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 $A_k\lt \infty$?
A real function question
-
0You 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... – 2010-11-13
3 Answers
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 $(\frac{|y+t|}{1+|y|})^k |g(t)|\le (1+|t|)^k|g(t)|\le B_kC_k.$
-
0The 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. – 2010-11-15
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.
-
0@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 i$f$ my statements are a bit hard, but this is an impression I am getting here. – 2010-11-13
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.
-
0The hint given in the book is to consider cases $|x|\le 2|y|$ and $|x|\ge 2|y|$. But I don't see how. – 2010-11-14