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We say a subset $\overline{D}$ of $\Bbb R^2$ is $x\textbf{-normal}$ is there exist $a,b\in \Bbb R$ with $a<b$ and continuous functions $\psi_1,\psi_2:[a,b]\to \Bbb R$ such that $\forall x\in ]a,b[(\psi_1(x)<\psi_2(x))$ (open interval as a domain) and $\overline{D}=\{(x,y)\in \Bbb R^2: a\leq x\leq b \ \wedge \psi_1(x)\leq y\leq \psi_2(x)\}$.

We say $\overline{D}$ is $y\textbf{-normal}$ if the set $D:=\{(x,y)\in \Bbb R^2:(y,x)\in \overline{D}\}$ is $x$-normal.

We say $\overline{D}$ is normal if it is both $x$-normal and $y$-normal.

We say $\overline{D}$ is a normal regular domain if it is normal and we require the inequality $\psi_1<\psi_2$ to also hold on $\{a,b\}$.

Question : I want to prove that any circle is a normal regular domain. I already proved that any circle is normal. However the regularity escapes me.

I suppose it's easy to generalize from the circles centered at the origin to any circle, so I'll write what I've done on that.

Suppose the circle $C:=\{(x,y)\in \Bbb R^2:x^2+y^2\leq r^2\}$ is a normal regular domain.

Then twe have $C=\{(x,y)\in \Bbb R^2: a\leq x\leq b \ \wedge \psi_1(x)\leq y\leq \psi_2(x)\}$ for some $a,b\in \Bbb R$ with $a<b$ and some continuous functions such that $\forall x\in [a,b](\psi_1(x)<\psi_2(x))$.

Since $(a,\psi_1(a)),(a,\psi_2(a)),(b,\psi_1(b)),(b,\psi_2(b))\in C$ we have $a^2+(\psi_1(a))^2\leq r^2$ and $a^2+(\psi_2(a))^2\leq r^2$. Subtracting the inequalities from each other we can see that $|\psi_1(a)|=|\psi_2(a)|$ and because of the inequality $\psi_1<\psi_2$ we must have $\psi_1( a) =-\psi_2(a)$.

I can't do anything from this nor do I see any other way.

À toa
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2 Answers2

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The definition of normal regular domain you give is not what I would expect. And the circle does not satisfy it. In your definition the boundary of a regular domain should contain some segment.

Emanuele Paolini
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    If it isn't a normal regular domain, can you please tell me functions $\psi_1, \psi_2$ that make a circle normal but not regular? – À toa Mar 09 '13 at 00:03
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Following up on your proof by contradiction, prove that $[a,b\textbf{]}=[-r,r\textbf{]}$.

$\subseteq$: Let $x\in [a,b\textbf{]}$. Since $C$ is regular $(x,\psi_1(x))\in C$. Therefore $x^2\leq x^2+(\psi_1(x))^2\leq r^2$ and $-r\leq x\leq r$ follows.

$\supseteq:$ Let $x\in [-r,r\textbf{]}$. Now note that $r^2-x^2\ge 0$ and

$$\begin{align} -r\leq x\leq r &\Longrightarrow &x^2\leq r^2 \\ &\Longrightarrow &x^2+r^2-x^2\leq r^2 \\ &\Longrightarrow &x^2+(\sqrt{r^2-x^2})^2\leq r^2\\ &\Longrightarrow &\Bigl(x,\sqrt{r^2-x^2}\Bigr)\in C \end{align}$$ Therefore $x\in [a,b\textbf{]}$.

From $[a,b\textbf{]}=[-r,r\textbf{]}$ conclude that $a=-r$ and use the fact that $(a,\psi_1(a)),(a,\psi_2(a))\in C$:

$$\begin{align} (a,\psi_1(a)),(a,\psi_2(a))\in C&\Longrightarrow a^2+(\psi_1(a))^2\leq r^2 \wedge a^2+(\psi_2(a))^2\leq r^2\\ &\Longrightarrow r^2+(\psi_1(a))^2\leq r^2 \wedge r^2+(\psi_2(a))^2\leq r^2\\ &\Longrightarrow \psi_1(a)=0=\psi_2(a) \end{align}$$

which contradicts the assumption that $\psi_1(a)<\psi_2(a)$.

$\therefore$ C isn't regular.

Git Gud
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