# Ngô Quốc Anh

## June 25, 2010

### Some operations on the Hölder continuous functions

Filed under: Giải Tích 1 — Tags: — Ngô Quốc Anh @ 4:37

This entry devotes the following fundamental question: if $u$ is Hölder continuous, then how about $u^\gamma$ for some constant $\gamma$? Throughout this entry, we work on $\Omega \subset \mathbb R^n$ which is not necessarily bounded.

Firstly, we have an elementary result

Proposition. If $f$ and $g$ are $\alpha$-Hölder continuous and bounded, so is $fg$.

Proof. The proof is simple, we just observe that $\displaystyle\left| {f(x)g(x) - f(y)g(y)} \right| \leqslant \left| {f(x) - f(y)} \right||g(x)| + \left| {g(x) - g(y)} \right||f(y)|$

which yields $\displaystyle\frac{{\left| {f(x)g(x) - f(y)g(y)} \right|}}{{{{\left| {x - y} \right|}^\alpha }}} \leqslant \frac{{\left| {f(x) - f(y)} \right|}}{{{{\left| {x - y} \right|}^\alpha }}}\left( {\mathop {\sup }\limits_\Omega |g(x)|} \right) + \frac{{\left| {g(x) - g(y)} \right|}}{{{{\left| {x - y} \right|}^\alpha }}}\left( {\mathop {\sup }\limits_\Omega |f(y)|} \right)$.

Consequently,

for any positive integer number $n$ and any $\alpha$-Hölder continuous and bounded function $u$, function $u^n$ is also $\alpha$-Hölder continuous and bounded.

Let us assume , $u$ is $\alpha$-Hölder continuous and bounded, $\gamma>0$ is a constant. Let $n =\left\lfloor \gamma \right\rfloor$. Since $\gamma \in \mathbb R$, we may assume $u$ is also bounded away from zero, that means there exist two constants $0 such that $m.

We now study the $\alpha$-Hölder continuity of $u^\gamma$. Observe that function $\displaystyle f(t) = {t^{\left\lceil \gamma \right\rceil }}, \quad t>0$

is sub-additive in the sense that $\displaystyle f(t_1+t_2) \leqslant f(t_1)+f(t_2), \quad \forall t_1,t_2>0$.

Indeed, by dividing both sides by $f(t_2)$ we arrive at $\displaystyle {\left( {1 + \frac{{{t_1}}}{{{t_2}}}} \right)^{\left\lceil \gamma \right\rceil }} \leqslant 1 + {\left( {\frac{{{t_1}}}{{{t_2}}}} \right)^{\left\lceil \gamma \right\rceil }}$.

Since ${\left\lceil \gamma \right\rceil } \leqslant 1$

the above inequality comes from the Bernoulli inequality. Therefore if there are some $0 then $\displaystyle {({t_1} + {t_2})^{1 - \left\lceil \gamma \right\rceil }}{\left( {{t_1} + {t_2}} \right)^{\left\lceil \gamma \right\rceil }} = {\left( {{t_1} + {t_2}} \right)^1} \leqslant t_1^1 + t_2^1 = t_1^{\left\lceil \gamma \right\rceil }t_1^{1 - \left\lceil \gamma \right\rceil } + {t_2}$

holds for any $m\leqslant t_1 \ne t_2 \leqslant M$. We now arrive at $\displaystyle {\left( {{t_1} + {t_2}} \right)^{\left\lceil \gamma \right\rceil }} \leqslant {\left( {\frac{{{t_1}}}{{{t_1} + {t_2}}}} \right)^{1 - \left\lceil \gamma \right\rceil }}t_1^{\left\lceil \gamma \right\rceil } + \frac{1}{{{{({t_1} + {t_2})}^{1 - \left\lceil \gamma \right\rceil }}}}{t_2} \leqslant t_1^{\left\lceil \gamma \right\rceil } + \frac{1}{{{{(2m)}^{1 - \left\lceil \gamma \right\rceil }}}}{t_2}$.

We now make use the above estimate to show that $u^\gamma$ is $\alpha$-Hölder continuous. Indeed, we have the following $\displaystyle \left| {u{{(x)}^{\left\lceil \gamma \right\rceil }} - u{{(y)}^{\left\lceil \gamma \right\rceil }}} \right| \leqslant \frac{1}{{{{(2m)}^{1 - \left\lceil \gamma \right\rceil }}}}\left| {u(x) - u(y)} \right|$.

Consequently, we get $\displaystyle \frac{{\left| {u{{(x)}^{\left\lceil \gamma \right\rceil }} - u{{(y)}^{\left\lceil \gamma \right\rceil }}} \right|}}{{{{\left| {x - y} \right|}^\alpha }}} \leqslant \frac{1}{{{{(2m)}^{1 - \left\lceil \gamma \right\rceil }}}}\frac{{\left| {u(x) - u(y)} \right|}}{{{{\left| {x - y} \right|}^\alpha }}} < \infty$.

Thus, from $\displaystyle\frac{{\left| {u{{(x)}^\gamma } - u{{(y)}^\gamma }} \right|}}{{{{\left| {x - y} \right|}^\alpha }}} \leqslant \frac{{\left| {u{{(x)}^{\left\lfloor \gamma \right\rfloor }} - u{{(y)}^{\left\lfloor \gamma \right\rfloor }}} \right|}}{{{{\left| {x - y} \right|}^\alpha }}}\left( {\mathop {\sup }\limits_\Omega {u^{\left\lceil \gamma \right\rceil }}} \right) + \frac{{\left| {u{{(x)}^{\left\lceil \gamma \right\rceil }} - u{{(y)}^{\left\lceil \gamma \right\rceil }}} \right|}}{{{{\left| {x - y} \right|}^\alpha }}}\left( {\mathop {\sup }\limits_\Omega {u^{\left\lfloor \gamma \right\rfloor }}} \right)$

we claim that $u^\gamma$ is $\alpha$-Hölder continuous.

Let us now turn to the case $\gamma<0$. For the sake of simplicity, let us denote $\gamma :=-\gamma$, i.e. the function under investigating is $\frac{1}{u^\gamma}$. Obviously $\displaystyle\left| {\frac{1}{{u(x)}} - \frac{1}{{u(y)}}} \right| = \frac{{\left| {u(x) - u(y)} \right|}}{{\left| {u(x)u(y)} \right|}} \leqslant \frac{1}{{{m^2}}}\left| {u(x) - u(y)} \right|$.

Thus $\frac{1}{u}$ is $\alpha$-Hölder continuous since $\displaystyle\frac{{\left| {\frac{1}{{u(x)}} - \frac{1}{{u(y)}}} \right|}}{{{{\left| {x - y} \right|}^\alpha }}} \leqslant \frac{1}{{{m^2}}}\frac{{\left| {u(x) - u(y)} \right|}}{{{{\left| {x - y} \right|}^\alpha }}} < \infty$

for any $x\ne y$. Consequently,

for any positive integer number $n$ and any $\alpha$-Hölder continuous and bounded function $u$, function $\frac{1}{u^n}$ is $\alpha$-Hölder continuous.

Similarly, we can prove $\displaystyle \left| {\frac{1}{{u{{(x)}^{\left\lceil \gamma \right\rceil }}}} - \frac{1}{{u{{(y)}^{\left\lceil \gamma \right\rceil }}}}} \right| \leqslant \frac{1}{{{{(2M)}^{1 - \left\lceil \gamma \right\rceil }}}}\left| {\frac{1}{{u(x)}} - \frac{1}{{u(y)}}} \right| \leqslant \frac{1}{{{{(2M)}^{1 - \left\lceil \gamma \right\rceil }}{m^{\frac{{\left\lceil \gamma \right\rceil }}{2}}}}}\left| {u(x) - u(y)} \right|$.

Thus $\displaystyle\frac{1}{{{u^{\left\lceil \gamma \right\rceil }}}}$

is $\alpha$-Hölder continuous. It is now easy to prove that $\frac{1}{u^\gamma}$ is also $\alpha$-Hölder continuous. So far we have proved the following

Theorem. If $u$ is $\alpha$-Hölder continuous, positive and bounded (i.e. $u). We assume in addition $u$ is bounded away from zero (i.e. $0). Then $u^\gamma$ is also $\alpha$-Hölder continuous for any constant $\gamma >0$.

The boundedness of $u$ plays an important role in our argument. In practice, if strictly positive function $u$ is locally $\alpha$-Hölder continuous on $\mathbb R^n$, we are the able to apply the theorem. We will discuss something related to this application later.

## 7 Comments »

1. Hi Anh,

I found a Holder continuity estimate which is potentially useful for me. $\displaystyle { \sup_{x,y\in \mathbb{R}^n} \left\{ \frac{\left| f(x)-f(y) \right|}{\left| x-y \right|^{1-n/p} } \right\} \leq C\|\nabla f\|_p }$

However, I have problem finding the book talking about it. Do you happen to know any relevant reference?

Thanks,
Chao

Comment by Chao — August 18, 2010 @ 16:53

• Hello Chao,

Where did you find such an estimate?

Comment by Ngô Quốc Anh — August 18, 2010 @ 17:50

• Hi Anh,

The book is
“Aspects of Sobolev Type Inequalities”, by Laurent Saloff-Coste.

I found the inequality is actually proved in the book. I will try to understand it. I am cs background. So reading math textbook is always a pain.

Will keep you updated.

Chao

Comment by Chao — August 18, 2010 @ 19:31

2. Sorry the formula should be $\displaystyle\sup_{x,y\in \mathbb{R}^n} \left\{ \dfrac{\left| f(x)-f(y) \right|}{\left| x-y \right|^{1-n/p} } \right\} \leq C\|\nabla f\|_p$

Comment by Chao — August 18, 2010 @ 16:53

3. I realize this might be a combination of Holder inequality and Soblev inequality, in the case when p>n, and n the dimension of the Euclidean space.

Comment by Chao — August 18, 2010 @ 17:23

• Hi Chao,

I just go through the proof on page 24 of the book. So what is your problem here? The proof is quite clear. What you need is prove the following $\displaystyle |f(x)-f(y)| \leqslant C|x-y|^{1-n/p}\|\nabla f\|_p$.

Comment by Ngô Quốc Anh — August 18, 2010 @ 19:52

• Hi Anh,

Let me think more before asking questions. I will come back later.

Chao

Comment by Chao — August 18, 2010 @ 20:24

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