Theorem 15.5.456C5E320D330

Proof. Normalize $\sup _ X\varphi =0$ and write $M=\| \varphi \| _\infty $. By proposition 5.29.13, the measure $\mu $ has a square-capacity bound with a fixed constant. Put

The weighted estimate of lemma 5.42.8 gives $\| R_\delta \varphi -\varphi \| _{L^1(\mu )}\leq C\delta ^\eta $.

Use the constants $A_0,K$ of lemma 5.42.1, taking $A_0\geq 1$, and define the unshifted transform

It differs from $\Phi _{b,\delta }$ by the constant $K(\delta ^2+\delta )$, so

when $\delta \leq 1$. Choose $\varepsilon =\delta ^ q$ and $b=(\varepsilon -2K\delta )/A_0$. For small $\delta $, $\varepsilon \leq 1/2$ and $\varepsilon /(2A_0)\leq b\leq \varepsilon /A_0$.

Set $T=(1-\varepsilon )F$. Then $\omega +\mathrm{dd}^{\mathrm{c}}T\geq \varepsilon ^2\omega $. The transform is continuous: its positive logarithmic penalty excludes radii near zero uniformly in the base point for each fixed $\delta ,b$, and the minimum of its continuous defining function on the resulting compact interval depends continuously on the base point. Take $B\geq M+2K+1$ and set $\psi =T-B\varepsilon $. Since $R_\delta \varphi \leq 0$, we have $-M-B\leq \psi \leq 0$. Moreover,

The weighted smoothing estimate and the fixed bound for $\mu (X)$ give $\int (\psi -\varphi )_+\, \mathrm d\mu \leq C\delta ^\eta $. Thus proposition 14.19.23, with a fixed competitor bound $H\geq M+B+1$, gives

Since $\varphi \leq 0$,

At a minimizing radius $t_0\in (0,\delta ]$, monotonicity of $R_ t\varphi +Kt^2$ implies

The constant $\kappa $ is independent of $\delta $ and the point. Applying monotonicity again gives $R_{\kappa \delta }\varphi -\varphi \leq 2C_1\delta ^ q$. With the lower bound $R_ r\varphi -\varphi \geq -Kr^2$ and $r=\kappa \delta $, this proves $\| R_ r\varphi -\varphi \| _\infty \leq Cr^ q$.

Finally lemma 5.42.2 gives $\| \nabla R_ r\varphi \| _\infty \leq C/r$. For points a distance $s$ apart,

Take $r=s^{1/(1+q)}$ for small $s$; boundedness handles larger distances. The exponent is $q/(1+q)=\beta $. $\square$