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$\newcommand\R{\Bbb R}$Let $F$ be the set of all functions of the form $\max(a,b,c)$, where $a,b,c$ are affine functions from $\R^2$ to $\R$ and the maximum is taken pointwise. Let $G$ be the set of all functions of the form $\sum_{i=1}^n c_i f_i$, where $n$ is a natural number, the $c_i$'s are nonnegative real numbers, and the $f_i$'s are in $F$.

Is it true that every convex function from $\R^2$ to $\R$ is the pointwise limit of a sequence of functions in $G$?

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  • $\begingroup$ If your claim is true, then I may pick $F$ as my set of test functions in my post mathoverflow.net/questions/478453/… $\endgroup$
    – GJC20
    Commented Sep 9 at 8:48
  • $\begingroup$ @GJC20 : Yes, that was the motivation. $\endgroup$ Commented Sep 11 at 22:04

2 Answers 2

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The Laplacian measure of a function in the closure of $G$ is the weak limit of linear combinations of the Laplacian measures of your elementary functions, whose support are "tripods", that is, unions of three rays emanating from a vertex.

In particular, the support of the Laplacian measure of a function in the closure of $G$ must be a union of tripods. Take a maximum of three independent linear functions that is lower bounded, and take the max of that and a positive number. This function has a Laplacian whose support cannot be expressed as a union of tripods.

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  • $\begingroup$ Thank you. Simple and clean. $\endgroup$ Commented Sep 9 at 16:16
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The answer is negative. Example: $$\max\{ |y|,|x|-1\}.$$ This is a convex piecewise linear function which is not a convex combination of your "tripods", and not a limit of such combinations.

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    $\begingroup$ Why is this a counterexample? And why do you need $0$ here? $\endgroup$ Commented Sep 9 at 12:46
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    $\begingroup$ there is a typo in the example. $\endgroup$
    – alesia
    Commented Sep 9 at 15:48
  • $\begingroup$ @Iosif Pinelis: thanks, the zero was unnecessary and I removed it. $\endgroup$ Commented Sep 10 at 0:50

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