Pick out the uniformly continuous function for $x \in (0,1)$

$$(1) \quad \quad\quad f(x)= \cos x \,\cos \frac {\pi}x$$

$$(2)\quad \quad \quad f(x) = \sin x \, \cos \frac {\pi}x$$

i was trying this question, i was think that $\sin x$ and $\cos x$ are periodic and continuous , so it is uniformly continuous function,, therefore from my point of both option 1 and option 2 both are true uniformly continuous. But i m not sure about my answer If anybody help me i would be very thankful to him thank

  • $\begingroup$ Is it "$\cos (\pi/x)$" or "$(\cos\pi)/x$"? $\endgroup$ – MPW Aug 10 '17 at 12:40
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    $\begingroup$ I reformatted your post fairly heavily. Please check to see if I introduced any errors. $\endgroup$ – lulu Aug 10 '17 at 12:43
  • $\begingroup$ thanks a lots ,, everything is ok @ lulu $\endgroup$ – lomber Aug 10 '17 at 12:47

In case (2) you can define it as $0$ for $x=0$. This turns it into a continuous function on $[0,1]$.

In fact, $\sin(x)\cos(\pi/x)$ is continuous on $(0,1]$ and $\lim_{x\to0^+}\sin(x)\cos(\pi/x)=0$.

Therefore, by Cantor's theorem, it is uniformly continuous.

In case (1), for $\epsilon=1/2$ we can find $x_n=\frac{1}{2k}$ and $y_n=\frac{1}{2k+1}$ which for $k$ large will have $|x_n-y_n|$ arbitrarily small, $\cos(x_n)$ and $\cos(y_n)$ very close to $1$, and $\cos(\pi/x_n)=1$ while $\cos(\pi/y_n)=-1$. Therefore $|\cos(x_n)\cos(\pi/x_n)-\cos(y_n)\cos(\pi/y_n)|>1/2=\epsilon$. Therefore, it is not uniformly continuous on $(0,1)$ or any neighborhood of $0$.

  • $\begingroup$ thanks a lots @ User470888 $\endgroup$ – lomber Aug 10 '17 at 13:35

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