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Let $p$ be a polynomial. Show that $p\colon\Bbb R \to \Bbb R$ is Lipschitz iff the degree of $p$ is less than $2$.

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What have you tried? –  Chris Eagle Oct 16 '12 at 17:38
    
I have started with p(X)= asub1+asub2*x,and tried to apply the defination of Lipscetz on it, but stuck with the step what to say about asub2 absolute (u-v)? could I take the constant 'a'sub 2 as lipschetz constant, or Ihave to apply some conditions on it? –  Athar Raheel Ahmad Oct 16 '12 at 17:51
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2 Answers

up vote 1 down vote accepted

We can use the following facts:

  • A $C^1$ function on $\Bbb R$ is Lipschitz continuous if and only if its derivative is bounded.
  • A polynomial is bounded if and only if it's a constant one (hence of degree $0$ or $-\infty$).
  • The derivative of a polynomial of degree $d\geq 1$ is of degree $d-1$.
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I take $C^1$ now (it's shorter), differentiable is enough. –  Davide Giraudo Oct 16 '12 at 19:29
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Now I learned something! I used to think of $C^1$ as continuously differentiable = differentiable. Certainly this is false and we have diff'able $\neq$ cont. diff'able! : ) –  Matt N. Oct 16 '12 at 19:35
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One direction: let $p$ have degree less than $2$. Then $p(x) = ax + b$ for some $a,b \in \mathbb R$. Then $\dot{p}(x) = a$ that is, the first derivative is bounded. Hence $p$ is Lipschitz continuous.

Here we have used the following theorem:


A continuous and everywhere differentiable function $f: \mathbb R \to \mathbb R$ is Lipschitz continuous if and only if its first derivative is bounded, that is, there is $K \in \mathbb R$ such that $|f'(x)| \leq K$ for all $x$ in $\mathbb R$.

Proof:

$\Longrightarrow$: Let $f$ be Lipschitz, that is, there is $K \in \mathbb R$ such that $\frac{|f(x) - f(y)|}{|x-y|} \leq K$ for all $x$ in $\mathbb R$. Then in particular, $\lim_{\delta \to 0} \frac{|f(x) - f(x + \delta)|}{|\delta|} = |f'(x)| \leq K$ for all $x$.

$\Longleftarrow$: Let $|f'(x)| \leq K$ for all $x$. By the mean value theorem we know that for any $x,y \in \mathbb R$ we have that there exists $x_0$ such that $ \frac{f(y) - f(x)}{y - x} = f'(x_0)$. Hence we have that for all $x,y \in \mathbb R$: $$ \frac{|f(y) - f(x)|}{|y - x|} = |f'(x_0)| \leq K$$ that is, $f$ is Lipschitz. $\Box$


Now try the other direction.

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Hope this helps. I'll add some more details later. –  Matt N. Oct 16 '12 at 17:56
    
Thanks Matt for you help... it works now.. –  Athar Raheel Ahmad Oct 16 '12 at 17:57
    
No problem! : ) –  Matt N. Oct 16 '12 at 17:58
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