 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 $$x$ approaches 0, $1/x$ goes through an odd multiple of $\frac{\pi}{2}$ to an even multiple of (\frac{\pi}{2}$$ 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.