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Show that the function $f\left(x\right) = \frac{\sin x} {x}, x >0$ is uniformly continuous.

Here what I have done :

Let $x_1, x_2 \in \mathbb R^+$

Then,

$$|f\left(x_1\right)-f\left(x_2\right) | = \left|\frac{\sin x_1} {x_1} - \frac{\sin x_2} {x_2}\right| \implies\frac{|x_2\sin x_1 - x_1\sin x_2|}{x_1 x_2}$$

Then, what to do??

AsukaMinato
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Bapon Das
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  • Try approaching this way: search for discontinuities. Sometimes, the raw definitions aren't very comfortable to work with. – Pritt Balagopal Jun 14 '17 at 13:50
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    Can you prove that a function continuous on $(-\infty,\infty)$ with limit $0$ at $\pm \infty$ is uniformly continuous? (That's a standard exercise.) If so, what prevents you from applying this to your function? And can that be circumvented? – Clement C. Jun 14 '17 at 13:53
  • The function is continous on (0, \infty) – Bapon Das Jun 14 '17 at 13:53
  • @BaponDas that doesn't answer the question. There are functions which are continuous on $(0,\infty)$ and not uniformly continuous there, which is exactly what the OP is asking. – Kernel_Dirichlet Jun 14 '17 at 14:04
  • Yes yes, it is true. – Bapon Das Jun 14 '17 at 14:09

3 Answers3

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We have $|\cos x_1-\cos x_2|\le |x_1-x_2|$ for all $x_1, x_2,$ because the absolute value of the derivative is $\le1$. Then $$\left|\frac{\sin x_1}{x_1}-\frac{\sin x_2}{x_2}\right|=\left|\int^1_0(\cos tx_1-\cos tx_2)\,dt\right|\le|x_1-x_2|\int^1_0t\,dt=|x_1-x_2|/2.$$

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The function extends by continuity to $x=0$ by setting $f(0)=1$. Thus $f$ is uniformly continuous on the compact interval $[0,1]$. It remains to show that $f$ is uniformly continuous on $[1,\infty)$. This can be done for instance by showing that the derivative on this interval is bounded.

Mikhail Katz
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Here is a way you can do this. For $x\in (1,\infty)$ First, note that if $f=\frac{sin(x)}{x}$, then the derivative is $f'(x)=\frac{cos(x)}{x}-\frac{sin(x)}{x^2}$. The derivative $f'$ is actually bounded, You can use the crude bound $|f'|<1$. If the derivative is bounded, the function $f$ is Lipschitz, which also means it is uniformly continuous.

for $x\in (0,1)$ there is more work to be done, as the derivative is not bounded, and there are uniformly continuous functions with unbounded derivatives.

For uniform continuity, we want to show that $\forall \epsilon>0$, we have a $\delta>0$ with $|f(x_1)-f(x_2)|<\epsilon$ when $|x_1-x_2|<\delta$. For $f$ to be uniformly continuous, the choice of $\delta$ cannot depend on $x$. In other words, there is a universal $\delta$ $\forall x$.

Also, use the fact that if $g,h$ are uniformly continuous, then their sum $f+g$ is uniformly continuous. Define $g(x)=\frac{cos(x)}{x}$ and $h(x)=\frac{-sin(x)}{x^2}$. For $g$ and $h$, choosing $\delta=\frac{\epsilon}{2}$ should do the trick (unless I was sloppy). Now just put $f=g+h$ and conclude the sum of two uniformly continuous functions are uniform to complete the proof.

Kernel_Dirichlet
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