This problem from TIFR 2013, Problem 19 discusses the example of a non-uniformly continuous function.

Question: TIFR 2013 problem 19


Every differentiable function \(f:(0,1)\to [0,1] \) is uniformly continuous.


\( sin(1/x) \) is not uniformly continuous. However, range is not [0,1]. But its a simple matter of scaling.


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, \(1/x\) 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.

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