Construction of a right uniformly continuous function which is not left uniformly continuous.

Question

Consider the locally compact topological group $$ G= \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} $$ Where $a>0$ and $b\in \mathbb{R}$. In the Book Measure Theory by Cohn (2nd edition, page 284, problem 12), it is asked to construct a function $f$ from $G$ to $\mathbb{R}$ that is right uniformly continuous but not left uniformly continuous. According to his definitions, a function $f$ will be left (right) uniformly continuous if for a given $\epsilon$, we can to find an open neighbourhood $U$ of $id$ such that $$| f(y)-f(x) |<\epsilon \,\,\text{whenever}\,\,y\in xU\,\, (y\in Ux).$$

As a hint, it has been told to find a suitable function $\phi :\mathbb{R}\rightarrow \mathbb{R}$ and define

$$f: \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \mapsto \phi(b).$$

I can’t find any such function $\phi$ with which the expressions of $x^{-1} y$ and $y x^{-1}$ are working. I can’t even find a function $\phi$ (not even the identity function) for which $f$ can be shown to be right uniformly continuous.

A little help will be very much appreciated. Thank you.

Answer 1

Consider an open $\epsilon$ neighbourhood of the identity $I_2$ $$B(I_2, \epsilon)=\Big\{\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}:|a-1|< \epsilon,\, |b|< \epsilon\Big\}.$$ Also consider the function $f:G\rightarrow \mathbb{R}$ sending $\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}$ to $\frac{b}{1+ |b|}.$

If we set $x_n = \begin{pmatrix} e^n & 0 \\ 0 & 1 \end{pmatrix}$ and $y_n = \begin{pmatrix} e^n & n \\ 0 & 1 \end{pmatrix}$, then $x_n ^{-1} y_n$ tends to $I_2$ but $|f(x_n) - f(y_n)|=\frac{n}{n+1}$ tends to $1$ as $n$ tends to $\infty$. So $f$ is not left uniformly continuous.

Now consider $x=\begin{pmatrix}x_1 & x_2 \\ 0 & 1 \end{pmatrix}$ and $y=\begin{pmatrix}y_1 & y_2 \\ 0 & 1 \end{pmatrix}$ such that $yx^{-1}\in B(I_2, \epsilon),\,\,1>\epsilon.$ Then it is easy to see the following:

1. $1-\epsilon <\frac{y_1}{x_1}$, $|x_2|<\frac{\epsilon}{1-\epsilon}$, $|y_2|<\epsilon$ and $\frac{|x_2 -y_2|}{1+|x_2|}<\epsilon$.

2. If $x_2 y_2 \geq 0$, then $x_2 |y_2| -|x_2| y_2=0$.

3. If $x_2 y_2 <0$, then $|\,x_2 |y_2| -|x_2| y_2\,| <2\frac{\epsilon ^2}{1-\epsilon}$.

4. $|f(x)-f(y)|\leq \frac{|x_2 -y_2|}{1+|x_2|} + |\,x_2 |y_2| -|x_2| y_2\,| $.

From these claims, it is easy to deduce that $f$ is right uniformly continuous.