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.


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