Function That is Differentiable but Not Uniformly Continuous
TIFR 2013 problem 19 | Non-uniformly continuous function
This problem from TIFR 2013, Problem 19 discusses the example of a non-uniformly continuous function. Question: TIFR 2013 problem 19 True/False? Every differentiable function $f:(0,1) \to [0,1]$ is uniformly continuous. Hint: $\sin(1/x)$ is not uniformly continuous. However, range is not $[0,1]$. But its a simple matter of scaling. Discussion: Let $f(x)=\sin(\frac{1}{x})$ for all $x\in(0,1)$. For simplicity, we first prove that this $f$ is not uniformly continuous. Then we will scale things down, which won't change the non-uniform continuity of $f$. Note that $f$ is differentiable. To show that $f$ is not uniformly continuous, we first note that as $x$ approaches $0$, $\frac{1}{x}$ goes through an odd multiple of $\frac{\pi}{2}$ to an even multiple of $\frac{\pi}{2}$ real fast. So in a very small interval close to $0$, I can find two such points which gives value $1$ and $0$. Let $x=\frac{2}{(2n+1)\pi}$ and $y=\frac{2}{2n\pi}$. Then $|x-y|=\frac{2}{2n(2n+1)\pi}$. Since the right hand side of above goes to zero as $n$ increases, given any $\delta > 0$ we can find $n$ large enough so that $|x-y|<\delta$. For these $x$ and $y$, $|f(x)-f(y)|=1$. Of course, choosing $\epsilon$ as any positive number less than $1$ shows that $f$ is not uniformly continuous. We have with us a function $f$ which is bounded, differentiable and not uniformly continuous. To match with the questions requirements, notice $-1\le f(x)\le 1$. So $0\le 1+f(x) \le 2$ And $0\le \frac{1+f(x)}{2} \le 1$. Define $g(x)= \frac{1+f(x)}{2}$ for all $x\in(0,1)$. Since sum of two uniformly continuous functions is uniformly continuous and a scalar multiple of uniformly continuous function is uniformly continuous, if $g(x)$ was uniformly continuous, then $f(x)=2g(x)-1$ would also be uniformly continuous. This proves that g is in fact not uniformly continuous. It is still differentiable, and range is $[0,1]$, which shows that the given statement is actually false.Some Useful Links:
This problem from TIFR 2013, Problem 19 discusses the example of a non-uniformly continuous function. Question: TIFR 2013 problem 19 True/False? Every differentiable function $f:(0,1) \to [0,1]$ is uniformly continuous. Hint: $\sin(1/x)$ is not uniformly continuous. However, range is not $[0,1]$. But its a simple matter of scaling. Discussion: Let $f(x)=\sin(\frac{1}{x})$ for all $x\in(0,1)$. For simplicity, we first prove that this $f$ is not uniformly continuous. Then we will scale things down, which won't change the non-uniform continuity of $f$. Note that $f$ is differentiable. To show that $f$ is not uniformly continuous, we first note that as $x$ approaches $0$, $\frac{1}{x}$ goes through an odd multiple of $\frac{\pi}{2}$ to an even multiple of $\frac{\pi}{2}$ real fast. So in a very small interval close to $0$, I can find two such points which gives value $1$ and $0$. Let $x=\frac{2}{(2n+1)\pi}$ and $y=\frac{2}{2n\pi}$. Then $|x-y|=\frac{2}{2n(2n+1)\pi}$. Since the right hand side of above goes to zero as $n$ increases, given any $\delta > 0$ we can find $n$ large enough so that $|x-y|<\delta$. For these $x$ and $y$, $|f(x)-f(y)|=1$. Of course, choosing $\epsilon$ as any positive number less than $1$ shows that $f$ is not uniformly continuous. We have with us a function $f$ which is bounded, differentiable and not uniformly continuous. To match with the questions requirements, notice $-1\le f(x)\le 1$. So $0\le 1+f(x) \le 2$ And $0\le \frac{1+f(x)}{2} \le 1$. Define $g(x)= \frac{1+f(x)}{2}$ for all $x\in(0,1)$. Since sum of two uniformly continuous functions is uniformly continuous and a scalar multiple of uniformly continuous function is uniformly continuous, if $g(x)$ was uniformly continuous, then $f(x)=2g(x)-1$ would also be uniformly continuous. This proves that g is in fact not uniformly continuous. It is still differentiable, and range is $[0,1]$, which shows that the given statement is actually false.Some Useful Links:
Knowledge Partner
Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta Academy
Aditya Birla Education Academy
Cheenta. Passion for Mathematics
Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
Academic Programs
Free Resources
Why Cheenta?
Close Menu
Source: https://www.cheenta.com/example-of-a-non-uniformly-continuous-function-tifr-2013-problem-19/