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Theorem 17.16.766E38862017F

Theorem 17.16.7.66E38862017F Suppose that $u_0\sim u_1$, and choose $C\geq 0$ with $u_0-C\leq u_1\leq u_0+C$. Then the Perron segment from $u_0$ to $u_1$ exists, every slice has the ordinary singularity type of $u_0$ and $u_1$, and

\[ u_0-Ct\leq u_ t^{\mathrm P}\leq u_0+Ct, \qquad u_1-C(1-t)\leq u_ t^{\mathrm P}\leq u_1+C(1-t). \]

Moreover, away from the common pole set,

\[ |u_ t^{\mathrm P}-u_ s^{\mathrm P}|\leq C|t-s|, \qquad 0\leq s,t\leq 1. \]

Proof. Consider the barrier

\[ w_ t=\max \{ u_0-Ct,\ u_1-C(1-t)\} . \]

It is a subgeodesic by proposition 17.16.2, and the bounded difference gives $w_0=u_0$ and $w_1=u_1$. Upper semicontinuity gives the joint upper limits at the endpoints. Off the common poles, $|w_ t-u_0|\leq Ct$ and $|w_ t-u_1|\leq C(1-t)$, so $w_ t$ runs from $u_0$ to $u_1$, and proposition 17.16.6 applies.

The lower bounds follow from $w_ t\leq u_ t^{\mathrm P}$, and the upper bounds from the chord estimate; together they give the asserted singularity type. Off the common poles, put $f(t)=u_ t^{\mathrm P}(x)$. For $0<s<t<1$, monotonicity of the slopes of secants of the convex function $f$ gives

\[ -C\leq \frac{f(s)-f(0)}s \leq \frac{f(t)-f(s)}{t-s} \leq \frac{f(1)-f(t)}{1-t}\leq C. \]

The two-sided estimates at the endpoints cover the cases $s=0$ and $t=1$, which proves the Lipschitz bound on $[0,1]$. $\square$

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