Hello

Can anyone tell me a good reference for holders condition?

- Oct 29th 2009, 09:21 PMaukieHolders Condition Reference?
Hello

Can anyone tell me a good reference for holders condition? - Oct 30th 2009, 05:57 PMJose27
Why did you delete the original question? If you want to ask another question simply make another thread.

- Oct 30th 2009, 11:20 PMaukie
Because nobody replied so i didn't think anyone was interested. More so I didn't want anyone to waste their energy (i have answered questions on this forum only to find the author of the post is no longer interested). For any one interested here is a proof to my original question (the title of the post).

$\displaystyle \sqrt{x}$ is lipschitz of order 1/2 on [0, +inf)

proof:

$\displaystyle |\sqrt{x} - \sqrt{y}| <= \sqrt{x}+ \sqrt{y}$ (by the triangle inequality)

$\displaystyle => |\sqrt{x} - \sqrt{y}|^2 <= |x - y|$ (some simple algebra)

$\displaystyle => |\sqrt{x} - \sqrt{y}| <= |x - y|^{1/2}$ (i.e. Lipschitz condition).

If anyone knows how to prove the more general case i.e. prove $\displaystyle x^r$, $\displaystyle 0<r<1$, satisfies the lipschitz conditon of some order. i would be greatfull if you could share your proof.

ps: lipschitz condition of order $\displaystyle \alpha$ is equivalent to saying holder's condition of order $\displaystyle \alpha$ - Oct 31st 2009, 05:37 AMLaurent
Hi,

I just devised the following simple proof:

Let $\displaystyle 0<r<1$. Let $\displaystyle 0<a\leq b$.

For any $\displaystyle 0\leq x\leq 1$, we have $\displaystyle x\leq x^r$, hence $\displaystyle x^r+(1-x)^r\geq x+(1-x)=1$.

Taking $\displaystyle x=\frac{a}{b}$ and multiplying both sides by $\displaystyle b^r$, we get $\displaystyle a^r+(b-a)^r\geq b^r$, hence $\displaystyle b^r-a^r\leq (b-a)^r$. QED. - Oct 31st 2009, 01:44 PMaukie
Hi Laurent, thanks for your input.

If I have interpreted your proof correctly. It does not prove the problem statement completely, i.e.

$\displaystyle |x^r - y^r| \leq K|x-y|^r, \forall x,y \in [0, \infty)$ and some constant K>0.

but your proof only proves this condition $\displaystyle \forall x,y \in \{ q \in \mathbb{Q}: 0 \leq q \leq 1 \}$, i.e rational numbers in the unit interval.

Here is what I have come up with so far. We want to prove the equivalent statement:

$\displaystyle \frac{|x^r - y^r|}{|x-y|^r} \leq K \label{eq1} $, for some constant $\displaystyle K>0$. (1)

If y = 0, it is obvious the lipschitz constant is K = 1. Thus assume $\displaystyle y>0$. Now if I consider the limit of the l.h.s as $\displaystyle x \rightarrow y$, L'Hopitals rule tells me

$\displaystyle \lim_{x \to y} \frac{x^r - y^r}{(x-y)^r} = 0$ ... (here i am assuming x > y, hopefully without loss of generality)

Now this tells me $\displaystyle \forall \epsilon >0, \exists \delta > 0$ , s.t. if $\displaystyle |x-y|<\delta$ then $\displaystyle \frac{x^r - y^r}{(x-y)^r} \leq \epsilon$. i.e. equation (1) holds in some neighbourhood of y.

But I am not quite sure on how or if I can extend this result to the the entire half line, $\displaystyle [0, \infty)$ ... ??? - Oct 31st 2009, 03:00 PMLaurent
Where did you see this ?? :confused::confused: I took any $\displaystyle 0<a\leq b$ and proved $\displaystyle b^r-a^r\leq (b-a)^r$... I don't know where you see rationals and unit interval. I needed a first inequality where $\displaystyle x\in[0,1]$, and then this is $\displaystyle \frac{a}{b}$ (which obviously is in $\displaystyle [0,1]$) that played the role of $\displaystyle x$.

Note also that $\displaystyle b^r-a^r\leq (b-a)^r$ implies $\displaystyle |b^r-a^r|\leq |b-a|^r$ since both sides are positive. The last inequality is symmetric in $\displaystyle a,b$ so that it holds even if $\displaystyle a>b$. - Oct 31st 2009, 09:37 PMaukie
hello laurent, spare me my blushes, for some reason i was assuming a, b were integers. very elegantly done ... thank you. i cant believe how over complicated i was making the proof in my attempt.