Let $f: \mathbb R^d \to \mathbb R^d$ be a measurable function such that $$ \sup_{x,y \in \mathbb R^d} \frac{|f(x) - f(y)|}{\max \{1, |x-y| \}} < \infty. $$

Are there functions $g,h: \mathbb R^d \to \mathbb R^d$ such that $f = g + h$ and $$ \begin{align*} \sup_{\substack{x,y \in \mathbb R^d \\x \neq y}} \frac{|g(x) - g(y)|}{|x-y|} + \sup_{x \in \mathbb R^d} |h(x)| &< \infty. \end{align*} $$ ?

Thank you so much for your elaboration!


1 Answer 1


Sure. Triangulate the unit cube $[0,1]^d$ into $d!$ simplices $S_1$, $\ldots$, $S_{d!}$, as in any of the answers to this question.

Let $g$ satisfy $g(\vec{n}) = f(\vec{n})$ for all $\vec{n} \in \mathbb{Z}^d$, and make it linear on $S_i + \vec{n}$ for every $1 \leq i \leq d!$ and $\vec{n} \in \mathbb{Z}$. Then $g$ is Lipschitz and $f - g$ is bounded.

  • 5
    $\begingroup$ One does not even need triangulation, since any scalar Lipschitz function on a subset of a metric space can be extended to a Lipschitz function on the full space with the same Lipschitz constant. Since the restriction of $f$ to say ${\bf Z}^d$ is already Lipschitz, just extend each component to get $g$. (Presumably the general theory of quasi-isometries would also handle this question, though I didn't see a particularly slick way to do so.) $\endgroup$
    – Terry Tao
    Feb 27 at 16:05
  • $\begingroup$ Yeah, good point. $\endgroup$
    – Nik Weaver
    Feb 27 at 16:18
  • 1
    $\begingroup$ Another approach (relying not on discretization, but on the vector space structure on ${\bf R}^d$) is to take $g$ to be a convolution of $f$ with a standard bump function of unit mass. Verifying the required estimates is then a nice undergraduate real analysis exercise. $\endgroup$
    – Terry Tao
    Feb 28 at 19:24

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