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).


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<m<M<\infty such that

m<u(x)<M \quad \forall x \in \Omega.

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 }}.


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

the above inequality comes from the Bernoulli inequality. Therefore if there are some 0<m<M<\infty 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|.


\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<M). We assume in addition u is bounded away from zero (i.e. 0<m<u). 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.


  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?


    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.


        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.


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

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at

%d bloggers like this: