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Q1: Are there some linear forms $a_ix+b_iy+c_iz$ and signs $\sigma_i\in\{1,-1\}$ such that

$$\max(x,y,z)=(a_0x+b_0y+c_0z)+\sum_{i\geq1}\sigma_i|a_ix+b_iy+c_iz|$$

for all $x,y,z\in\mathbb R$ ?

Q2: Same question for $\max(|x|,|y|,|z|)$.

For the analogous questions with only two variables, the answers are

$$\max(x,y)=\frac{x+y}{2}+\left|\frac{x-y}{2}\right|$$

and

$$\max(|x|,|y|)=\left|\frac{x+y}{2}\right|+\left|\frac{x-y}{2}\right|$$

Show that the $\max{ \{ x,y \} }= \frac{x+y+|x-y|}{2}$.

And the two-variable case can be applied to the three-variable case:

$$\max(x,y,z)=\max(\max(x,y),z)$$ $$=\frac{\frac{x+y+|x-y|}{2}+z+\left|\frac{x+y+|x-y|}{2}-z\right|}{2}$$ $$=\frac{x+y+2z}{4}+\left|\frac{x-y}{4}\right|+\left|\frac{x+y-2z}{4}+\left|\frac{x-y}{4}\right|\right|$$

But this has nested absolute values, which I don't want.

I did find an approximate formula, which is exact when one of the variables is $0$ or when all are equal:

$$\max(|x|,|y|,|z|)\approx$$ $$-\tfrac13\Big(|x|+|y|+|z|\Big)\\+\tfrac16\Big(|x+y|+|x-y|+|x+z|+|x-z|+|y+z|+|y-z|\Big)\\+\tfrac16\Big(|x+y+z|+|x+y-z|+|x-y+z|+|x-y-z|\Big)$$

daw
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mr_e_man
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    Compare https://math.stackexchange.com/q/13253/42969 – Martin R Mar 01 '23 at 18:47
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    Yes, but this question is more specific. – mr_e_man Mar 01 '23 at 18:52
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    There is one answer arguing “why there is no simple such formula” – Martin R Mar 01 '23 at 18:54
  • That argument would seem to imply that real square roots are not sufficient, but this is contradicted by the formula with nested absolute values, since $|x|=\sqrt{x^2}$. – mr_e_man Mar 01 '23 at 19:00
  • About the "functional analysis" tag - There is an infinite-dimensional vector space of functions $\mathbb R^3\to\mathbb R$ spanned by linear forms and their absolute values. The question is whether $\max$ is contained in this space. (Actually, I don't see how to prove that it's infinite-dimensional...) – mr_e_man Mar 01 '23 at 19:19
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    @mr_e_man A quick proof that it's infinite-dimensional is that, any finite-dimensional space of such functions has finitely many points of non-differentiability in $x$ for fixed $y=y_0$ and $z=z_0$ (points at which some function in the subspace is not differentiable), but we can find a linear form which is zero at $(x,y_0,z_0)$ for any real $x$ we wish (and its absolute value is thus non-differentiable at that $x$). – Carl Schildkraut Mar 01 '23 at 21:14
  • @daw - Did you not see my comment about the tag? You could at least explain why you disagree that it's relevant. – mr_e_man Mar 02 '23 at 19:49

2 Answers2

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The answer to Q1 is negative.

Let $x\in {\bf R}$ and $\alpha \in [0,1]$. $$ max(x,\alpha x,0) = max(x,0) = {1\over 2} x + {1 \over 2} |x|. $$ On the other hand, assuming Q1 true, $$ max(x,\alpha x,0) = ({a_0+\alpha b_0})\ x + C(\alpha)\ |x|, $$ where $C(\alpha)$ does not depend on $x$. Since the two functions $x\mapsto x$ and $x \mapsto |x|$ are linearly independent, we obtain $$\forall \alpha \in [0,1], \ a_0+b_0\alpha = 1/2$$ and so $a_0 = 1/2$ and $b_0 = 0$. Repeating the same argument with $max(\alpha x, x, 0)$, we end with $a_0 =0$, $b_0 = 1/2$, a contradiction.

coudy
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  • No, it doesn't work for Q2. $$\max(|x|,|\alpha x|,0)=|x|$$ $$=(a_0+b_0\alpha)x+C(\alpha)|x|$$ $$\forall\alpha\in[0,1],\quad a_0+b_0\alpha=0$$ This gives only $a_0=b_0=0$, with no contradiction. – mr_e_man Mar 01 '23 at 22:26
  • And for Q1, why doesn't this proof work in the two-variable case? – mr_e_man Mar 01 '23 at 22:32
  • @mr_e_man in the two variable case, what would you put in the max? $max(x,\alpha x)$ would give $\frac{1+\alpha}{2}x+\frac{1-\alpha}{2}|x|$ – Zoe Allen Mar 01 '23 at 22:36
  • @ZoeAllen - Right. For some reason I thought the third input $0$ wouldn't change anything. – mr_e_man Mar 01 '23 at 22:47
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For Q2:

Consider the set of points where it isn't differentiable. First, make sure there's no duplication in your sum by making sure (wlog) none of the terms are scalar multiples of each other. This guarantees the set where it isn't differentiable is just the union of the planes $a_i x + b_i y + c_i z = 0$. This would have to be equal to the set where $max(|x|,|y|,|z|)$ is not differentiable, which is equivalent to $$(x,y,z) \in S \iff |x|=|y| \ge |z| \space \vee|x|=|z| \ge |y| \space \vee |y|=|z| \ge |x|$$ But this can't be made as a union of planes. Take a point where $|x| \ne |y| \ne |z|$. Clearly this isn't in $S$. But any plane through the origin will contain a point like this unless it is the plane $x=y$ or $x=z$ or $y=z$. And one can check that these planes aren't contained in $S$ either. E.g. $(0,0,1) \notin S$ so $x=y$ doesn't work.

So the sum would have to be empty. And clearly that doesn't work.

This also proves the answer to Q1, but @coudy already handled that.

Zoe Allen
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