# Is anything known about $w^*(x)=\sup_y w(x+y)/w(y)$ for measurable functions w on $R^n$

In my recent studies (fourier multipliers on weighted Lp spaces) I have to deal with this kind of transformation: if w is a measurable function on $R^n$, define $w^*(x)=\sup_y \frac{w(x+y)}{w(y)}$. Has anyone seen this transformation before? Has this interesting properties? In particular I'm interested in functions such that w=w* or $w^*\leq cw$, like the exponential. Thanks in advance.

UPDATE: I'm interested in $w$ continuous, positive and even

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Thanks for tour suggestion, i've edited the title following your hint – Nicolò Jun 30 '10 at 13:24
$w=w^*$ might be strong enough to imply $w(x) = a^x$ for $n = 1$. At least it implies $w(0) = 1$. Do you have an example of a function with $w = w^*$ other than $a^x$? – Helge Jun 30 '10 at 14:23
You can take, for example, $w(x)= e^{A|x|^\varepsilon}$, with A greater than 0 and $0<\varepsilon\leq 1$ – Nicolò Jun 30 '10 at 15:04
Certainly expressions like this have come up in various papers which deal with "weighted $L^1$-convolution algebras" - objects that have cropped up in Banach-algebra papers as sources of examples. Unfortunately I don't know of any definitive or convenient reference off the top of my head. – Yemon Choi Jun 30 '10 at 18:49

Sort of a random thought: if $w = w^*$, then $w$ must be positive. So take logs on both side. Since log is monotonic, it commutes with sup. Then we have $a = \log w$, and $$a(x) = \sup_y a(x+y) - a(y)$$ which implies that $$a(x) + a(y) \geq a(x+y)$$ so $a$ must be sub-additive. Helge's comment tells us that $a(0) = 0$.

This looks necessary and sufficient. (As for sub-additive functions $a$ such that $a(0) = 0$, $a(x) + a(0) = a(x+0)$ is satisfied, so the sup is attained.)

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Just, a basic property one establish, when one assumes that $$w^*(x) \leq C w(x)$$ is that $w(x)$ grows subexponentially, if furthermore $w(x)$ is continuous.

Define $A = C \cdot \sup_{|x| \leq 1} |w(x)|$. Then $$|w(x)| \leq A^{|x|}.$$

Proof:

Let $n = \lfloor |x| \rfloor$. Then we can write $x = n y + z$, where $y = \frac{x}{|x|}$ so $|y|, |z| \leq 1$. By assumption, we have for $|u| \leq 1$ that $C w(x) \geq w(x+u)/ w(u)$ so that $$w(x+u) \leq C w(x) w(u) \leq A w(x).$$ Applying this $n$ times yields the claim. q.e.d.

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I would guess that a converse also holds: Assume $w$ has subexponential growth and is 'monotone' then this property holds. Idea of proof: 'subexponential growth' => I can restrict to a compact set. On compact set $w*(x)$ and $w(x)$ are continuous, so one has $w*(x) \leq C w(x)$ for some $C$... But I didn't try to make this rigorous. – Helge Jun 30 '10 at 19:34
Oh hey, looks like your answer is more or less in a similar vein to mine. Cool. – Willie Wong Jun 30 '10 at 19:57
Very interesting, thank you. – Nicolò Jul 1 '10 at 12:06

Following Willie Wong's answer, I think I've found a characterization of the positive real functions w such that $w\leq w^\*$ or $w^\*\leq w$.

Proposition: Let $w:R\rightarrow R$ a positive real function, and let $f(x)=\log(w(x))$. Then

1. $w^*\leq w \iff f \text{ subadditive}$
2. $w\leq w^\* \iff w(0) \leq 1$

In particular $w=w^\*\iff f \text{ subadditive and } w(0) \leq 1$

Proof: 1) ($\Rightarrow$) We have that

$f(x)=\log w(x) \geq \log w^\*(x)=\sup _y\left\{\log w(x+y)-\log w(y)\right\}=\sup _y\left\{f(x+y)-f(y)\right\}$ which implies that f is subadditive, like Willie Wong suggested.

($\Leftarrow$) f is subadditive, so

$w^\*(x) = \sup_y \frac{w(x+y)}{w(y)}=\sup_y \frac{e^{f(x+y)}}{e^{f(y)}}\leq \sup_y\frac{e^{f(x)+f(y)}}{e^{f(y)}}=e^{f(x)}=w(x)$.

2) ($\Rightarrow$) $w(0)\leq w^\*(0)=1$

($\Leftarrow$) $w^\*(x)=\sup_y\frac{w(x+y)}{w(y)}\geq\frac{w(x)}{w(0)}\geq w(x)$

Noting that, if C is a costant, that $(Cw)^\*=w^\*$ the following corollary is trivial:

Corollary:

1. $w^*\leq C w \iff f + \log C \text{ subadditive}$
2. $w\leq C w^\* \iff w(0) \leq C$
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This is a long comment rather than an answer.

In this paper of Klartag and Milman, the following operation on functions is defined and called the Asplund sum (as far as I can tell, that name has not been used elsewhere):

$$f \star g (x) = \sup_y f(y)g(x-y).$$

So if $w^-(x) = w(-x)$, then your function is $w^* = w\star (1/w^-)$. Some properties of the $\star$ operation, including its relationship with the Legendre transform, are discussed in the paper. However, the class of functions of interest there is not closed under taking reciprocals, so the usefulness in your setting may be limited.

Trivial update: since Nicolò has clarified that he is interested in even functions $w$, of course the above simplifies to $w^* = w\star (1/w)$.

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thanks for the reference, I will take a look – Nicolò Jul 1 '10 at 12:04