Pointwise maximum: translation and scaling

Question

From this question, I am getting that $2\times \max(a,b) = \max(2a,2b)$, and more generally, $c\times \max(a,b) = \max(c\times a, c\times b)$ for $c\geq0$, since we have $$ \begin{align*} a + b + |a-b| &= a+b + \max(a-b, b-a) \\ &= \max\big[(a+b) + a-b),\; (a+b) + b-a\big] \\ &= \max(2a,2b))\\ a + b + |a-b| &= \big[\max(a,b) + \min(a,b)\big] + \big[\max(a,b) - \min(a,b)\big] \\ &= 2\max(a,b) \end{align*} $$

Is this correct? It feels odd to move $(a+b)$ into the max in the first line. This seems to indicate that for $f:\mathbb{R}^n\to \mathbb{R}$ defined by the pointwise max $$f(x) := \max(x_1,x_2,\ldots,x_n)$$ satisfies $f(c\times x + d) = cf(x) + d$ for $c\geq0$ and $d$ a (specific?) linear combination of $x$. For $d$ a constant scalar, we have that $f$ some translation and scaling properties (not sure what this is called), which is intuitive, but is this right for $d = d(x)$ some linear combination of $x$?

What operations are allowed for moving terms in and out of $\max(\cdot)$?

Answer 1

You are making things too complicated.

It is simply that $a\ge b$ if and only if $2a\ge 2b$; so

when $a\ge b$, $$ LHS=2\cdot a=2a=\max(2a,2b) $$

and when $a<b$, $$ LHS=2\cdot b=2b=\max(2a,2b) $$

If you want to apply the formula that $$ \max(x,y)=\frac{x+y-|x-y|}{2} $$ then it is simply $$ 2\max(x,y)=\frac{2x+2y-2|x-y|}{2}=\frac{2x+2y-|2x-2y|}{2}=\max(2x,2y) $$

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Added.

Okay, after fixing the typos, in your approach, you have $$ \begin{align} a+b+|a-b| &=a+b+\max(a-b,b-a)\\ &=\max(a-b+(a+b),b-a+(a+b))\\ &=\max(2a,2b) \end{align} $$

In particular, it is correct that $$ a+b+\max(a-b,b-a) =\max(a-b+(a+b),b-a+(a+b)) $$ because $$ \max(x,y)+L=\max(x+L,y+L) $$ since $x\ge y$ if and only if $x+L\ge y+L$.

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In general, if you have $x=(x_1,x_2,\cdots,x_2)$ and $c\ge 0$, you have $$ c\max(x_1,\cdots,x_n)=\max(cx_1,\cdots,cx_n) $$

So $f(cx)=cf(x)$.

On the other hand, for any $L\in\mathbb{R}$, $$ \max(x_1+L,\cdots,x_n+L)=\max(x_1,\cdots,x_n)+L $$ Combining together, $$ \max(cx_1+L,\cdots,cx_n+L)=c\max(x_1,\cdots,x_n)+L $$