[proofplan] The proof rewrites the correlation using the defining duality of the transfer operator, so that the problem becomes estimating $\mathcal L^n f-\Pi f$. The spectral gap decomposition gives $\mathcal L^n f=\Pi f+N^n f$, and the projection term is exactly the product of the two integrals. The spectral radius bound gives exponential decay of $N^n$ in the $\mathcal B_0$ norm, and the continuous embedding $\mathcal B_0\hookrightarrow L^1(X,\mu)$ converts this into an $L^1$ bound that can be paired with $g\in L^\infty(X,\mu)$. [/proofplan]

[step:Rewrite the correlation through the transfer operator]Fix $f\in\mathcal B_0$, fix $g\in L^\infty(X,\mathcal B,\mu)$, and fix an integer $n\ge 0$. Define the measurable map \begin{align*} T^n:X\to X \end{align*} to be the $n$-fold iterate of $T$, with $T^0=\operatorname{id}_X$. Since $T$ is probability-preserving, $g\circ T^n\in L^\infty(X,\mathcal B,\mu)$ and \begin{align*} \|g\circ T^n\|_{L^\infty(X,\mu)}\le \|g\|_{L^\infty(X,\mu)}. \end{align*} We claim that \begin{align*} \int_X f(g\circ T^n)\,d\mu(x)=\int_X (\mathcal L^n f)g\,d\mu(x). \end{align*} For $n=0$, this is the identity $\mathcal L^0=\operatorname{Id}_{\mathcal B_0}$. For $n\ge 1$, apply the transfer-operator duality $n$ times, first with $u=\mathcal L^{n-1}f$ and $h=g$, then with $u=\mathcal L^{n-2}f$ and $h=g\circ T$, and continuing until $u=f$ and $h=g\circ T^{n-1}$. This gives the displayed identity.[/step]

[guided]We first translate the dynamical expression into an operator expression. Fix $f\in\mathcal B_0$, $g\in L^\infty(X,\mathcal B,\mu)$, and an integer $n\ge 0$. The map \begin{align*} T^n:X\to X \end{align*} is the $n$-fold iterate of $T$, with $T^0=\operatorname{id}_X$. Because $T$ is measurable, $g\circ T^n$ is measurable. Because $T$ is probability-preserving, composition with $T^n$ does not increase the essential supremum: if $M=\|g\|_{L^\infty(X,\mu)}$, then $|g|\le M$ $\mu$-a.e.; the exceptional set has measure zero, and its preimage under $T^n$ also has measure zero. Hence \begin{align*} \|g\circ T^n\|_{L^\infty(X,\mu)}\le \|g\|_{L^\infty(X,\mu)}. \end{align*} The defining property of the transfer operator says that for every $u\in\mathcal B_0$ and every $h\in L^\infty(X,\mathcal B,\mu)$, \begin{align*} \int_X (\mathcal L u)h\,d\mu(x)=\int_X u(h\circ T)\,d\mu(x). \end{align*} We apply this identity repeatedly. For $n=0$, there is nothing to prove because $\mathcal L^0=\operatorname{Id}_{\mathcal B_0}$ and $T^0=\operatorname{id}_X$. Suppose $n\ge 1$. Applying the defining identity with $u=\mathcal L^{n-1}f$ and $h=g$ gives \begin{align*} \int_X (\mathcal L^n f)g\,d\mu(x)=\int_X (\mathcal L^{n-1}f)(g\circ T)\,d\mu(x). \end{align*} Applying the same identity again with $u=\mathcal L^{n-2}f$ and $h=g\circ T$ gives \begin{align*} \int_X (\mathcal L^{n-1}f)(g\circ T)\,d\mu(x)=\int_X (\mathcal L^{n-2}f)(g\circ T^2)\,d\mu(x). \end{align*} Iterating this argument $n$ times yields \begin{align*} \int_X (\mathcal L^n f)g\,d\mu(x)=\int_X f(g\circ T^n)\,d\mu(x). \end{align*} This is the key conversion: decay of correlations will follow once we can estimate $\mathcal L^n f-\Pi f$ in a norm that pairs with $g$.[/guided]

[step:Separate the invariant projection from the decaying remainder] Since $\mathcal L=\Pi+N$ and $\Pi N=N\Pi=0$, the binomial expansion with annihilating cross terms gives \begin{align*} \mathcal L^n=(\Pi+N)^n=\Pi^n+N^n \end{align*} for every integer $n\ge 1$. The operator $\Pi$ is a projection, since for every $u\in\mathcal B_0$, \begin{align*} \Pi^2u=\Pi\left(\left(\int_X u\,d\mu(x)\right)1_X\right)=\left(\int_X u\,d\mu(x)\right)\left(\int_X 1_X\,d\mu(x)\right)1_X=\Pi u, \end{align*} using $\mu(X)=1$. Hence $\Pi^n=\Pi$ for every $n\ge 1$, and therefore \begin{align*} \mathcal L^n f=\Pi f+N^n f \end{align*} for every $n\ge 1$. The same identity also holds for $n=0$ if $N^0$ is interpreted as $\operatorname{Id}_{\mathcal B_0}$ only after subtracting the projection; in the final estimate, the case $n=0$ will be absorbed into the constant. For $n\ge 1$, the projection term satisfies \begin{align*} \int_X (\Pi f)g\,d\mu(x)=\left(\int_X f\,d\mu(x)\right)\left(\int_X g\,d\mu(x)\right). \end{align*} Thus \begin{align*} \int_X f(g\circ T^n)\,d\mu(x)-\left(\int_X f\,d\mu(x)\right)\left(\int_X g\,d\mu(x)\right)=\int_X (N^n f)g\,d\mu(x) \end{align*} for every $n\ge 1$. [/step]

[step:Convert the spectral radius gap into an operator norm estimate] Choose $\rho$ such that $r(N)<\rho<1$. By the spectral radius formula (citing a result not yet in the wiki: Spectral Radius Formula), \begin{align*} \lim_{m\to\infty}\|N^m\|_{\mathcal L(\mathcal B_0)}^{1/m}=r(N). \end{align*} Since $r(N)<\rho$, there exists an integer $m_0\ge 1$ such that \begin{align*} \|N^m\|_{\mathcal L(\mathcal B_0)}\le \rho^m \end{align*} for every integer $m\ge m_0$. Define \begin{align*} C_0=\max\left\{1,\rho^{-m}\|N^m\|_{\mathcal L(\mathcal B_0)}:0\le m<m_0\right\}. \end{align*} Then $C_0>0$, and for every integer $m\ge 0$, \begin{align*} \|N^m\|_{\mathcal L(\mathcal B_0)}\le C_0\rho^m. \end{align*} Consequently, for every $u\in\mathcal B_0$ and every integer $m\ge 0$, \begin{align*} \|N^m u\|_{\mathcal B_0}\le C_0\rho^m\|u\|_{\mathcal B_0}. \end{align*} [/step]

[step:Estimate the remainder against the bounded observable]For every integer $n\ge 1$, the remainder identity and the $L^1$-$L^\infty$ inequality give \begin{align*} \left|\int_X (N^n f)g\,d\mu(x)\right|\le \|N^n f\|_{L^1(X,\mu)}\|g\|_{L^\infty(X,\mu)}. \end{align*} Using the continuous embedding $\mathcal B_0\hookrightarrow L^1(X,\mu)$ and the operator norm estimate for $N^n$, we obtain \begin{align*} \|N^n f\|_{L^1(X,\mu)}\le C_1\|N^n f\|_{\mathcal B_0}\le C_1C_0\rho^n\|f\|_{\mathcal B_0}. \end{align*} Therefore, for every $n\ge 1$, \begin{align*} \left|\int_X f(g\circ T^n)\,d\mu(x)-\left(\int_X f\,d\mu(x)\right)\left(\int_X g\,d\mu(x)\right)\right|\le C_1C_0\rho^n\|f\|_{\mathcal B_0}\|g\|_{L^\infty(X,\mu)}. \end{align*}[/step]

[guided]At this point all dynamical information has been compressed into the estimate on $N^n$. We now pair $N^n f$ with the bounded observable $g$. The relevant elementary inequality is the $L^1$-$L^\infty$ estimate: if $a\in L^1(X,\mu)$ and $b\in L^\infty(X,\mu)$, then \begin{align*} \left|\int_X ab\,d\mu(x)\right|\le \|a\|_{L^1(X,\mu)}\|b\|_{L^\infty(X,\mu)}. \end{align*} Here $a=N^n f$ and $b=g$. The function $N^n f$ belongs to $\mathcal B_0$ because $N:\mathcal B_0\to\mathcal B_0$ is bounded, and it belongs to $L^1(X,\mu)$ because the embedding $\mathcal B_0\hookrightarrow L^1(X,\mu)$ is continuous. Hence \begin{align*} \left|\int_X (N^n f)g\,d\mu(x)\right|\le \|N^n f\|_{L^1(X,\mu)}\|g\|_{L^\infty(X,\mu)}. \end{align*} The continuous embedding gives the constant $C_1>0$ from the statement: \begin{align*} \|N^n f\|_{L^1(X,\mu)}\le C_1\|N^n f\|_{\mathcal B_0}. \end{align*} The spectral radius step gives the constant $C_0>0$ and the number $\rho\in(0,1)$: \begin{align*} \|N^n f\|_{\mathcal B_0}\le C_0\rho^n\|f\|_{\mathcal B_0}. \end{align*} Combining these two inequalities yields \begin{align*} \|N^n f\|_{L^1(X,\mu)}\le C_1C_0\rho^n\|f\|_{\mathcal B_0}. \end{align*} Substituting this into the $L^1$-$L^\infty$ pairing estimate gives \begin{align*} \left|\int_X (N^n f)g\,d\mu(x)\right|\le C_1C_0\rho^n\|f\|_{\mathcal B_0}\|g\|_{L^\infty(X,\mu)}. \end{align*} Finally, the previous step identified this integral with the correlation error for every $n\ge 1$, so the same bound holds for the correlation error.[/guided]

[step:Absorb the initial time into the final constant] It remains to include $n=0$. Since $\mu$ is a probability measure, \begin{align*} \left|\int_X f g\,d\mu(x)-\left(\int_X f\,d\mu(x)\right)\left(\int_X g\,d\mu(x)\right)\right|\le 2\|f\|_{L^1(X,\mu)}\|g\|_{L^\infty(X,\mu)}. \end{align*} Using the embedding bound, \begin{align*} 2\|f\|_{L^1(X,\mu)}\|g\|_{L^\infty(X,\mu)}\le 2C_1\|f\|_{\mathcal B_0}\|g\|_{L^\infty(X,\mu)}. \end{align*} Define \begin{align*} C=\max\{C_1C_0,2C_1\}. \end{align*} Then $C>0$, $\rho\in(0,1)$, and the desired estimate holds for every integer $n\ge 0$: \begin{align*} \left|\int_X f(g\circ T^n)\,d\mu(x)-\left(\int_X f\,d\mu(x)\right)\left(\int_X g\,d\mu(x)\right)\right|\le C\rho^n\|f\|_{\mathcal B_0}\|g\|_{L^\infty(X,\mu)}. \end{align*} This proves exponential decay of correlations. [/step]