Question

Because of Theorem 5.47 any function that is continuous on (0, 1) but unbounded cannot be uniformly continuous there. Give an example of a continuous function on (0, 1) that is bounded, but not uniformly continuous.

Answer 1

Answer:

$f: (0,1) \to \mathbb{R}$

$f(x) = \sin(1/x)$

Step-by-step explanation:

f is continuous because is the composition of two continuous functions:

$g(x) = \sin(x)$ (it is continuous in the real numbers)

$h(x) = 1/x$ (it is continuous in the domain (0,1))

It is bounded because $-1 \leq \sin(\theta) \leq 1$

And it is not uniformly continuous because we can take $\varepsilon = 1$ in the definition. Let $\delta > 0$ we will prove that there exist a pair $x,y\in \mathbb{R}$ such that $|x-y|< \delta$ and $|f(x) -f(y)|> \varepsilon = 1$.

Now, by the archimedean property we know that there exists a natural number N such that

$\frac{1}{N} < 2\pi \delta$

$\Rightarrow \frac{1}{2\pi N} < \delta$.

Let's take $x = \frac{1}{2\pi N + \pi/2}$ and $y = \frac{1}{2\pi N + 3\pi/2}$. We can see that

$|x-y| = \frac{1}{2\pi N + \pi/2}-\frac{1}{2\pi N + 3\pi/2}$

And also:

$|f(x)- f(y)| = |f(2\pi N + \pi/2) - f(2\pi N + 3\pi/2)| = |1 - (-1)| = 2 > \varepsilon$

And we conclude the proof.