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Math Help - [SOLVED] Lipschitz

  1. #1
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    [SOLVED] Lipschitz

    Problem:
    Suppose that p:R \rightarrow R is a polynomial. Show that p:R \rightarrow Ris Lipschitz if and only if the degree of the polynomial is less than 2.

    Attempt:
    By definition of Lipschitz, then we need to prove that the function p is Lipschitz for degree less than 2 or equivalently \leq 1 then
    | p(x_{1}) - p(x_{2}) | \leq K |x_{1} - x{2}| for  K \geq 0 and for all  x, x_{1} \in R

    Here I'm not sure how to approach this. Since p is a polynomial, then it must be continuous. I was thinking of showing that if p is a polynomial of less than 2, then its derivative is a constant. Thus, its derivative is bounded and therefore a Lipschitz function.

    Thank you for your time.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    Problem:
    Suppose that p:R \rightarrow R is a polynomial. Show that p:R \rightarrow Ris Lipschitz if and only if the degree of the polynomial is less than 2.
    obviously every polynomial of degree at most 1 is Lipschitz. so we only need to show that a polynomial of degree

    at least 2 is not Lipschitz. so suppose p(x) is a polynomial of degree at least 2. see that \lim_{x\to\infty}p'(x) = \pm \infty. \ \ \ (1)

    now suppose p(x) is Lipschitz. then for some constant K \geq 0: \ \left|\frac{p(t)-p(x)}{t-x}\right| \leq K, for all t \neq x. taking limit of

    both sides as t \rightarrow x gives us |p'(x)| \leq K, which will contradict (1) if we let x \rightarrow \infty. \ \ \ \square
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  3. #3
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    Thank you NonCommAlg. I was planning to prove by contradiction, but didn't know how to show it mathematically. By dividing both sides by |t-x| and taking the limit, we get the definition of a derivative.
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