I know that there is a theorem that says that "Every continuous function on a bounded closed interval $[a,b]$ is uniformly continuous therein."
But I know that the function $f(x) = 1/x$ defined on (0,1) is not uniformly continuous (I know how to prove this). Does the previous theorem is saying that $f(x) = 1/x$ defined on [0.1,1] is uniformly continuous? why? what should I change in my proof of being not uniform continuous.
Here is my proof of not being uniformly continuous:
Let $\delta > 0,$ take $\epsilon = 1$ and $x = \min \{1, \delta\}$ and $a = \frac{x}{2}.$ then $|x – a | = |x – x/2| = \frac{x}{2} < \delta$ but $|\frac{1}{a} – \frac{1}{x}| = |\frac{2}{x} – \frac{1}{x}| = \frac{1}{x} > 1.$ And so $f$ is not uniformly continuous.
Also, my justification that it is continuous because it is the division of 2 polynomials (and in this justification I do not see any use of open or closed intervals)
Could anyone help me in answering my questions, please?
Thanks in advance!
Best Answer
Uniform Continuous function
To understand it clearly, you should think what uniform continuity looks like. Let me illustrate it.
Given function is:
$f(x) = \frac{1}{x}$
Claim: In the interval (0, 1), you cannot find $\delta$ corresponding to any $\epsilon$.
To understand this, fix $\epsilon = 1$, and choose whatever $\delta$ you think will satisfy the definition. You will find that this will not work. Reduce $\delta$ by half. You will find this will also not work. Keep on dividing it by half. You will find that no such $\delta$ works.
This happens because $f$ grows too fast in the neighbourhood of 0 so that no such $\delta$ sized sub-interval of (0, 1) can capture the growth of $f$.
In the case of interval [0.1, 1], there is no such neighborhood where $f$ grows rapidly (which is true for continuous functions on closed and bounded intervals), therefore you can find a $\delta>0$ corresponding to every $\epsilon>0$.
How to find $\delta$ corresponding to $\epsilon$ in the interval [0.1, 1]:
Choose an arbitrary $\epsilon > 0$.
$|\frac{1}{x} - \frac{1}{y}| = \frac{|x-y|}{|xy|}$
Using the fact that $x, y \geq 0.1$,
$|\frac{1}{x} - \frac{1}{y}| = \frac{|x-y|}{|xy|} \leq 100 |x-y| < \epsilon$
Clearly, $\delta = \frac{\epsilon}{100}$