# Thread: Another Uniform Continuity Problem

1. ## Another Uniform Continuity Problem

Prove that $f(x)=cos^3x$ is either uniform continuous or not.

I claim that this function is not, since its derivative is not bounded. But how do I show that using the definition?

Proof so far:

Let $\epsilon = 1$, for every $\delta > 0$, I need to find $x,y \in \mathbb {R} , |x-y| < \delta$ implies $|f(x)-f(y)| = |cos^3x - cos^3y| \geq 1$

What kind of x and y should I pick?

Thanks.

2. You made a mistake. The derivative is bounded.

3. Theorem: a continuous periodic function is uniformly continuous.

Indeed, it is uniformly continuous on two periods $[0,2T]$ by Heine's Theorem (a continuous function defined on a compact is uniformly continuous), and by periodicity this implies the uniform continuity on $\mathbb{R}$ (choose $\delta so that, if $|x-y|<\delta$, then $x$ and $y$ lie in the same interval of length $T$ and the situation matches a similar situation inside $[0,2T]$).

And if you don't know this theorem by Heine, you can write $|\cos^3 x -\cos^3 y|=|\cos x - \cos y||\cos^2 x + \cos x \cos y + \cos^2 x|\leq 3|\cos x-\cos y|$ $=6|\sin\frac{x+y}{2}\sin\frac{x-y}{2}|\leq 6|\sin\frac{x-y}{2}|\leq 3|x-y|$.

4. I understand all but this last part, how is $6|sin \frac {x-y}{2}| \leq 3|x-y|$?

I understand all but this last part, how is $6|sin \frac {x-y}{2}| \leq 3|x-y|$?
Because $|\sin \alpha| \leq |\alpha|$.

6. Another question, how does $3| \cos x- \cos y| \leq 6| \sin \frac {x+y}{2} \sin \frac {x-y}{2}|$? Thanks.

Another question, how does $3| \cos x- \cos y| \leq 6| \sin \frac {x+y}{2} \sin \frac {x-y}{2}|$? Thanks.
It is an identity,
$\cos A - \cos B = 2 \sin \frac{A + B}{2} \sin \frac{A-B}{2}$

8. Originally Posted by ThePerfectHacker
It is an identity,
$\cos A - \cos B = 2 \sin \frac{A + B}{2} \sin \frac{A-B}{2}$
Actually, it is $\cos A - \cos B = - 2 \sin \frac{A + B}{2} \sin \frac{A-B}{2}$ (which in absolute value is the same).