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Is the function $$f(x)= {\sin x \over x}$$ Uniformly continuous over $R$

How do i approach this ? I need some hints.

leonbloy
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Aman Mittal
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    One thing would be to compute the derivative and see that that is bounded. (And for completeness, explicitly say that $f(0) = 1$.) – Daniel Fischer Oct 01 '13 at 20:05
  • @DanielFischer - what is the principle that we should be applying ? like the theorems related to closed intervals etc cannot be used. – Aman Mittal Oct 01 '13 at 20:15
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    Self-duplicate. What's next? A third instance of the same question by the same user? – Did Feb 06 '14 at 20:46

3 Answers3

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One way to see that $f$ is uniformly continuous (assuming the removable singularity in $0$ removed) is to note that it is differentiable, and its derivative is bounded. Then use the fact that any differentiable function with bounded derivative is uniformly continuous (even Lipschitz continuous), since by the mean value theorem we have

$$\left\lvert \frac{f(y) - f(x)}{y-x}\right\rvert = \lvert f'(\xi)\rvert \Rightarrow \lvert f(y) - f(x)\rvert \leqslant M\cdot\lvert y-x\rvert,$$

where $M$ is a bound for the derivative.

Another way to see it is to use the fact that $\lim\limits_{\lvert x\rvert\to\infty} f(x) = 0$, and that every continuous function on a compact interval is uniformly continuous. For a given $\varepsilon > 0$, choose a $K > 0$ with $\lvert x\rvert \geqslant K \Rightarrow \lvert f(x)\rvert < \varepsilon/3$. By the uniform continuity of $f$ restricted to the compact interval $[-K-1,K+1]$, there is a $\delta > 0$ such that $\lvert x\rvert, \lvert y\rvert \leqslant K+1, \lvert y-x\rvert < \delta \Rightarrow \lvert f(y)-f(x)\rvert < \varepsilon$. If $\delta$ is chosen $< 1$, that $\delta$ works then on all of $\mathbb{R}$.

Javier
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Daniel Fischer
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    Wouldn't it be (slightly) more direct to use the MVT for derivatives rather than integrals? – dfeuer Oct 01 '13 at 21:06
  • Indeed. Don't know why I didn't think of that. – Daniel Fischer Oct 01 '13 at 22:09
  • Uh, embarrasing. Thanks, @Javier. – Daniel Fischer Oct 01 '13 at 22:36
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    In order to show how the given function is bounded and differentiable over $\mathbb{R}$ if i proceed this way ,

    I re-define the function as $f(x)={\sin x \over x}$ if $x\ne0$ and $f(x)=1$ for $x=0$ Then

    $$Rf'(0)=\lim_{x\rightarrow0}{{sinh\over x}-1 \over x}$ $=\lim_{x\rightarrow0} {-x^2\over 3!}+{x^4\over 5!}- ...=0=Lf'(0)$$

    Again, $$f'(x)={x\cos x-\sin x \over x^2}$$

    $$|f'(x)|={|{x\cos x-\sin x \over x^2}|}$$

    $$|f'(x)|={|{{\cos x\over x}-{\sin x \over x^2}}|}\le {2\over x}<2$$

    Does it look good ?

     – Aman Mittal Oct 02 '13 at 02:52
  • @AmanMittal Pretty good, but one the one hand, it should be $\frac{2}{\lvert x\rvert}$, and on the other, that doesn't give a global bound since for $x \to 0$, $2/\lvert x\rvert \to \infty$. You can, for $\lvert x\rvert < 1$, for example use the Taylor series to get a bound, or argue abstractly ($f'$ is continuous, $[-1,1]$ is compact, hence $f'$ is bounded on $[-1,1]$). – Daniel Fischer Oct 02 '13 at 09:03
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$\mathbb{R} = (-\infty, - M ) \cup [-M,M] \cup (M,+\infty)$

Now,$\sin x \over x$ is continuous in $ [-M,M] $and hence uniformly continuous with a $\delta$ =$\delta_1$ .

Over the other intervals show that $\sin x \over x$ has a bounded derivative and hence uniformly continuous with $\delta$ = $\delta_2$ and $\delta_3$ respectively.

Then the universally $\delta$ = min ($\delta_1$,$\delta_2$,$\delta_3$)

MathMan
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  • You've used this answer three times in a row to duplicate questions. (Copy and paste.) Do not go looking for duplicates to answer each with an identical answer. If you find a duplicate question, you should flag it as such. – amWhy Feb 06 '14 at 20:51
  • okay, will keep that in mind. – MathMan Feb 06 '14 at 21:06
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Wouldn't it be easier to pick an interval, say $[-1,1]$, and say that $f(x)=\begin{cases} \frac{\sin x}x\,, & x\ne 0 \\ 1 \,, & x=0\end{cases}$ is continuous, hence uniformly continuous on $[-1,1]$, then argue that $f$ is uniformly continuous on $|x|\ge 1$, and then make a "standard" argument that allows us to glue together? (Hint: You need $\min(\delta_1,\delta_2,\delta_3)$, not just two of 'em.)

Ted Shifrin
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