Proof verification: Does $\alpha$-Hölder condition on an open interval imply its uniform continuity and $f$ is constant when $\alpha>1$?

Question

Let $f:(a,b)\to\mathbb{R}$ and $f$ satisfies the $\alpha$-Hölder condition

If $\alpha>0$, $\forall u,x\in(a,b)$ there exists some constant $H$ such that $$|f(u)-f(x)|\le H|u-x|^{\alpha}$$

1. Does this imply its uniform continuity on $(a,b)$?
2. What would happen if $\alpha>1$?

My Attempt:

1. $f(x)$ is continuous since for each $\epsilon>0$ there exists $\delta=(\frac{\epsilon}{H})^{1/\alpha}$ s.t. $|u-x|<\delta\implies|f(u)-f(x)|<\epsilon.$

   Let $(u_n)$, a sequence in $(a,b)$ s.t. $(u_n)\to a$ and $f(u_n)$ converges so that we can define $\widetilde{f}=\lim f(u_n)$ (the expansion). Since the $\alpha$-Hölder condition restricted the values of $f$ so that $f(u_n)\le H|u-x|^{\alpha}<\infty$ for all $n$. Therefore it can be expanded continuously to $[a,b]\implies$ it's uniformly continuous.

2. When $\alpha>1$, Let $y=H|u-x|^{\alpha}$. So $y'=0$ if $u-x=0$ and this holds for all $x\in(a,b)$ which implies that $f'(x)=0,\forall x\in(a,b)$. Hence $f$ is constant.

Need Help:

I wonder if my logic in the proof is flawed, if so can someone point it out. And my proof for the second part of this question seems to lack in details，I wonder if my argument makes sense.

Any help is appreciated, thanks in advanced.

Answer 1

1. You have some incomplete idea of uniform convergence. The epsilon-delta definition is directly satisfied by Hölder continuity, using the same epsilon-delta relation you used. For the point-wise argument you would have to argue that the function values $f(x_n)$ form a Cauchy sequence to conclude convergence to $\tilde f$.

2. One way is to use for any two fixed $x,y\in(a,b)$ subdivisions $x_k=x+kd$, $x_n=y$, $d=\frac{y-x}n$, and the triangle identity to find $$ |f(y)-f(x)|\le \sum_{k=1}^n|f(x_k)-f(x_{k-1})| \le n\cdot H\,|d|^\alpha=H|y-x|\,|d|^{\alpha-1} $$ which goes to zero for $n\to\infty$, $d=\frac{y-x}n\to 0$.