Proving $ f(x) = x^2 $ is not uniformly continuous on the real line
In this question, for the proof that a function is not uniformly continuous on a given domain. How did the responder come up with the idea to use $y = x + \frac{\delta}{2}$?
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I understand the idea behind the proof and logic. But how did the respond, essentially out of "thin air" come up with $y = x + \frac{\delta}{2}$? – Timothy Frisch May 12 '15 at 02:35
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Indeed, you can choose $y = x + C\delta$ for any $C$, The main point is that the $x^2$ is cancelled so that the term is "more obvious" to be unbounded. – May 12 '15 at 02:50
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So say, instead of my initial function from the problem being $f(x) = x^2$, I had instead $f(x) = \frac{1}{x}$. Can the same exact disproof approach be taken? Or does this happen to be unique to the function. – Timothy Frisch May 12 '15 at 02:52
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2I would say that is a bit peculiar to that $x^2$. $x^2$ and $\frac 1x$ are quite different: although both of them are not uniform continuous, $x^2$ fails to be uniform continuous when $x\to \infty$, while $\frac 1x$ fails when $x\to 0$. (In this case, you need to choose $y = x/2$ for example. – May 12 '15 at 03:01
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I understand, thank you for your help sir. If you would like to write a reply I can mark it as an answer. – Timothy Frisch May 12 '15 at 03:10