The $Hol$ operator is a continuous function?

Question

Let $\Omega$ be a compact space, and consider $C(\Omega)$ the space of the continuous functions over $\Omega$, consider also, $C^\gamma(\Omega)$ the space of all $\gamma$-holder continuous functions, i.e, $f\in C(\Omega)$ s.t $$ Hol(f):=\sup_{x\neq y}\dfrac{f(x)-f(y)}{d(x,y)^{\gamma}}<\infty $$

My problem is the following: Let $(f_n)\subset C^\gamma(\Omega)$ be a sequence of Holder continuous functions s.t $f_n$ converge uniformly to a constant function $f\equiv F.$

It is clearly that $Hol (f)=0$, my question: It is truth that $Hol(f_n)\to 0$?

Answer 1

For $\gamma = .$ NO. Let $\Omega =[0,1].$ For $n\in N,$ for $x\in [1/n,1]$ let $f_n(x)=0$ and for $x\in [0,1/n)$ let $f_n(x)=1/n-x.$ Then Hol$_1(f_n)=1$ but $(f_n)_n$ converges uniformly to $0.$

Answer 2

As it is written in user254665's post, the answer is negative. But under some additional assumptions this convergence takes place.

A simple sufficient condition is that $\Omega$ is compact (or just totally bounded) and $\{f_n,n\ge 1\}$ is bounded in $C^{\gamma'}(\Omega)$ with $\gamma'>\gamma$, i.e. $\sup_{n\ge 1} \mathrm{Hol}_{\gamma'}(f_n)<\infty$. Indeed, in this case the embedding $C^{\gamma'}(\Omega)\subset C^{\gamma}(\Omega)$ is compact, so $\{f_n,n\ge 1\}$ is precompact in $C^{\gamma}(\Omega)$. But since $f_n$ converges in $C(\Omega)$, it must converge in $C^\gamma(\Omega)$ as well.