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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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    @James: What have you tried? For example, what difficulty do you find in showing $g$ is bounded above?2011-05-31
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    @James, what is $\displaystyle\lim_{x\to\pm\infty} g(x)$ ?2011-05-31
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    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").2011-05-31
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    @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?2011-05-31
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    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.2011-05-31
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    @James, you still need to know what happens at infinity. Hence my suggestion above.2011-05-31
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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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