Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
@James: What have you tried? For example, what difficulty do you find in showing $g$ is bounded above? –  JavaMan May 31 '11 at 1:07
1  
@James, what is $\displaystyle\lim_{x\to\pm\infty} g(x)$ ? –  lhf May 31 '11 at 1:25
    
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 1:26
    
@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 1:31
1  
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 8:05

1 Answer 1

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 $x$.

share|improve this answer
    
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 1 '11 at 1:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.