# Is $f(x)= \cos(e^x)$ uniformly continuous?

As in the topic, my quest is to check (and prove) whether the given function $$f(x)= \cos(e^x)$$is uniformly continuous on $\left\{\begin{matrix}x \in (-\infty;0] \\ x \in [0; +\infty) \end{matrix}\right.$ .

My problem is that I have absolutely no idea how to do it. Any hints will be appreciated and do not feel offended if I ask you a question which you consider stupid, but such are the beginnings. Thank you in advance.

• Do you know what "uniformly continuous" means? – Chris Eagle Mar 2 '13 at 23:05
• On $(-\infty,0]$, use the mean value theorem to show is it $1$-Lipschitz. – Julien Mar 2 '13 at 23:05
• that i can "draw it without hettin goff pencil" – fdhd Mar 2 '13 at 23:06
• No, that's (somehow) continuous. – Julien Mar 2 '13 at 23:06
• I know Lipschitz but what is 1-Lipschitz? – fdhd Mar 2 '13 at 23:08

## 4 Answers

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.

• Are you sure the distance of the $f$-values of $x_n$ and $y_n$ is always $2$, and not $1$? Isn't $\cos (2n\pi)=0$ and $\cos(2n\pi+\pi)=-1$ for all $n$? – T. Eskin Mar 2 '13 at 23:21
• @ThomasE. Really? $\cos (2n\pi)=0$? – Julien Mar 2 '13 at 23:24
• @julien. Ah, shoot. It's $1$ :) made a silly mistake when drawing the circle. Then their difference is $2$ as wanted. – T. Eskin Mar 2 '13 at 23:27

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 $[a,\infty)$. $_\square$

Note that $f$ is differentiable on $\mathbb{R}$ and $f'(x)=-e^x\sin{(e^x)}.$

Let $x_n=\ln{\left(\dfrac{\pi}{6}+2\pi n\right)}, \;\; y_n= \ln{\left(\dfrac{\pi}{3}+2\pi n\right)}.$ Then $$|x_n-y_n| = {\ln{\left(\dfrac{\pi}{3}+2\pi n\right)} -\ln{\left(\dfrac{\pi}{6}+2\pi n\right)}}=\ln{\dfrac{\dfrac{\pi}{3}+2\pi n}{{\dfrac{\pi}{6}+2\pi n}}}=\\ = \ln{\left(1+\dfrac{1}{1+12 n }\right)}\sim \dfrac{1}{1+12 n }, \;\; n\to \infty .$$
By the mean value ( Lagrange's) theorem $$|f(x_n)-f(y_n|=|f'(\xi_n)|\cdot|x_n-y_n|=e^{\xi_n}|\sin(e^{\xi_n})|\cdot|x_n-y_n|\geqslant \\ \geqslant e^{x_n}\cdot\dfrac{1}{2}\cdot\dfrac{1}{1+12 n }=\dfrac{\dfrac{\pi}{6}+2\pi n}{2+24n} \underset{n\to\infty}{\rightarrow} \dfrac {\pi}{12},$$ which proves that $f$ is not uniformly continuous on $[0,\;+\infty)$.

• If by Lagrange theorem you mean the mean value theorem, how do you do that? – Julien Mar 2 '13 at 23:21
• Ah, ok. I just thought it was much more trouble than the explicit estimate by Michael Touitou. Thanks for the edit, though. – Julien Mar 3 '13 at 0:52

It is uniformly continuous on the interval $(-\infty,0]$. To see that, use the mean value theorem twice.

Added: Exploiting the identies $|\cos(t) - \cos(z)|\leq | t-z |$ and $|e^x - e^y| \leq | {x}-{y} |$ on $(-\infty,0]$, which they can be proved by the mean value theorem, we have

$$| \cos(e^{x}) - \cos(e^{y})|\leq | e^{x}-e^{y} |\leq| x-y |< \epsilon=\delta.$$

• Why twice? I think once is enough. – Julien Mar 2 '13 at 23:07
• @julien: he needs to bound $|e^x-e^y|$ too. – Mhenni Benghorbal Mar 2 '13 at 23:08
• @MhenniBenghorbal. He only needs bound for the derivative of $f$. – T. Eskin Mar 2 '13 at 23:09
• $|f'(x)|=|e^x\sin (e^x)|\leq 1$ so (mvt, once) $f$ is $1$-Lipschitz, so $f$ is uniformly continuous. – Julien Mar 2 '13 at 23:10
• I don't forget it... I did the chain rule. To find $\|f'\|_\infty\leq 1$ on $(-\infty,0)$ so $|f(x)-f(y)|\leq |x-y|$. Mvt, once... – Julien Mar 2 '13 at 23:14