# Decompose a function into a bounded part and a Lipschitz part

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!

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.
• 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.) Feb 27 at 16:05
• 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. Feb 28 at 19:24