Global Lipschitz implies bounded in coefficient

Consider $g:\mathbb{R}^2\to \mathbb{R}$ of the form $g(x,y)=p(x)q(y).$

Assume $g$ is uniformly Lipschitz in $x,y$ in the sense that there exists $K>0$ such that for any $(x_1,y_1),(x_2,y_2)\in \mathbb{R}^2$ we have $$|g(x_1,y_1)-g(x_2,y_2)|\le K(|x_1-x_2|+|y_1-y_2|).$$ Then does this global Lipschitz assumption implies $p,q$ must be bounded?

I think it's true, since the lipshcitz assumption intuitively means for each component, the function is bounded by a linear function uniformly in the other component. Hence $p,q$ can't be unbounded. May I know whether my intuition is correct?

• Consider $p(x)=x$, $q(y)=1$. – David C. Ullrich May 30 '16 at 15:00
• Think (loosely) of the Lipschitz constant as an upper bound on the derivative (which need not exist, of course). Just because the derivative is bounded does not mean that the function is bounded. – copper.hat May 30 '16 at 15:42

It's not quite true, as David C. Ullrich points out in the comments. If one of the two factors is constant, the other has no reason to be bounded. But for $g$ of the given specific form, that is the only thing that allows unboundedness of one of the factors.
If $p$ is not constant, we choose $x_1,x_2$ with $p(x_1) \neq p(x_2)$, and then find
$$\lvert q(y)\rvert \leqslant \frac{K\cdot\lvert x_1 - x_2\rvert}{\lvert p(x_1) - p(x_2)\rvert}$$
for all $y\in \mathbb{R}$, which shows $q$ must be bounded. Symmetrically, if $q$ is not constant, then $p$ must be bounded.
So a function of the form $g(x,y) = p(x)q(y)$ is uniformly Lipschitz continuous on $\mathbb{R}^2$ if and only if
2. $p$ and $q$ are both uniformly Lipschitz continuous on $\mathbb{R}$, and either one of the factors is constant, or both factors are bounded.