I was given $f: <1,+\infty>\times<1,+\infty>\rightarrow <0,+\infty> $ defined with $f(x,y)=\ln x+\ln y$ and metric on both spaces is induced by taxicab norm. I need to prove this function is not uniformly continuous,.
I tried taking $\epsilon =1$ so for $d((x_1,y_1),(x_2,y_2))=|x_2-x_1|+|y_2-y_1|<\delta$ there must be $d(f(x_1,y_1),f(x_2,y_2))=|\ln x_1+\ln y_1 -\ln x_2 - \ln y_2 |=\ln \frac{x_1y_1}{x_2y_2}\geq1 $ so I must find points $(x_1,y_1),(x_2,y_2)$ that depend only on $\delta$ and $\frac{x_1y_1}{x_2y_2}\geq e$. But I find this suspicious and can't find the way to do this. On the other hand, I also can't find proof that this IS uniformly continuous. Any help on how to find this points and what would be strategy for other functions?
It is true that $f$ is uniformly continuous on $[1,\infty)\times[1,\infty)$.
We have
$$d[f(x_1,y_1),f(x_2,y_2)]= |\ln x_2 + \ln y_2 - \ln x_1 - \ln y_1| \\\leqslant |\ln x_2 - \ln x_1| + |\ln y_2 - \ln y_1|.$$
Note that $|\ln b - \ln a| \leqslant |b-a| $ for $a,b \in [1,\infty)$. To show this assume without loss of generality that $b \geqslant a \geqslant 1$. Then we have
$$|\ln b - \ln a| = \ln\left(\frac{b}{a}\right)= \ln\left(1+\frac{b-a}{a}\right)\leqslant \frac{|b-a|}{|a|}\leqslant |b-a|.$$
Hence, for any $\epsilon > 0$,
$$d[f(x_1,y_1),f(x_2,y_2)] \leqslant | x_2 - x_1| + | y_2 - y_1|< \epsilon,$$
when
$$d[(x_1,y_1),(x_2,y_2)] = | x_2 - x_1| + | y_2 - y_1| < \delta = \epsilon.$$