Define $f:\left(0,+\infty \right)\rightarrow \mathbb{R}$ as $$f(x)=x\cos \frac{1}{x}$$
Is $f$ Hölder continuous?
I found it uniformly continuous because it is continuous in the interval $(0,1] $ and lipschitz continuous in the interval $ \left[ 1,+\infty \right)$
In fact, we can define $f(0)=0$, so $f$ is continuous in the interval $[0,1]$ .Then it makes sense.
I wonder whether this function which shows infinite concussion near $0$ is Hölder continuous. But I found it hard to prove or disprove.
Any ideas fot that? Thanks a lot!
Added: I guess this function is Hölder continuous, specificly I guess: $$f \in C^{0.5}\left[0,+\infty\right)$$ That's because when finding a counter-example that proves $f$ not Hölder continuous, given $a_n=\frac{1}{2n\pi}$ and $b_n=\frac{1}{(2n+1)\pi}$ I tried to get $\lim_{n\to\infty}\frac{a_n\cos \frac{1}{a_n}-b_n\cos\frac{1}{b_n}}{(a_n-b_n)^\alpha}=+\infty$ but it's not true when $\alpha=0.5$.