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How do I prove $\sin x$ is uniformly continuous on $\mathbb R$ with delta and epsilon?

I proved geometrically that $\sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|\sin x_1 - \sin x_2|\le|\sin x_1|+|\sin x_2|<|x_1|+|x_2|$$

But this doesn't help me much finding a delta...

Thanks for any help!

P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).

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    $\begingroup$ I changed $sinx_1$, etc., to $\sin x_1$. That is standard TeX usage. $\endgroup$ Commented Jan 1, 2013 at 19:35
  • $\begingroup$ This is a particular case of (at least) two more general results. First: a periodic continuous function on $\mathbb{R}$ is uniformly continuous on $\mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $\sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$. $\endgroup$
    – Julien
    Commented Jan 1, 2013 at 19:41

6 Answers 6

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Let $\epsilon>0$ and $x,y\in \mathbb{R}$. We want $$\left|f(x)-f(y)\right|<\epsilon\implies \left|\sin x-\sin y\right|<\epsilon\implies \left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|$$ Because $$\left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|\le 2\left|\sin\frac{x-y}2\right|$$ it suffices $$2\left|\sin\frac{x-y}2\right|<\epsilon$$ when $$\left|x-y\right|<\delta\implies \left|\frac{x-y}2\right|<\delta$$ SInce $\left|\sin x\right|\le \left|x\right|$, $$2\left|\sin\frac{x-y}2\right|\le 2\left|\frac{x-y}2\right|<2\delta$$

Choosing $\delta=\frac{\epsilon}{2}>0$ will do the trick. Because $\delta$ doesn't depend on $x,y$, the continuity is uniform

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    $\begingroup$ Hi, thanks for the fast response! Why $$\left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|\le 2\left|\sin\frac{x-y}2\right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again! $\endgroup$
    – Harold
    Commented Jan 1, 2013 at 19:32
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    $\begingroup$ because $\cos \alpha \leqslant 1 $, $\forall \alpha \in \mathbb{R}$. $\endgroup$ Commented Jan 1, 2013 at 19:34
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    $\begingroup$ Because $\left|\cos(\bullet)\right|\le 1$. $\endgroup$ Commented Jan 1, 2013 at 19:34
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    $\begingroup$ @Harold $\sin x$ is odd $\endgroup$
    – Nameless
    Commented Jan 1, 2013 at 19:48
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    $\begingroup$ @Harold You want $\left|\sin y\right|\le \left|y\right|\iff -\sin y\le -y\iff \sin y\ge y$ for $y<0$ near $0$. But $\sin x\le x$ for $x>0$. Multipliying by $-1$ gives $-\sin x\ge -x\iff \sin (-x)\ge -x\iff \sin y\ge y$ $\endgroup$
    – Nameless
    Commented Jan 1, 2013 at 20:10
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By Mean Value Theorem,

$$ |\sin{x}- \sin{y}| \leq |x-y| |\cos{\xi}| \leq |x-y|, \quad x\leq\xi \leq y.$$

Hence, you may choose $\epsilon=\delta$.

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    $\begingroup$ This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving. $\endgroup$
    – wdacda
    Commented Feb 1, 2016 at 19:55
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    $\begingroup$ @wdacda How is that circular reasoning? $\endgroup$
    – 3x89g2
    Commented Sep 5, 2016 at 5:27
  • $\begingroup$ How is that circular? I can't see it. $\endgroup$ Commented Dec 7, 2020 at 10:30
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Since $\sin x$ is a periodic continuous function with a period $2\pi$, it suffices to prove that it is uniformly continuous on $[0, 2\pi]$. Since $[0, 2\pi]$ is compact, this follows from the well-known theorem.

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There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, y\in(-\pi, \pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|\sin(x) - \sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case $$|\sin(x) - \sin(y) | \le |x - y|.$$

This gives us uniform continuity on $(-\pi, \pi]$, so by periodciity the sine function is uniformly continuous on the entire line.

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  • $\begingroup$ Another way to prove this identity: math.stackexchange.com/questions/620305/… $\endgroup$
    – GinKin
    Commented Dec 28, 2013 at 10:08
  • $\begingroup$ This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions. $\endgroup$ Commented Dec 29, 2013 at 19:56
  • $\begingroup$ Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense. $\endgroup$
    – GinKin
    Commented Dec 29, 2013 at 20:03
  • $\begingroup$ What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique. $\endgroup$ Commented Dec 30, 2013 at 0:46
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Theorem: Real-valued differentiable function on R with bounded derivative is uniformly continuous.

So $f'(x) = cosx$ and $|cosx| \leq 1$ for every $x \in \mathbb R$. Hence $f$ is uniformly continuous.

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We have to show $\ f(x) $= $ sin(x)$ is is uniformly continuous on $\mathbb R$. Let c $\in \Bbb R$. Then for all x $\in \Bbb R$
$|f(x)-f(c)|=|sin(x)-sin(c)|=2|cos(\frac{x+c}{2})||sin(\frac{x-c}{2})| \le 2|sin(\frac{x-c}{2})| \le 2\frac{|x-c|}{2}$, since $|sin(x)| \le |x|$ for all $x \in \Bbb R$. Let us choose $\epsilon > 0$. Then for all $x \in \Bbb R$ satisfying $|x-c|<$ $\epsilon$, $|f(x)-f(c)| \lt \epsilon$.
This shows $f$ is uniformly continuous on $\Bbb R$.

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