_g(E_t)\leq V/2$, the finite-perimeter Cheeger lower bound gives \begin{align*} P_g(E_t) \geq h(M)\operatorname{vol}_g(E_t) \end{align*} for $\mathcal{L}^1$-almost every $t>0$. Therefore \begin{align*} \int_M |\nabla u|_g\, d\operatorname{vol}_g &\geq h(M)\int_0^\infty \operatorname{vol}_g(E_t)\, d\mathcal{L}^1(t). \end{align*} Finally, the [layer-cake formula](/page/Layer-Cake%20Formula) for nonnegative [measurable functions](/page/Measurable%20Functions) gives \begin{align*} \int_0^\infty \operatorname{vol}_g(E_t)\, d\mathcal{L}^1(t) = \int_M u\, d\operatorname{vol}_g. \end{align*} Combining these identities proves the claimed inequality.[/step]

[guided]The point of this step is to turn the isoperimetric lower bound encoded by $h(M)$ into an analytic inequality for functions. Let \begin{align*} u: M &\to [0,\infty) \end{align*} be Lipschitz and suppose its positive set has volume at most half of $M$: \begin{align*} \operatorname{vol}_g(\{x \in M : u(x)>0\}) \leq \frac{V}{2}. \end{align*} For each $t>0$, define the superlevel set \begin{align*} E_t := \{x \in M : u(x)>t\}. \end{align*} Since $E_t \subset \{u>0\}$, each $E_t$ also satisfies \begin{align*} \operatorname{vol}_g(E_t) \leq \frac{V}{2}. \end{align*} This is exactly why we needed the half-volume hypothesis: it allows the Cheeger definition to be applied to every superlevel set. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Now apply the [coarea formula](/page/Coarea%20Formula) for Lipschitz functions on Riemannian manifolds to the map $u:M\to[0,\infty)$. It states that the superlevel sets $E_t$ have finite perimeter for $\mathcal{L}^1$-almost every $t>0$ and that \begin{align*} \int_M |\nabla u|_g\, d\operatorname{vol}_g = \int_0^\infty P_g(E_t)\, d\mathcal{L}^1(t), \end{align*} where $P_g(E_t)$ denotes the Riemannian perimeter of $E_t$. The definition of $h(M)$ is stated for smooth domains, so we must explain why it applies to these finite-perimeter superlevel sets. If $E\subset M$ has finite perimeter and $\operatorname{vol}_g(E)\leq V/2$, the [strict approximation theorem for finite-perimeter sets](/page/Finite%20Perimeter%20Set) on compact Riemannian manifolds applies because $M$ is compact and $E$ has finite perimeter. It gives smooth domains $E_j\subset M$ such that \begin{align*} \operatorname{vol}_g(E_j)&\to\operatorname{vol}_g(E),\\ \mathcal{H}^{n-1}_g(\partial E_j)&\to P_g(E). \end{align*} For each $j$, the defining Cheeger inequality gives \begin{align*} \mathcal{H}^{n-1}_g(\partial E_j) \geq h(M)\min\{\operatorname{vol}_g(E_j),\operatorname{vol}_g(M\setminus E_j)\}. \end{align*} Passing to the limit gives \begin{align*} P_g(E) \geq h(M)\min\{\operatorname{vol}_g(E),\operatorname{vol}_g(M\setminus E)\}. \end{align*} Because $\operatorname{vol}_g(E)\leq V/2$, the minimum is $\operatorname{vol}_g(E)$, and hence \begin{align*} P_g(E)\geq h(M)\operatorname{vol}_g(E). \end{align*} Applying this finite-perimeter extension to $E_t$, using $\operatorname{vol}_g(E_t)\leq V/2$, gives \begin{align*} P_g(E_t) \geq h(M)\operatorname{vol}_g(E_t) \end{align*} for $\mathcal{L}^1$-almost every $t>0$. Substituting this lower bound into the coarea identity gives \begin{align*} \int_M |\nabla u|_g\, d\operatorname{vol}_g &= \int_0^\infty P_g(E_t)\, d\mathcal{L}^1(t)\\ &\geq h(M)\int_0^\infty \operatorname{vol}_g(E_t)\, d\mathcal{L}^1(t). \end{align*} The remaining integral over $t$ reconstructs the integral of $u$ itself. Indeed, by the [layer-cake formula](/page/Layer-Cake%20Formula), \begin{align*} \int_0^\infty \operatorname{vol}_g(E_t)\, d\mathcal{L}^1(t) = \int_M u\, d\operatorname{vol}_g. \end{align*} Therefore \begin{align*} \int_M |\nabla u|_g\, d\operatorname{vol}_g \geq h(M)\int_M u\, d\operatorname{vol}_g. \end{align*} This is the analytic form of the Cheeger isoperimetric lower bound.[/guided]

[step:Apply the half-volume inequality to the squared positive and negative parts] Because $M$ is compact and $v=f-m$ is smooth, $|\nabla v|_g=|\nabla f|_g$ is bounded on $M$; hence $v$ is Lipschitz. The maps $r\mapsto \max\{r,0\}$ and $r\mapsto \max\{-r,0\}$ from $\mathbb{R}$ to $[0,\infty)$ are Lipschitz, so $v_+$ and $v_-$ are Lipschitz. Since $v_+$ and $v_-$ are bounded on compact $M$, the squares $v_+^2$ and $v_-^2$ are Lipschitz maps from $M$ to $[0,\infty)$. Their positive sets have volume at most $V/2$ by the median property. Applying the half-volume $L^1$ inequality to \begin{align*} u_+: M &\to [0,\infty)\\ x &\mapsto v_+(x)^2 \end{align*} gives \begin{align*} h(M)\int_M v_+^2\, d\operatorname{vol}_g \leq \int_M |\nabla(v_+^2)|_g\, d\operatorname{vol}_g. \end{align*} Since $v_+$ is Lipschitz and $\nabla(v_+^2)=2v_+\nabla v_+$ almost everywhere, while $\nabla v_+=\nabla f$ almost everywhere on $\{v>0\}$ and $\nabla v_+=0$ almost everywhere on $\{v\leq 0\}$, we obtain \begin{align*} \int_M |\nabla(v_+^2)|_g\, d\operatorname{vol}_g = 2\int_M v_+|\nabla v_+|_g\, d\operatorname{vol}_g \leq 2\int_M v_+|\nabla f|_g\, d\operatorname{vol}_g. \end{align*} Thus \begin{align*} h(M)\int_M v_+^2\, d\operatorname{vol}_g \leq 2\int_M v_+|\nabla f|_g\, d\operatorname{vol}_g. \end{align*} The same argument applied to \begin{align*} u_-: M &\to [0,\infty)\\ x &\mapsto v_-(x)^2 \end{align*} gives \begin{align*} h(M)\int_M v_-^2\, d\operatorname{vol}_g \leq 2\int_M v_-|\nabla f|_g\, d\operatorname{vol}_g. \end{align*} Adding the two inequalities and using $v_+v_-=0$ pointwise yields \begin{align*} h(M)\int_M |v|^2\, d\operatorname{vol}_g \leq 2\int_M |v|\,|\nabla f|_g\, d\operatorname{vol}_g. \end{align*} [/step]

[step:Use Cauchy-Schwarz to obtain the Rayleigh quotient bound]Apply the [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality) in the [Hilbert space](/page/Hilbert%20Space) $L^2(M,d\operatorname{vol}_g)$ to the functions $|v|$ and $|\nabla f|_g$. This gives \begin{align*} \int_M |v|\,|\nabla f|_g\, d\operatorname{vol}_g \leq \left(\int_M |v|^2\, d\operatorname{vol}_g\right)^{1/2} \left(\int_M |\nabla f|_g^2\, d\operatorname{vol}_g\right)^{1/2}. \end{align*} Combining this with the previous step, \begin{align*} h(M)\int_M |v|^2\, d\operatorname{vol}_g \leq 2 \left(\int_M |v|^2\, d\operatorname{vol}_g\right)^{1/2} \left(\int_M |\nabla f|_g^2\, d\operatorname{vol}_g\right)^{1/2}. \end{align*} Since $f$ is nonconstant, $v=f-m$ is not identically zero, so \begin{align*} \int_M |v|^2\, d\operatorname{vol}_g>0. \end{align*} Dividing by its square root and squaring gives \begin{align*} \int_M |\nabla f|_g^2\, d\operatorname{vol}_g \geq \frac{h(M)^2}{4} \int_M |f-m|^2\, d\operatorname{vol}_g. \end{align*}[/step]

[guided]The previous step produced the estimate \begin{align*} h(M)\int_M |v|^2\, d\operatorname{vol}_g \leq 2\int_M |v|\,|\nabla f|_g\, d\operatorname{vol}_g. \end{align*} To turn this into an eigenvalue estimate, we need a square of the gradient norm rather than the mixed product on the right. We apply the [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality) in the Hilbert space $L^2(M,d\operatorname{vol}_g)$ to the two square-integrable functions \begin{align*} x &\mapsto |v(x)|,\\ x &\mapsto |\nabla f(x)|_g. \end{align*} They are square-integrable because $M$ is compact, $v=f-m$ is smooth, and $|\nabla f|_g$ is continuous. Cauchy-Schwarz gives \begin{align*} \int_M |v|\,|\nabla f|_g\, d\operatorname{vol}_g \leq \left(\int_M |v|^2\, d\operatorname{vol}_g\right)^{1/2} \left(\int_M |\nabla f|_g^2\, d\operatorname{vol}_g\right)^{1/2}. \end{align*} Substituting this into the preceding estimate yields \begin{align*} h(M)\int_M |v|^2\, d\operatorname{vol}_g \leq 2 \left(\int_M |v|^2\, d\operatorname{vol}_g\right)^{1/2} \left(\int_M |\nabla f|_g^2\, d\operatorname{vol}_g\right)^{1/2}. \end{align*} Because $f$ is nonconstant, subtracting the constant $m$ cannot make $v=f-m$ identically zero. Hence \begin{align*} \int_M |v|^2\, d\operatorname{vol}_g>0. \end{align*} We may divide by the positive factor $\left(\int_M |v|^2\, d\operatorname{vol}_g\right)^{1/2}$. This gives \begin{align*} \frac{h(M)}{2} \left(\int_M |v|^2\, d\operatorname{vol}_g\right)^{1/2} \leq \left(\int_M |\nabla f|_g^2\, d\operatorname{vol}_g\right)^{1/2}. \end{align*} Both sides are nonnegative, so squaring preserves the inequality. Since $v=f-m$, we obtain \begin{align*} \int_M |\nabla f|_g^2\, d\operatorname{vol}_g \geq \frac{h(M)^2}{4} \int_M |f-m|^2\, d\operatorname{vol}_g. \end{align*}[/guided]

[step:Take the infimum over smooth nonconstant functions] For every $c \in \mathbb{R}$, the definition of the infimum gives \begin{align*} \inf_{a \in \mathbb{R}}\int_M |f-a|^2\, d\operatorname{vol}_g \leq \int_M |f-c|^2\, d\operatorname{vol}_g. \end{align*} Taking $c=m$, the previous step implies \begin{align*} \int_M |\nabla f|_g^2\, d\operatorname{vol}_g \geq \frac{h(M)^2}{4} \inf_{a \in \mathbb{R}}\int_M |f-a|^2\, d\operatorname{vol}_g. \end{align*} Therefore every smooth nonconstant $f:M\to\mathbb{R}$ satisfies \begin{align*} \frac{\int_M |\nabla f|_g^2\, d\operatorname{vol}_g} {\inf_{a \in \mathbb{R}}\int_M |f-a|^2\, d\operatorname{vol}_g} \geq \frac{h(M)^2}{4}. \end{align*} Taking the infimum over all smooth nonconstant functions in the variational characterization of $\lambda_1(M)$ fixed at the beginning of the proof gives \begin{align*} \lambda_1(M)\geq \frac{h(M)^2}{4}. \end{align*} This proves Cheeger's inequality. [/step]