Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be uniformly continuous. I want to show that $$ \lim_{h\rightarrow0}\sup_{x\in\mathbb{R}}|f(x+h)-f(x)|=0.\tag{1} $$
By the uniform continuity, given $\varepsilon >0$, there is $\delta >0$ such that whenever $|h| < \delta$ we get $$ | f ( x + h ) - f ( x ) | < \varepsilon \quad \forall x \in \mathbb{R}. $$ It follows that whenever $|h| < \delta$ $$ \sup_{ x \in \mathbb{R} } | f ( x + h ) - f ( x ) | \leq \varepsilon. $$ Therefore $$ \limsup_{ h \rightarrow 0} \sup_{ x \in \mathbb{R} } | f ( x + h) - f ( x)| \leq \varepsilon, $$ and as $\varepsilon\downarrow 0$, we get $(1)$.
My questions are:
- Are all the steps above correct?
- Is there some well-known result which can be applied here immediately?