Prove that the function $f: (0,1) \to \Bbb{R}$ defined by $f(x) = sin(\frac{1}{x})$ is not uniformly continuous.

On the contrary we assume there is a $\delta$ and we derive a contradiction. $a_n = \frac{1}{2n\pi}$ and $b_n = \frac{1}{(2n+\frac{1}{2})\pi}$. We observe that $|\frac{1}{2n\pi} - \frac{1}{(2n+\frac{1}{2})\pi}| = |\frac{1}{2n\pi} - \frac{2}{(4n+1)\pi}| = \frac{1}{2n\pi(4n+1)}$ can be made arbitrarily small as the sequence $\{\frac{1}{2n\pi(4n+1)}\}$ converges to 0. Hence $|\frac{1}{2n\pi} - \frac{1}{(2n+\frac{1}{2})\pi}|$ can be made less that $\delta$ for large $n$. Now, for such $n$, $|f(\frac{1}{2n\pi})-f(\frac{1}{2n+\frac{1}{2})\pi})|$ should be less than $\epsilon = \frac{1}{2}$. But $$|f(\frac{1}{2n\pi})-f(\frac{1}{2n+\frac{1}{2})\pi})| = |sin(2n\pi) - sin((2n+\frac{1}{2})\pi)| = |0-1|=1$$ contradiction!.

The general strategy is, to prove a function is not uniformly continuous, find two sequences $a_n$ and $b_n$ such that $|a_n - b_n|$ is arbitrarily small for large $n$ but $|f(a_n) - f(b_n)|$ is a fixed constant for all $n$. Why it is enough to prove such sequences is explained above using $\epsilon$ and $\delta$ argument.

**Proof**: A function $f : (0,1) \to \Bbb{R}$ is said to be uniformly continuous if given $\epsilon > 0$ there exists a $\delta>0$ such that $|x-y| < \delta$ implies that $|f(x) - f(y)| < \epsilon$**for all**$x,y \in (0,1)$. We will show that, for $\epsilon = \frac{1}{2}$ finding such a $\delta$ (which works**for all**$x,y \in (0,1)$) is not possible.On the contrary we assume there is a $\delta$ and we derive a contradiction. $a_n = \frac{1}{2n\pi}$ and $b_n = \frac{1}{(2n+\frac{1}{2})\pi}$. We observe that $|\frac{1}{2n\pi} - \frac{1}{(2n+\frac{1}{2})\pi}| = |\frac{1}{2n\pi} - \frac{2}{(4n+1)\pi}| = \frac{1}{2n\pi(4n+1)}$ can be made arbitrarily small as the sequence $\{\frac{1}{2n\pi(4n+1)}\}$ converges to 0. Hence $|\frac{1}{2n\pi} - \frac{1}{(2n+\frac{1}{2})\pi}|$ can be made less that $\delta$ for large $n$. Now, for such $n$, $|f(\frac{1}{2n\pi})-f(\frac{1}{2n+\frac{1}{2})\pi})|$ should be less than $\epsilon = \frac{1}{2}$. But $$|f(\frac{1}{2n\pi})-f(\frac{1}{2n+\frac{1}{2})\pi})| = |sin(2n\pi) - sin((2n+\frac{1}{2})\pi)| = |0-1|=1$$ contradiction!.

The general strategy is, to prove a function is not uniformly continuous, find two sequences $a_n$ and $b_n$ such that $|a_n - b_n|$ is arbitrarily small for large $n$ but $|f(a_n) - f(b_n)|$ is a fixed constant for all $n$. Why it is enough to prove such sequences is explained above using $\epsilon$ and $\delta$ argument.

**Share to your groups:****SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO SUBSCRIBE TO MY BLOG**
## No comments:

## Post a Comment