[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.
[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]
[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*}
[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]
[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]
