Schweitzer 2019

Since $S$ is compact with nonempty interior, choose a point $c$ and radii $0<r\le R$ with $B(c,r)\subseteq S\subseteq B(c,R)$. We show that any $\alpha<r/R$ works.

Let $P$ be a point; if $P\in S$ there is nothing to prove. So let $d=\operatorname{dist}(P,S)>0$, and suppose, for contradiction, that $\operatorname{dist}(P,H_i)<\alpha d$ for every $i$.

A test point. Choose $s\in(0,1)$ with $\dfrac{s}{1-s}=\dfrac{\alpha d}{r}$, and let $Y=(1-s)P+sc$, a point on the segment from $P$ toward the center.

$Y$ lies in every $H_i$. Each $H_i$ contains $S$, hence the ball $B(c,r)$, and it contains a point $P_i$ with $|P-P_i|<\alpha d$. Write $$Y=(1-s)P_i+s\Big(c+\tfrac{1-s}{s}(P-P_i)\Big).$$ The second point is within $\frac{1-s}{s}\alpha d=r$ of $c$, so it lies in $B(c,r)\subseteq H_i$. As $H_i$ is convex, $Y\in H_i$. Therefore $Y\in\bigcap_iH_i=S$.

But $Y$ is outside $S$. If $Y\in S$, then $d\le|P-Y|=s\,|P-c|$. The nearest point of $S$ to $P$ lies within $R$ of $c$, so $|P-c|\le d+R$, and $$d\le s(d+R)\ \Longrightarrow\ d\le\frac{s}{1-s}R=\frac{\alpha dR}{r}\ \Longrightarrow\ r\le\alpha R,$$ which contradicts $\alpha<r/R$. So some $H_i$ has $\operatorname{dist}(P,H_i)\ge\alpha\operatorname{dist}(P,S)$.

Let $n=\deg P$. A constant $P$ cannot give a Jordan curve, and for $n=1$ there are no critical points, so let $n\ge2$. By the Jordan curve theorem, the complement of $L$ has two components: a bounded inside $\Omega$ and an unbounded outside $E$.

Step 1: $\Omega$ is exactly where $|P|<1$. Neither component meets $L$, so $|P|\ne1$ on each, and each is connected; hence $|P|<1$ on all of it or $|P|>1$ on all of it. On $E$, $|P|\to\infty$ at infinity, so $|P|>1$ there. If also $|P|>1$ on $\Omega$, then $P$ would have no zeros at all (a zero is not on $L$ either). So $|P|<1$ on $\Omega$, and all $n$ zeros of $P$ lie in $\Omega$.

Step 2: a proper map. $P$ maps $\Omega$ into the unit disc $D$. If points of $\Omega$ leave every compact subset of $\Omega$, they approach $L$ (as $\Omega$ is bounded), so $|P|\to1$ along them. Thus $P:\Omega\to D$ is proper. A proper nonconstant holomorphic map has a degree: every point of $D$ has the same number of preimages, counted with multiplicity. Counting the preimages of $0$, the degree is $n$.

Step 3: Riemann–Hurwitz. For a proper holomorphic map of degree $n$ between plane domains, $\chi(\Omega)=n\,\chi(D)-b$, where $b$ is the number of critical points in $\Omega$ counted with multiplicity. The inside of a Jordan curve is simply connected, so $\chi(\Omega)=\chi(D)=1$ and $b=n-1$. But $P'$ has only $n-1$ zeros in the whole plane, counted with multiplicity. Hence all of them lie in $\Omega$, the interior of $L$.

The hypothesis doesn’t change if we translate $f$, so it is enough to show that $f$ is integrable on some interval around $0$.

Step 1: many good shifts. The function $(x,t)\mapsto f(x+t)-f(x)$ is measurable on the plane. By Tonelli’s theorem, $\Phi(t)=\int_{-2}^{2}|f(x+t)-f(x)|\,dx$ is a measurable function of $t$, finite for every $t$ by hypothesis. The sets $E_N=\{t\in[-1,1]:\Phi(t)\le N\}$ increase to $[-1,1]$, so some $E=E_N$ has positive measure. By Steinhaus’ theorem, $E-E$ contains an interval $(-\delta,\delta)$.

Step 2: all small shifts are good. Let $s=t-t'$ with $t,t'\in E$. For every $x$, $$f(x+s)-f(x)=\big[f(x-t'+t)-f(x-t')\big]-\big[f(x)-f(x-t')\big].$$ For $x\in[-1,1]$, the point $x-t'$ lies in $[-2,2]$, so $$\int_{-1}^{1}|f(x+s)-f(x)|\,dx\le\Phi(t)+\Phi(t')\le2N\qquad(|s|<\delta).$$

Step 3: a good base point. Integrate over $s\in(0,\delta)$ and swap the integrals: $$\int_{-1}^{1}\Big(\int_0^\delta|f(x+s)-f(x)|\,ds\Big)dx\le2N\delta .$$ So for almost every $x_0\in[-1,1]$ the inner integral is finite, and since $f(x_0)$ is a real number, $\int_{x_0}^{x_0+\delta}|f|<\infty$. The same argument with $s\in(-\delta,0)$ gives $\int_{x_0-\delta}^{x_0}|f|<\infty$ for almost every $x_0$. Pick one $x_0$ with $|x_0|<\delta/2$ that is good for both. Then $f$ is integrable on $(x_0-\delta,x_0+\delta)$, which contains $(-\delta/2,\delta/2)$.