Is the multiplication on $\mathbb{R}$ a Lipschitz function?

Question

This question is related to this unanswered question on the site. My question is specifically about the special case when $n=1$:

Let $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ be the function defined as $f(x,y)=xy$. Is $f$ Lipschitz?

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An unsuccessful attempt and some thoughts:

By definition, IF it is Lipschitz, one has a uniform estimate like $$ |xy-uv|\le K|(x,y)-(u,v)|=K\sqrt{(x-u)^2+(y-v)^2}\tag{1} $$ It suffices to prove a squared version of (1): $$ (xy)^2+(uv)^2-2(xyuv)\le K \big(x^2+y^2+u^2+v^2-2(xu+yv)\big)\tag{2} $$ I see no hope to get an $K$ for (2).

On the other hand, if we fix a value of $y$ and consider the map $x\mapsto xy$, then this map is Lipschitz with a Lipschitz constant depending on the magnitude of $y$. This seems to suggest that $f$ cannot be Lipschitz.

Answer 1

Not globally, no. You can compare $|x y_1-xy_2|=|x||y_1-y_2|$ while $\sqrt{(x-x)^2+(y_2-y_1)^2}=|y_2-y_1|$ so any Lipschitz constant would need to be at least as large as the largest possible value of $x$. Same for $y$.

Answer 2

If it were Lipschitz, it would also be uniformly continuous, but it isn't. Just take sequences $x_n=(\sqrt{n},\sqrt n)$ and $y_n=(\sqrt{n+1},\sqrt{n+1}).$ Then $|y_n-x_n|\to 0,$ while $|f(y_n)-f(x_n)|\nrightarrow 0$