Let $g:\mathbb R\to\mathbb R$ be continuous and define $f:\mathbb R^2\to\mathbb R$ by $f(x_1,x_2)=g(x_1x_2)$. Show that $f$ is uniformly continuous only if $g$ is a constant function.
I'm not sure whether the following is right. I don't use the continuity of $g$, so I guess something in the argument must be wrong.$\newcommand{\norm}[1]{\lVert#1\rVert}$
Proof. Suppose that $g$ is not constant. Then there exist $\gamma_0,\gamma_1\in\mathbb R$ such that $g(\gamma_0)\neq g(\gamma_1)$. Let $\{a_n:=(n,\gamma_0/n)\}_{n\in\mathbb N}$ and $\{b_n:=(n,\gamma_1/n)\}_{n\in\mathbb N}$ be sequences on $\mathbb R^2$. Then $$\norm{a_n-b_n}=\norm{\left(0,\frac{\gamma_0-\gamma_1}n\right)}\to\norm{(0,0)}=0$$ as $n\to\infty$, but $$\norm{f(a_n)-f(b_n)}=\norm{g(\gamma_0)-g(\gamma_1)}$$ is constant and non-zero. Therefore $f$ cannot be uniformly continuous.