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Let $g(t)$ be a real function of a real variable such that $$\sup |p(t)g(t)|\lt \infty$$ for any polynomial $p$. Is it true that for any nonngetive integer $k$, one has $$\sup_{(x,y)\in\mathbb{R}^2}(\frac{|x|}{1+|y|})^k |g(x-y)|\le A_k, $$ for some positive constant $A_k\lt \infty$?

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Do you mean to ask whether there is a uniform bound (depending only on $k$) that works for all $x$ and $y$? –  Jonas Meyer Nov 13 '10 at 21:03
    
You need to be more careful with your expressions. For any continuous function, say, your expression is finite for all $x,y$. So you probably want a $\sup$ somewhere in there... –  Willie Wong Nov 13 '10 at 21:08
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3 Answers

up vote 3 down vote accepted

By change of variable $x=y+t$, we need to bound the expression $$(\frac{|y+t|}{1+|y|})^k |g(t)|.$$ Since $$\frac{|y+t|}{1+|y|}\le \frac{|y|+|t|}{1+|y|}\le 1+|t|,$$ and $$\frac{(1+|t|)^k}{1+t^{2k}}$$ and $$(1+t^{2k})|g(t)|$$ are bounded, say by $B_k$ and $C_k$, we find that $$(\frac{|y+t|}{1+|y|})^k |g(t)|\le (1+|t|)^k|g(t)|\le B_kC_k.$$

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Nice. I am sorry I misjudged you. –  Marek Nov 14 '10 at 1:25
    
Thanks. I understand. –  TCL Nov 14 '10 at 1:58
    
The hint given in the book where this problem is taken from says consider cases $|x|\le 2|y|$ and $|x|\ge 2|y|$. I don't see how this would work. –  TCL Nov 15 '10 at 14:25
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Hint 1: What kind of function must $g(t)$, satisfying the first inequality, be? Can you think of some simple such function yourself?

Hint 2: What can you say about the function $y \mapsto \left({1 \over 1 + |y|}\right)^k$? Can it spoil the second inequality?

Hint 3: Let $x = y + t$. Triangle inequality is your friend.

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A sup was missing from my original post. So your hints appear to be not good. –  TCL Nov 13 '10 at 21:27
    
@TCL: actually, that's precisely what I had in mind (otherwise the inequality would be trivially true) and I am certain that these hints should help you solve the problem as it is stated right now. Anyway, why do you think the hints are not good? Did you really think about them? –  Marek Nov 13 '10 at 21:40
    
A constant $A_k$ was missing. So your hints are not good. Sorry for the confusion. –  TCL Nov 13 '10 at 21:57
    
@TCL: ok, this is getting tiresome. For one thing, I am not sure you are really trying to solve the problem (because the hints still work. The reason being that they are just general hints and not a complete solution) and also I am not sure you are even trying to state the problem correctly (is this really the last edit?). Why should we care to solve something you don't even bother to ask well? I apologize if my statements are a bit hard, but this is an impression I am getting here. –  Marek Nov 13 '10 at 22:08
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Trying to be systematic here... your condition is equivalent to saying $|g(x)| \leq C_n{1 \over 1 + |x|^n}$ for all $n$. So if your inequality is going to be true, we'd expect for some $l$ depending on $k$ we would have the following for all $(x,y)$: $${|x|^k \over 1 + |y|^k}{1 \over 1 + |x - y|^l} \leq C_k$$ And if we can show this then your inequality follows. A natural place to start would be choosing $l = k$, in which case what is needed is $$|x|^k \leq C_k (1 + |y|^k)(1 + |x - y|^k)$$ Notice though that since $x = y + (x-y)$, $|x| \leq |y| + |x-y|$ and thus $|x|^k \leq A_k(|y|^k + |x - y|^k)$ by an easy argument. These two terms are part of the right hand side of the above, so that's all you need.

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The hint given in the book is to consider cases $|x|\le 2|y|$ and $|x|\ge 2|y|$. But I don't see how. –  TCL Nov 14 '10 at 2:30
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