# Prove that the function is uniformly continuous in $\mathbb{R}$.

$f(x)=(\sin x+\cos x)^{100}$ in $\mathbb{R}$.

(My attempt)

I first thought that it is not uniformly continuous.

Using $\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi }{4}\right)$,

then $f(x)=2^{50}\left(\sin\left(\frac{\pi}{4}\right)\right)^{100}$

Let $\epsilon > 0$ be given.

For every $\delta > 0$, let $x=2n\pi -\frac{\pi }{4}+\frac{\delta }{2}$

and $y=2n\pi -\frac{\pi }{4}-\frac{\delta }{2}$ for any $n\in \mathbb{N}$.

That is $\left \lvert x-y \right\rvert< \frac{\delta }{2}< \delta$. Then $\left \lvert f(x)-f(y) \right \rvert=2^{50}\left \lvert \sin\left(x+\frac{\pi }{4}\right)-\cos\left(x+\frac{\pi }{4}\right) \right \rvert\left \lvert \left(\sin\left(x+\frac{\pi }{4}\right)\right)^{99}+ \dots +\left(\cos(x+\frac{\pi }{4}) \right)^{99} \right \rvert=2^{51}\left \lvert \sin\left(\frac{\delta }{4}\right) \right \rvert\left \lvert 100\left(\sin\left(\frac{\delta }{4}\right)\right)^{99} \right |=200\cdot 2^{50}\left \lvert \left(\sin\left(\frac{\delta }{4}\right)\right)^{100}\right \rvert> 1=\epsilon$.

So, it is contradiction. Hence, $f$ is not uniformly continous in $\mathbb{R}$.

But, the answer to this problem is that $f(x)$ is uniformly continuous $\mathbb{R}$.

I don't know why this is uniformly continuous.

• How did you get $|f(x)-f(y)|=....$? I mean, the expression you obtained right after. – Frank Lu May 24 '18 at 6:15
• Formatting hint: if you prefix the trigonometric functions' names with a backslash, they will become special LaTeX symbols. That will make them to render in an upright font and look like functions' symbols: \sin x+\cos x → $\sin x+\cos x$, instead of ordinary maths italic font: sin x+cos x → $sin x+cos x$, which looks like a product of four variables $s\cdot i\cdot n\cdot x$. – CiaPan May 24 '18 at 6:25

The function $f$ is continuous Therefore, $f|_{[0,2\pi]}$ is uniformly continuous and, since $f$ is also periodic, with period equal to $2\pi$, it is not hard to deduce that $f$ is indeed uniformly continuous.
1. $a^n-b^n=(a-b)\left(\sum_{j=0}^{n-1}a^jb^{n-j-1}\right)$. Thus \begin{aligned}&|\sin^{100}(x+\pi/4)-\sin^{100}(y+\pi/4)|\\ =&|\sin(x+\pi/4)-\sin(y+\pi/4)|\left|\sum_{j=0}^{99}\sin^{j}(x+\pi/4)\sin^{99-j}(y+\pi/4)\right|\\ \leq &100|\sin(x+\pi/4)-\sin(y+\pi/4)|.\end{aligned}
1. Next, use the identity $$\sin a-\sin b=2\cos\frac{a+b}{2}\sin\frac{a-b}{2}.$$
2. Finally , use the fact that $$|\sin x|\leq|x|$$ for all real number $x$.
$|f'(x)|\leq 200(2^{99})$ so $f$ is uniformly continuous by MVT.