ScalingStacks

Proof. By Lemma 4.3, $\bar{\phi}$ is convex, and by Prop. 4.1 it has an $L^{1}$ bound in the $x^{m_{i}}$ coordinates:

$$\int|\bar{\phi}|dx^{m_{1}}\ldots dx^{m_{n}}\leq C.$$

Clearly $\bar{\phi}$ is also bounded above, so for the argument we may pretend $\bar{\phi}\leq 0$ upon shifting by a bounded constant.

We claim $\bar{\phi}(x)$ is bounded from below for $x$ in a shrinked interior region. The ball $B(x,2r)$ is contained in the coordinate chart, with $r$ bounded below by a positive constant. For $y$ in the annulus $B(x,2r)\setminus B(x,r)$, we have $2\bar{\phi}(\frac{x+y}{2})\leq\bar{\phi}(x)+\bar{\phi}(y)$, so upon integration

$$\int|\bar{\phi}|\gtrsim\int 2|\bar{\phi}(\frac{x+y}{2})|dy\geq\int|\bar{\phi}(x)|+|\bar{\phi}(y)|dy,$$

which bounds $|\bar{\phi}(x)|$. Thus on a slightly shrinked $x$-domain the oscillation is bounded:

$$\text{osc }\bar{\phi}=(\sup-\inf)\bar{\phi}\leq C,$$

and the Lipschitz bound follows again by convexity. ∎