# 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

$\displaystyle f(t_1+t_2) \leqslant f(t_1)+f(t_2), \quad \forall t_1,t_2>0$.