Generalization of Lipschitz continuity to higher order polynomials?
Generalization of Lipschitz continuity to higher order polynomials?
Lipschitz continuity of a function $xto f(x) $ is defined as: $$exists K,forall x_1,x_2 : fracleq K$$
This can be viewed with pairs of lines bounding the function in the graph
A line is a first order polynomial. Does there exist any higher order polynomials which we can define other kinds of continuity with? Would there be anything to gain from trying to define it? For example we could imagine second order polynomial (parabola) bounding at each point instead of lines. Something like this (?)
$$exists K,forall x_1,x_2: frac(x_1-x_2)^2leq K$$
$begingroup$
On the other hand, if instead of higher degree polynomials you consider $|x-y|^alpha$ with $alpha in (0, 1]$, you have rediscovered Hölder's spaces.
$endgroup$
– Giuseppe Negro
Sep 12 '18 at 7:30
$begingroup$
Ah yes you are right @KaviRamaMurthy it means something else as stated now doesn't it.
$endgroup$
– mathreadler
Sep 12 '18 at 23:40
$begingroup$
@GiuseppeNegro Ah I see you were faster than me. :)
$endgroup$
– mathreadler
Sep 12 '18 at 23:56
2 Answers
2
If $f$ is differentiable you can consider Lipschitz continuity of its derivative. If $f'(x)$ is Lipschitz continuous with constant $L$, then it is true that $$forall x,y inmathbbRquad f(x)+f'(x)(y-x)-fracL2|y-x|^2leqslant f(y)leqslant f(x)+f'(x)(y-x)+fracL2|y-x|^2,$$
so you have global quadratic bounds. Similarly, Lipschitz continuity of higher-order derivatives implies polynomial bounds of higher degree. The same holds true in the multivariate case, which is heavily used in numerical optimization.
$begingroup$
Yes it is something like this I was fishing for. Great that you managed to see it so easily.
$endgroup$
– mathreadler
Sep 20 '18 at 13:43
If :
$$ frac(x_1-x_2)^2leq K$$
Then :
$$fracleq K |x_1-x_2| undersetx_2 to x_1to 0$$
Then $f$ has a derivative, which is zero everywhere, thus $f$ is a constant function.
Note that this remains true if you replace $2$ by any $alpha>1$.
$begingroup$
Yes, so that would imply that it is... stronger or weaker than Lipschitz?
$endgroup$
– mathreadler
Sep 12 '18 at 7:08
$begingroup$
Way stronger, since every constant function is (trivially) Lipschitz.
$endgroup$
– nicomezi
Sep 12 '18 at 7:09
Thanks for contributing an answer to Mathematics Stack Exchange!
But avoid …
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Required, but never shown
Required, but never shown
By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.
$begingroup$
$forall x_1,x_2 exists K$ is not right. You should write $exists K forall x_1,x_2$
$endgroup$
– Kavi Rama Murthy
Sep 12 '18 at 7:24