Sin 1 x Continuous on 0 1 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:
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