15.5 Hölder continuity from bounded stabilityBDF9BACE965B

Proof. After adding a constant we may assume $-C_0\leq u\leq 0$. Choose

For small $\delta >0$ set

The curvature estimate of lemma 5.42.1 gives $v_\delta \in \mathrm{PSH}(X,\omega )$. Evaluating the infimum defining $\Phi _{c,\delta }$ at $t=\delta $ gives $\Phi _{c,\delta }\leq R_\delta u\leq 0$. On the other hand,

since $R_ tu+Kt^2\geq u$, $Kt\geq 0$ and $-c\log (t/\delta )\geq 0$ for $0<t\leq \delta $. For small $\delta $ this lower bound is at least $r$, and therefore

The Hermitian form $\omega $ is semipositive and big, so theorem 15.4.1 applies in the background $\omega $. Together with the $L^1$ estimate of lemma 5.42.2, it gives

Since $e\leq C\delta ^\alpha $ and $u-r\leq C_0+1$, it follows that

We now bound the minimizing radius from below. Fix $z$ and a radius $t=t(z)\in (0,\delta ]$ at which the infimum is attained. The lower estimate at this radius and (15.5.2) give

Thus $t(z)\geq a\delta $ with the constant $a=\mathrm{e}^{-C}>0$, which is independent of $z$ and $\delta $. Since $R_ su+Ks^2$ is increasing in $s$,

Combining this with (15.5.2) and the lower bound $R_ su-u\geq -Ks^2$, and setting $s=a\delta $, we obtain

for every sufficiently small $s>0$.

Finally, let $x,y$ be points at a small distance $d$. The gradient estimate $\| \nabla R_ su\| _\infty \leq C/s$ gives

With $s=d^{1/(1+\alpha )}$, both terms are bounded by a constant times $d^{\alpha /(1+\alpha )}$, and larger distances are controlled by the bound on the oscillation. This proves Hölder continuity with exponent $\alpha /(1+\alpha )$. Letting $\eta $ approach its upper endpoint gives every $\beta $ in the theorem; smaller exponents follow on compact $X$ by changing the constant. $\square$

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 $. $\square$