So, the objective is to determine if $$f(x)= \frac {\sin x}{x}$$ is Uniformly Continuous in the Interval $(0,\infty)$ or not.
What I tried is, that I decomposed the given interval into $(0,p]\cup[p,q]\cup[q,\infty)$ where $p,q\in (0,\infty)$. For $[p,q]$,
it is known that $\sin x$ and $\frac1x$ are continuous for all finite values of $x$, so it is trivial to show that in the closed interval, $f(x)$ would be uniformly continuous.
The difficult part is for the other two intervals. For$(0,p)$, I took a sequence $$x_n=\frac1n$$ which converges to 0 as $n\to \infty$.
Then I tried simplifying $$\lvert f(x_m)-f(x_{m+1})\rvert$$ to get a product involving $\lvert x_m-x_m+1\rvert$ . But this does not work as I get stuck in $sinm$ terms.
Now regarding $[q,\infty]$, I am not able to think in any direction.
Kindly provide a solution. Thanks!
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What have you tried? – Mark Jul 15 '21 at 13:04
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I took two points $x,y \in (0,\infty)$ and then tried to simplify $$\lvert f(x)-f(y)\rvert$$, to put it in the form of the known definition for uniform convergence. But that does not seem to work. – Vaibhav Dixit Jul 15 '21 at 13:21
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What are the standard ways? Tell us why they don’t work.Give us some concrete indication that you have made some effort. – Thomas Andrews Jul 15 '21 at 13:22
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1Put your work in the question, not in comments, @VaibhavDixit4-YrBTechMinin – Thomas Andrews Jul 15 '21 at 13:22
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1Can you answer the question on any smaller interval? – Thomas Andrews Jul 15 '21 at 13:27
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@ThomasAndrews I apologize. I am still a beginner on the platform. Would certainly give a better description in any question I ask in future. Thanks for your guidance. – Vaibhav Dixit Jul 15 '21 at 13:59
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More generally, you could show that a continuous function which vanishes at $-\infty$ and $\infty$ is in fact uniformly continuous. – csch2 Jul 15 '21 at 14:08
1 Answers
$f(x)=\frac{\sin x}x$ can be extended to $[0,\infty)$ by defining $f(0)=1,$ so that it is continuous on $[0,\infty).$
Any continuous function is uniformly continuous on any closed bounded interval, so $f$ is uniformly continuous on $[0,1].$
Now we will outline how to show that $f$ is uniformly continuous ion $[1,\infty).$
Use that:
$$f(x)-f(y)=\sin x\left(\frac1x-\frac1y\right)+\frac1y(\sin x-\sin y)$$
So for $x,y\in[1,\infty)$ $$\begin{align}|f(x)-f(y)|&\leq |\sin x|\left|\frac1x-\frac1y\right| +\frac{1}{y}|\sin x-\sin y|\\&\leq \left|\frac1x-\frac1y\right| +|\sin x-\sin y| \end{align}$$
Then show that $g(x)=\sin x$ and $h(x)=\frac1x$ are both uniformly continuous on $[1,\infty)$ to show that $f$ is uniformly continuous on $[1,\infty).$
More generally, you can use that $f(x)\to 0$ as $x\to\infty.$
Given any $\epsilon>0$ there is some $N$ such that $|f(x)|<\epsilon/4$ for all $x\geq N.$
Use that $f$ is uniformly continuous on $[0,N]$ to conclude that there is a $\delta>0$ that works for all of $[0,\infty).$
More generally you can show similarly:
If for some real $a,L,$ $f$ is continuous on $[a,\infty)$ and $f(x)\to L$ as $x\to\infty,$ then $f$ is uniformly continuous on $[a,\infty).$
A third approach is that if $f$ is differentiable on $[0,\infty)$ and $f’$ is bounded then $f$ is uniformly continuous.
This is because $f(x)-f(y)=(x-y)f’(z)$ for some $z$ between $x,y$ so:
$$|f(x)-f(y)|\leq B|x-y|$$ where $|f’(x)|<B.$
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Thanks. Edited it to “closed bounded” rather than “compact” because some calculus texts might cover compactness after uniform convergence. @principal-ideal-domain – Thomas Andrews Jul 15 '21 at 15:00
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