Is $f(x)= \cos(e^x)$ uniformly continuous?
Solution 1:
There is uniform continuity for $x \in (-\infty,0]$, because the derivative $f'(x)=-e^x \sin(e^x)$ is bounded there ($|f'(x)| \le 1$ there):
Let $\varepsilon>0$ be given, and suppose $|x_1-x_2|<\delta=\varepsilon$. Apply Lagrange's mean value theorem: $|f(x_1)-f(x_2)|=|f'(\xi)| |x_1-x_2| \le 1 \cdot |x_1-x_2|<\varepsilon$. ($\xi$ is some number between $x_1$ and $x_2$).
There is no uniform continuity on $[0,+\infty)$ though. To see this look at the pairs of points $x_n=\ln (2\pi n),y_n=\ln((2n+1) \pi)$, they get closer than any $\delta$, yet the distance between their f-value is always 2.
Solution 2:
The derivative is not needed, you can replace $\cos$ with any nonconstant periodic continuous function $g$:
Let $g\colon \mathbb R\to\mathbb R$ be
- continuous
- non-constant
- periodic.
Then for any $a\in \mathbb R$, the function $f(x):=g(e^x)$ is
- uniformly continuos on $(-\infty,a]$
- not uniformly continuous on $[a,\infty)$
Proof: Every continuous function is continous on compact intervals, therefore every periodic continous function is uniformly continuous
Let $\epsilon>0$ be given. Then there exists $\delta>0$ such that $|g(x)-g(y)|<\epsilon$ if $|x-y|<\delta$. Wlog. $\delta<e^a$. Then $0<1-\delta e^{-a}<1$ ad hence $\delta':=-\ln(1-\delta e^{-a})>0$. Now consider $x,y<a$ with $|x-y|<\delta'$. Wlog. $x\le y$. Then $$e^y-e^x=e^y(1-e^{x-y})<e^a(1-e^{-\delta'})=\delta$$ and hence $$ |f(x)-f(y)|=|g(e^x)-g(e^y)|<\epsilon.$$ Thus $f$ is uniformly continuous on $(-\infty,a]$.
As $g$ is not constant, there exist $x_1, x_2$ with $g(x_1)\ne g(x_2)$. Let $\epsilon=|g(x_1)-g(x_2)|>0$. Let $p>0$ be a period of $g$. No matter how small we choose $\delta>0$, for sufficiently big $k\in\mathbb Z$ (especially with $x_i+kp>e^a>0$), we have $|\ln(x_1+kp)-\ln(x_2+kp)|<\delta$ because $$\ln(x_1+kp)-\ln(x_2+kp)=\ln\frac{x_1/k+p}{x_2/k+p}\to\ln1=0$$ as $k\to+\infty.$ But then $$|f(\ln(x_1+kp))-f(\ln(x_2+kp))|=|g(x_1+kp)-g(x_2+kp)|=|g(x_1)-g(x_2)|=\epsilon $$ so we se ethat $f$ is not uniformly continuous on $[a,\infty)$. $_\square$