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let $D:=\left\{z\in\mathbb{C}: \lvert z\rvert <1\right\}$ and define $P\colon D\times D\to \mathbb{R}$ by $$ P(x,y)=\begin{cases}\frac{1-\lvert x\rvert^2}{\lvert x-y\rvert^2}, & \text{ if }x\neq y\\0, & \text{ if }x=y\end{cases}. $$

Is $P(x,\cdot)$ continous in the second component?

I think that means: Is $y\mapsto P(x,y)$ for any $x\in D$ continous?

As far as I see for $x=0$ we have that $y\mapsto P(x,y)$ is not continous in $y=0$.

But I do not see more cases like this.

So is my suggestion right, that $$ y\mapsto P(x,y)\text{ is continous in }D\setminus \left\{0\right\}\text{ for any }x\in D? $$

mathfemi
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1 Answers1

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Take an arbitrary $x<1$. For $P(x,\cdot)$ to be continuous, it must be true that the limit of $P(x,y)$ as $y$ approaches $x$ is equal to the value $P(x,x)$, i.e.

$\lim_{y\rightarrow x}P(x,y) = P(x,x)=0.$

But the denominator in the expression for $P$ gets arbitrarily close to $0$ for $y$ tending towards $x$, and thus the left hand side limit is infinity.

So I'd say that $P(x,\cdot)$ is continuous except for $y=x$.

Mankind
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