Square root x satisfies holders condition of alpha >= 1/2 ... How?

**aukie** · October 29th 2009, 09:21 PM

Hello
Can anyone tell me a good reference for holders condition?

**Jose27** · October 30th 2009, 05:57 PM

Why did you delete the original question? If you want to ask another question simply make another thread.

**aukie** · October 30th 2009, 11:20 PM

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).

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

proof:

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

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

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

ps: lipschitz condition of order $\alpha$ is equivalent to saying holder's condition of order $\alpha$

**Laurent** · October 31st 2009, 05:37 AM

Originally Posted by aukie

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

Hi,
I just devised the following simple proof:

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

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

Taking $x=\frac{a}{b}$ and multiplying both sides by $b^r$ , we get $a^r+(b-a)^r\geq b^r$ , hence $b^r-a^r\leq (b-a)^r$ . QED.

**aukie** · October 31st 2009, 01:44 PM

Hi Laurent, thanks for your input.

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

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

but your proof only proves this condition $\forall x,y \in \{ q \in \mathbb{Q}: 0 \leq q \leq 1 \}$ , 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:

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

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

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

Now this tells me $\forall \epsilon >0, \exists \delta > 0$ , s.t. if $|x-y|<\delta$ then $\frac{x^r - y^r}{(x-y)^r} \leq \epsilon$ . 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, $[0, \infty)$ ... ???

**Laurent** · October 31st 2009, 03:00 PM

Originally Posted by aukie

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

Where did you see this ??

I took any $0<a\leq b$ and proved $b^r-a^r\leq (b-a)^r$ ... I don't know where you see rationals and unit interval. I needed a first inequality where $x\in[0,1]$ , and then this is $\frac{a}{b}$ (which obviously is in $[0,1]$ ) that played the role of $x$ .

Note also that $b^r-a^r\leq (b-a)^r$ implies $|b^r-a^r|\leq |b-a|^r$ since both sides are positive. The last inequality is symmetric in $a,b$ so that it holds even if $a>b$ .

**aukie** · October 31st 2009, 09:37 PM

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.

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