Is the function $f\colon \mathbb{R} \to \mathbb{R}$ defined by $f(x)=x+\sin(x)$ uniformly continuous on $\mathbb{R}$

Question

I know how to show not uniformly or how to show it is uniformly continuous but not how to differentiate when to know which one to use

Answer 1

$|f'(x)|=|1+cosx|\leq 2$, and thus $f$ is Lipschitz continuous and thus uniformly continuous.

Answer 2

$|f(x)-f(y)|\\=|x+\sin x-y-\sin y|\\\le|x-y|+|\sin x-\sin y|\\=|x-y|+2\left|\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}\right|\\\le\left|x-y\right|+2\left|\sin\dfrac{x-y}{2}\right|\\\le|x-y|+2\left|\dfrac{x-y}{2}\right|\\=2|x-y|$

Thus $f$ is Lipschitz continuous on $\mathbb R$ and hence uniformly continuous on $\mathbb R.$

Answer 3

A sum of uniformly continuous functions is uniformly continuous.

Let $f$ and $g$ be uniformly continuous on the same domain. For $\varepsilon>0$ there exist $\delta_1>0$ and $\delta_2>0$ such that

• if $|x-y|<\delta_1$, then $|f(x)-f(y)|<\varepsilon/2$
• if $|x-y|<\delta_2$, then $|g(x)-g(y)|<\varepsilon/2$

Take $\delta=\min\{\delta_1,\delta_2\}$; then, for $|x-y|<\delta$ we have

$$ |f(x)+g(x)-f(y)-g(y)|\le|f(x)-f(y)|+|g(x)-g(y)|< \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon $$

I suppose you know that $f(x)=x$ and $g(x)=\sin x$ are uniformly continuous. For $f$ it is obvious; for $g$ it suffices to remember that it is continuous and periodic, so by the Heine-Borel theorem it is uniformly continuous.