Set $\mathbb T=\mathbb R/2\pi\mathbb Z$. And let $\text{Lip}_\alpha(\mathbb T)$, $0<\alpha<1$ denote the subspacce of $C(\mathbb T)$ consisting of the functions $f$ for which $$ \sup_{t\in\mathbb T,\ h\ne 0}\frac{|f(t+h)-f(t)|}{|h|^\alpha}<\infty $$ with the norm $$ \|f\|_{\text{Lip}_\alpha}=\sup_t|f(t)|+\sup_{t\in\mathbb T,\ h\ne 0}\frac{|f(t+h)-f(t)|}{|h|^\alpha}. $$ Show that $$ \text{lip}_\alpha(\mathbb T)=\left\{ f:\lim_{h\to 0}\sup_t \frac{|f(t+h)-f(t)|}{|h|^\alpha}=0\right\} $$ is the subspace of $\text{Lip}_\alpha(\mathbb T)$ such that $\tau\mapsto f_\tau=f(t-\tau)$ is a continuous $\|\ \|_{\text{Lip}_\alpha(\mathbb T)}$-valued function.
It is sufficient to show that $\tau\mapsto f_\tau$ is continuous at $\tau=0$. I got stuck on this problem for hours, yet I can only show one direction is correct. I have no idea on showing that $f\in\text{lip}_\alpha(\mathbb T)$ is necessary for $\tau\mapsto f_\tau$ being continuous.
My attempt:
Suppose the converse, i.e., there exists $f\in\text{Lip}_\alpha(\mathbb T)\setminus\text{lip}_\alpha(\mathbb T)$ such that $\tau\mapsto f_\tau$ is continuous in $\text{Lip}_\alpha(\mathbb T)$-norm. Since $f\notin\text{lip}_\alpha(\mathbb T)$, we can find a positive integer $n$, and sequences $\{h_k\}_{k=1}^\infty\subset\mathbb R$, $\{t_k\}_{k=1}^\infty\subset\mathbb T$ such that $\lim_{k\to\infty}h_k=0$ and $$ \frac{|f(t_k+h_k)-f(t_k)|}{|h_k|^\alpha}\ge\frac{1}{n},\quad\forall k\in\mathbb Z_{> 0}. $$ Since $\tau\mapsto f_\tau$ is continuous, we know that for arbitrary but fixed $\varepsilon>0$, there exists $\delta>0$ such that whenever $|\tau|<\delta$, then $$ \|f-f_\tau\|_{\text{Lip}_\alpha(\mathbb T)}=\sup_t|f(t)-f_\tau(t)|+\sup_{t\in\mathbb T,\ h\ne 0}\frac{|f(t+h)-f_\tau(t+h)-(f(t)-f_\tau(t))|}{|h|^\alpha}<\varepsilon. $$ The above inequality implies: $$ \sup_{t\in\mathbb T,\ h\ne 0}\frac{|f(t+h)-f_\tau(t+h)-(f(t)-f_\tau(t))|}{|h|^\alpha}<\varepsilon $$ which implies $$ \sup_{t\in\mathbb T,\ h\ne 0}\left|\frac{|f(t+h)-f(t)|}{|h|^\alpha}-\frac{|f_\tau(t+h)-f_\tau(t)|}{|h|^\alpha}\right|<\varepsilon. $$ Notice that if $\varepsilon$ is small enough, then $$ \frac{|f_\tau(t_k+h_k)-f_\tau(t_k)|}{|h|^\alpha} $$ should be arbitraly close to $$\frac{|f(t_k+h_k)-f(t_k)|}{|h_k|^\alpha} $$ and in particular, $$ \frac{|f_\tau(t_k+h_k)-f_\tau(t_k)|}{|h|^\alpha}\ge\frac{1}{2n} $$ whenever $\tau<\delta$