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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.

User Asle
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1 Answer

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Answer:


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


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

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


(1)/(N) < 2\pi \delta


\Rightarrow (1)/(2\pi N) < \delta.

Let's take
x = (1)/(2\pi N + \pi/2) and
y = (1)/(2\pi N + 3\pi/2). We can see that


|x-y| = (1)/(2\pi N + \pi/2)-(1)/(2\pi N + 3\pi/2)<(1)/(2\pi N) <\delta

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.

User Arxo Clay
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