Let $(\Omega,d)$ be a non complete metric space, and
consider the space $C^{\gamma}_b(\Omega, \mathbb{R})$, $0<\gamma<1$
of all bounded Hölder continuous functions.
Is $C^{\gamma}_b(\Omega, \mathbb{R})$ endowed with the norm
$$
\|f\|_{\gamma}=\|f\|_0+\sup_{x\neq y} \dfrac{|f(x)-f(y)|}{d(x,y)^{\gamma}}
$$
a Banach Space?
Could someone provide me some reference?
Best Answer
Yes. Let $$\|f\|_{C^\gamma_b} = \sup\limits_{x \in \Omega}|f(x)| + \sup\limits_{x,y \in \Omega, x \neq y} {|f(x) - f(y)| \over d(x,y)^\gamma}.$$ You can easily check that this is a norm. To see that it is complete, suppose that $f_n \in C^\gamma_b$ is a Cauchy sequence. Then $f_n$ is Cauchy with respect to the $\sup$ norm, so there exists $f \in C_b$ such that $f_n$ converges uniformly to $f$. We want to show that in fact $f \in C_b^\gamma$, and that $f_n \to f$ in $C_b^\gamma$ norm. Let $\epsilon > 0$, and choose $N>0$ such that, if $n,m \geq N$, $\|f_n - f_m\|_{C_b^\gamma} \leq \epsilon/3$. Let $n \geq N$, and let $x,y \in \Omega$, $x \neq y$. Choose $L > n$ such that $|f_L(x) - f(x)| \leq {\epsilon \over 3 d(x,y)^\gamma}$ and $|f_L(y) - f(y)| \leq {\epsilon \over 3 d(x,y)^\gamma}$. We then have $${|(f_n(x) - f(x)) - (f_n(y) - f(y))| \over d(x,y)^\gamma} \leq {|(f_n(x) - f_L(x)) - (f_n(y) - f_L(y))| \over d(x,y)^\gamma}$$ $$+ {|f_L(x) - f(x)| \over d(x,y)^\gamma} + {|f_L(y) - f(y)| \over d(x,y)^\gamma} \leq \epsilon.$$ Since this holds for every $x$ and $y$, we have $$\sup\limits_{x,y \in \Omega, x \neq y} {|(f_n - f)(x) - (f_n - f)(y)| \over d(x,y)^\gamma} \leq \epsilon,$$ and therefore $f_n \to f$ in $C_b^\gamma$. The above also clearly implies that $f \in C_b^\gamma$, since $$\sup\limits_{x,y \in \Omega, x \neq y} {|f(x) - f(y)| \over d(x,y)^\gamma} \leq \|f - f_n\|_{C_b^\gamma} + \sup\limits_{x,y \in \Omega, x \neq y} {|f_n(x) - f_n(y)| \over d(x,y)^\gamma} < \infty.$$