Continuity of distance function

Question

I wonder if this is obvious because it does not appear to me obvious at all:

Reference: [Hormander: An introduction to Complex Analysis in Several Variables], page 37: Here is the quote

Now, let $\delta$ be an arbitrary continuous function in $\mathbb{C}^n$ such that $\delta>0$ except at $0$, and $$\delta(tz)=|t|\delta(z)$$ for all $t\in\mathbb{C}$ and $z\in\mathbb{C}^n$. Set $$\delta(z,\Omega^c)=\inf_{w\in\Omega^c}\delta(z-w).$$ It is obvious that this is continuous.

Why should this be obvious?? First I know it should be clear that $\delta(z,\Omega)$ is upper semicontinuous becase it is defined to be the infinimum over a family of continuous functions. But it is not yet clear to me that this should be lower semicontinuous as well. Any hints?

NOTE: I have no idea here what kind of open set $\Omega$ is as it is not explicitly stated.

Answer 1

I guess that $\Omega^c$ is the complement of $\Omega$.

1. The conditions on $\delta$ implies that distances $\delta(\cdot)$ and $|\cdot|$ are equivalent, i.e. $$ m|z|\le\delta(z)=\delta\left(\frac{z}{|z|}\right)|z|\le M|z| $$ where $M>0$ and $m>0$ are the largest, resp. the smallest, values of $\delta$ on the unit circle.
2. For fixed $z_1$, $z_2\in\Omega$ consider two sets $\Omega_k=\{w\in\Omega^c\colon\ \delta(z_k-w)\le\delta(z_k,\Omega^c)+1\}$, $k=1,2$. They are nonempty and compact due to 1 and closeness of $\Omega^c$. Thus the infimum is attained and $\exists w_1\in\Omega_1$, $w_2\in\Omega_2$ such that $$ \delta(z_1,\Omega^c)=\delta(z_1-w_1),\qquad \delta(z_2,\Omega^c)=\delta(z_2-w_2). $$ Note that it is possible to bound \begin{eqnarray*} \delta(z_2-w_2)&\le&\delta(z_2-w_1)\le M|z_2-w_1|\le M|z_2-z_1|+M|z_1-w_1|\le\\ &\le&M|z_2-z_1|+\frac{M}{m}\delta(z_1-w_1).\tag1 \end{eqnarray*}
3. Let us show that $w_2$ cannot be far away from $w_1$ if $z_2$ is close to $z_1$. Assume that $|z_1-z_2|\le 1$ and perform quite boring estimation (using $(1)$ at the end) \begin{eqnarray*} |w_1-w_2|&\le&|w_1-z_1|+|z_1-z_2|+|z_2-w_2|\le \frac{1}{m}\delta(z_1-w_1)+1+\frac{1}{m}\delta(z_2-w_2)\le\\ &\le&\frac{1}{m}\left(1+\frac{M}{m}\right)\delta(z_1,\Omega^c)+1+\frac{M}{m}=C. \end{eqnarray*} It means that if infimum for $z_1$ is attained at $w_1$ and $z\in B_{z_1}$ (the unit ball) then the infimum for $z$ is attained at $w$ that for sure belongs to the ball around $w_1$ with radius $C$. Therefore, it is enough to consider the compact set $\Omega_\text{tmp}=\{w\in\Omega^c\colon\ |w_1-w|\le C\}$.
4. Due to compactness of the set $B_{z_1}-\Omega_\text{tmp}$ the function $\delta$ is uniformly continuous on it, i.e. $\forall\epsilon\ \exists\nu\le 1$ such that $$ -\epsilon\le\delta(z_1-w)-\delta(z-w)\le\epsilon,\quad\forall w\in\Omega_\text{tmp},\quad |z_1-z|=|(z_1-w)-(z-w)|\le\nu. $$
5. The right inequality gives $$ \delta(z_1,\Omega^c)\le\delta(z_1-w)\le \delta(z-w)+\epsilon,\quad\Rightarrow \quad \delta(z_1,\Omega^c)\le\delta(z,\Omega^c)+\epsilon $$ since the infimum for $w$ is attained inside $\Omega_\text{tmp}$.
6. Similarly the left one gives $$ \delta(z,\Omega^c)-\epsilon\le\delta(z_1,\Omega^c). $$
7. Combining 5 and 6 gives continuity at $z_1$.

UPDATE: Added more details to the item 2 and the new item 3.

Answer 2

Let me denote $\delta=d$. Let $M=sup_{|x|=1}(\delta(x))$, and $N=inf_{|x|=1}(\delta(x))$. These are finite by compactness. We may assume that $N>0$, since if it is not, we may add $(|N|+1)|x|$, which will not effect continuity or the desired properties. Now note. $$d(x+y)\le |x+y|d((x+y)/|x+y|)\le M|x+y|\le M|x|+M|y|$$ Let $x\in \mathbb{C}^n$. Now let $z\in B(1, x)$. Let $A=sup_{z\in B(1, x)}d(z)$. $$d(z, w)-d(z, y)\le Md(z)+Md(w)-N|y-z|\le MA+Md(w)-N|y-z|$$ Now $MA+Md(w)/N<|y-z|$ for $|y|>1+MA+Md(w)/N$ so that for $y$ this large, $d(z, w)<d(z, y)$ for all $y$. Thus for $z\in B(1, x)$, we may assume that $\Omega^C\subseteq B(1+MA+Md(w)/N, x)$, or that $\Omega^C$ is compact.

Now the infinum of a continuous function in two variables over a compact set is always continuous, so we get the result.

Answer 3

Similarly to the other answers, we will need to compare the usual distance of $\mathbb{C}^n$ and $\delta$ (which, as you noted, does not induce a distance). Let $m=\inf\left\{\delta(z):\Vert z\Vert=1\right\}$ and $M=\sup\left\{\delta(z):\Vert z\Vert=1\right\}$. The condition on $\delta$ implies $$m\Vert z\Vert\leq\delta(z)\leq M\Vert z\Vert$$ for all $z\in \mathbb{C}^n$. Note that the first inequality implies that whenever $(z_n)$ is a sequence in $\mathbb{C}^n$ with $\Vert z_n\Vert\to\infty$, we also have $\delta(z_n)\to\infty$ (these are sometime called proper functions, I believe).

Lemma: For every $z\in\mathbb{Z}^n$, there exists $w\in\Omega^c$ such that $\delta(z,\Omega^c)=\delta(z-w)$. Moreover, $\Vert w\Vert\leq(1/m)\delta(z,\Omega^c)+\Vert z\Vert$.

Proof: Let $w_n$ be a sequence in $\Omega^c$ with $\delta(z-w_n)\to \delta(z,\Omega^c)$. Then $(w_n)$ is bounded by the comment above, so we can take a converging subsequence and obtain such $w$. Then $$\Vert w\Vert\leq\Vert z-w\Vert+\Vert z\Vert\leq (1/m)\delta(z-w)+\Vert z\Vert=(1/m)\delta(z,\Omega^c)+\Vert z\Vert.QED$$

Now let $z\in \mathbb{C}^n$ be fixed. As you noted, $\delta(\cdot,\Omega^c)$ is lower semicontinuous, so there exists $r>0$ such that $\delta(z',\Omega^c)<\delta(z,\Omega^c)+1$ whenever $\Vert z'-z\Vert\leq r$. For all such $z'$, the lemma gives us $w'\in\Omega^c$ with $\delta(z',\Omega^c)=\delta(z'-w')$ and $\Vert w'\Vert\le(1/m)\delta(z',\Omega^c)+\Vert z'\Vert\leq(1/m)\delta(z,\Omega^c)+1+\Vert z\Vert+ r$.

Let $\overline{B_r(z)}$ be the closed ball of radius $r$ around $z$ and $K=\Omega^c\cap\left\{w:\Vert w\Vert\leq(1/m)\delta(z,\Omega^c)+1+\Vert z\Vert\right\}$. The previous paragraph implies that $$\delta(z',\Omega^c)=\inf_{w\in K}\delta(z'-w),\qquad z'\in \overline{B_r(z)}$$

Now things are easy, since we can work in the compact $K$ instead of all $\Omega^c$. Precisely, we just need to prove the following proposition:

Proposition: Let $X$ and $Y$ be compact metric spaces and $f:X\times Y\to \mathbb{R}$ be continuous. Define $F(x)=\inf_{y\in Y}f(x,y)$ for all $x\in X$. Then $F$ is continuous.

Our problem follows from this by letting $X=\overline{B_r(z)}$, $Y=K$ and $f(x,y)=\delta(x-y)$, so that $\delta(\cdot,\Omega^c)=F$ in $\overline{B_r(z)}$.

Proof of proposition: Since $f$ is a continuous function on a compact, it is uniformly continuous. Given $\epsilon>0$, there exists $\nu>0$ such that $d(x,x')<\nu$ and $d(y,y')<\nu$ implies $|f(x,y)-f(x',y')|<\epsilon$. Suppose $x,x'\in X$ with $d(x,x')<\nu$, and we will show that $|F(x)-F(x')|<\epsilon$.

Choose $y$ and $y'$ in $Y$ with $f(x,y)=F(x)$ and $f(x',y')=F(x')$. Then the choice of $\nu$ implies $|f(x',y)-f(x,y)|<\epsilon$ and $|f(x,y')-f(x',y')<\epsilon$, thus $$F(x')=f(x',y')>f(x,y')-\epsilon\geq F(x)-\epsilon$$ and $$F(x)=f(x,y)>f(x',y)-\epsilon\geq F(x')-\epsilon$$ therefore $|F(x')-F(x)|<\epsilon$, as we wanted. QED

This shows that $\delta(\cdot,\Omega^c)$ is continuous on $\overline{B_r(z)}$, so in particular it is continuous at $z$.