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Suppose $f \in \mathbb{R}[x]$ and define $g \colon \mathbb{R} \to \mathbb{R}$ by $$g(x) = \frac{f(x)^2}{(x^2+1)^{d+1}}, \text{where } d = \deg(f)$$

I'm looking for a quick proof as to why $g$ is bounded above and Lipschitz.

Edit: $g$ is not proper as mentioned below.

James Rohal
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  • @James: What have you tried? For example, what difficulty do you find in showing $g$ is bounded above? – JavaMan May 31 '11 at 01:07
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    @James, what is $\displaystyle\lim_{x\to\pm\infty} g(x)$ ? – lhf May 31 '11 at 01:25
  • Intuitively all of these seem obvious. For example, $g$ should be bounded above by $\max g(c)$ where $c$ is a critical point of $g$. I feel like all of the proofs could be done by contradiction but I was looking for something more explicit (i.e., "this is the upper bound", or "this is the Lipschitz constant"). – James Rohal May 31 '11 at 01:26
  • @James: All zeros of $f$ are critical points of $g$ (but there may be others). Do you expect to be able to find all critical points of $g$ explicity? – lhf May 31 '11 at 01:31
  • Not necessarily, but it turns out (almost always) that $g$ has finitely many critical points on $f \neq 0$. Plus I know each connected component of $f \neq 0$ has at least one local max. – James Rohal May 31 '11 at 01:35
  • @James, you still need to know what happens at infinity. Hence my suggestion above. – lhf May 31 '11 at 01:49
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    Let $\eta:=\max_{x\in{\mathbb R}} g(x)$. If $f$ has a real zero then $g({\mathbb R})=[0,\eta]$, so $g$ is not proper. – Christian Blatter May 31 '11 at 08:05

1 Answers1

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Hints: to show $g(x)$ is

  • Bounded Above: $g$ is everywhere continuous (why?) and $\displaystyle\lim_{x \to \pm \infty} g(x)$ are finite.

  • Lipschitz: It would be enough to show that $g'(x)$ is bounded for every $x$.

JavaMan
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  • By the way, I just realized the the hint for the Lipschitz part only works if $f$ is differentiable. Otherwise, you may just need to compute $\frac{g(x)- g(y)}{x-y}$ by hand. – JavaMan Jun 01 '11 at 01:11