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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.

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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.2011-05-31

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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 $x$.

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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.2011-06-01