5

Question: If $f:\mathbf R\to\mathbf R_+$ is a Holder continuous function with exponent $0<\alpha<1$, how about the function $f^p$ with $p>1$? Does it still possesses the Holder continuity?
- If yes, then what's the Holder exponent of it?
- If no, what does the counterexample look like?
- Can we add some assumptions such that it is still Holder continuous?

If $|f(x)^p-f(y)^p|$ can be dominated by $|f(x)-f(y)|^p$, then the answer is obviously affirmative. But then the question amount to compare $|a^p-b^p|$ with $|a-b|^p$ for $a,b\in\mathbf R_+, p>1$. Still I got no idea... And I'm not able to give a counterexample to myself...

Any hints or comments will be appreciated. TIA...

Dreamer
  • 2,102
  • I know if $p$ is integer then $$a^p - b^p = (a-b)(.................)$$ – Red shoes Mar 01 '18 at 05:43
  • @Redshoes Is there some similar equality for non-integer $p$? – Dreamer Mar 01 '18 at 05:48
  • similar identity would be very luxury for noninteger exponent ! But I'm optimistic that I know if $p$ is integer then $a^p - b^p$ can be controlled by $(a-b)$ – Red shoes Mar 01 '18 at 06:09
  • That's absolutely not true. Take $a = b+1$ and send $b\to \infty$. Then $a-b = 1$ but $a^p - b^p \to \infty$. – User8128 Mar 01 '18 at 06:10
  • @User8128 control means $$a^p - b^p = O (a-b)$$

    I didn't claim that solves the problem.

     – Red shoes Mar 01 '18 at 06:20
  • Yes I know. If that were true, we would have $\lvert a^p - b^p\rvert \le C \lvert a -b \rvert$ for some fixed constant $C > 0.$ Now take $a = b+1$ and send $b \to \infty$ to see that no such $C$ exists. – User8128 Mar 01 '18 at 06:22
  • $a$ must be close enough to the fix and chosen $b$

    $$\lim_{a \to b} \frac{a^p - b^b}{a -b} = p b^{p-1}$$

    This is of course a local property and probably using this we can prove local holderian continuity and then global holder continuity on any given large compact interval . (remember that local lipschitizan implies global lipchizian on any compact subset) @User8128

     – Red shoes Mar 01 '18 at 06:34
  • Of course this would work on a compact set but it will not work on $\mathbb R$ because when $b^{p-1}$ becomes arbitrarily large, even taking $a$ arbitrarily close to $b$, we will never be able to have $a^p -b^p$ bounded by $a-b$ for precisely the reason you've pointed out. If the problem was on a compact set, it would be quite a bit easier since, for example, $C^{0,\beta} (X) \subset C^{0,\alpha}(X)$ when $X\subset \mathbb R$ is compact and $\beta \ge \alpha$. – User8128 Mar 01 '18 at 06:53
  • 1
    @Q.Huang For $\alpha \in (0,1)$, the function $f(x) = \lvert x\rvert^\alpha$ for $x \in \mathbb R$ is $\alpha$-Holder continuous but is not $\beta$-Holder continuous for any $\beta \in (0,1)$, $\beta \neq \alpha$. This shows that the best possible result you could hope for is that $f^p \in C^{0,\alpha p}(\mathbb R)$ when $1 < p < \alpha^{-1}$ – User8128 Mar 01 '18 at 07:07
  • @User8128 Thanks for all your comments. Take a look at my answer below... – Dreamer Mar 02 '18 at 04:53
  • @Redshoes Thanks for all your comments. Take a look at my answer below... – Dreamer Mar 02 '18 at 04:54

1 Answers1

12

I post an answer of myself...

We need some lemmas at first.

Lemma 1. Let $k\in\mathbf N_+$, $a,b\in\mathbf R$, then $$a^k-b^k=(a-b)\left[\sum_{i=1}^k\left(\begin{array}{c}i\\k\end{array}\right) b^{k-i}(a-b)^{i-1}\right].$$

Lemma 2. Let $0<p\le 1$, $a,b\in\mathbf R$, then $$\big||a|^p-|b|^p\big|\le|a-b|^p.$$

Lemma 3. Let $p\ge q>0$, $a,b\in\mathbf R_+$, $c_{p,q}:=\left(\frac{q}{p}\right)^{\frac{1}{p-q}}$.
(1). If $a\wedge b\ge c_{p,q}$, then $$|a^p-b^p|\ge|a^q-b^q|;$$ (2). If $a\vee b\le c_{p,q}$, then $$|a^p-b^p|\le|a^q-b^q|.$$

Proof. Lemma 1 is an immediate consequence of the binomial theorem. See here for Lemma 2. To prove Lemma 3, we set $a\ge b$ with out loss of generality. Define $F:\mathbf R_+\to\mathbf R$ by $F(x)=x^p-x^q$. Then it's easy to see that $F$ is decreasing on $[0,c_{p,q}]$ and increasing on $[c_{p,q},\infty)$. If $a\ge b\ge c_{p,q}$, then $F(a)\ge F(b)$, i.e., $a^p-b^p\ge a^q-b^q$. If $c_{p,q}\ge a\ge b$, then $F(a)\le F(b)$, i.e., $a^p-b^p\le a^q-b^q$.

Now we get the position to investigate the Holder continuity of the power of Holder continuous functions. Let $\Omega\subset\mathbf R^d$ be a domain.

Theorem 1. Let $0<\alpha<1$, $f\in C^\alpha(\Omega)$, i.e., $f$ is a locally Holder continuous function on $\Omega$ with exponent $\alpha$.
(1). If $0<p\le 1$, then $|f|^p\in C^{\alpha p}(\Omega)$;
(2). If $p>1$, then $|f|^p\in C^\alpha(\Omega)$.

Proof. Choose $D\subset\subset\Omega$ and $x\ne y\in D$. Then there exists $M>0$ such that $|f|\le M$ on $D$.
(1). By Lemma 2, $$\big||f(x)|^p-|f(y)|^p\big|\le|f(x)-f(y)|^p\le [f]_{\alpha;D}^p|x-y|^{\alpha p}.$$ (2). Define $c_{r,s}:=\left(\frac{s}{r}\right)^{\frac{1}{r-s}}$. Then $\frac{c_{p,\lfloor p\rfloor}}{M}|f|\le c_{p,\lfloor p\rfloor}$ on $D$. By Lemma 1 and 3.(2), \begin{align} \left(\frac{c_{p,\lfloor p\rfloor}}{M}\right)^p \big||f(x)|^p-|f(y)|^p\big| &\le \left(\frac{c_{p,\lfloor p\rfloor}}{M}\right)^{\lfloor p\rfloor} \big||f(x)|^{\lfloor p\rfloor}-|f(y)|^{\lfloor p\rfloor}\big| \\ &\le C(M,p,\|f\|_{0;D})||f(x)|-|f(y)|| \\ &\le C(M,p,\|f\|_{0;D})|f(x)-f(y)| \\ &\le C(M,p,\|f\|_{0,\alpha;D}) |x-y|^\alpha. \end{align}

Theorem 2. Let $0<\alpha<1$, $f\in C^\alpha(\bar\Omega)$, i.e., $f$ is a uniformly Holder continuous function on $\Omega$ with exponent $\alpha$.
(1). If $0<p\le 1$, then $|f|^p\in C^{\alpha p}(\bar\Omega)$;
(2). If $p>1$, then $|f|^p\in C^\alpha(\Omega)$;
(3). If $p>1$, and there exists $M>0$ such that $|f|\le M$ on $\Omega$, then $|f|^p\in C^\alpha(\bar\Omega)$;
(4). If $p>1$, and there exists $\delta>0$ such that $|f|\ge\delta$ on $\Omega$, then $|f|^p\in C^\alpha(\bar\Omega)$.

Proof. (1) is same as Theorem 1.(1). Note that $C^\alpha(\bar\Omega)\subset C^\alpha(\Omega)$, (2) is a corollary of Theorem 1.(2). The proof of (3) is just a revise of the proof of Theorem 1.(2) by replacing $D$ with $\Omega$. Note that $\frac{c_{\lceil p\rceil,p}}{\delta}|f|\ge c_{\lceil p\rceil,p}$, the proof of (4) is similar to Theorem 1.(2) by applying Lemma 3.(1).

Dreamer
  • 2,102