Suppose $f$ is continuous on $\mathbb{R}$ and $f$ has a fixed point. I am working on a problem, and I wondered if it was true that if we set $\epsilon > 0$ would there exists $\delta > 0$ such that $$f((l - \delta, l + \delta )) \subset (l-\epsilon, l + \epsilon)$$

If yes, how is the $\delta$ in relation with $\epsilon$, I think it would smaller than $\epsilon$. Any theorems or proofs on this?

  • $\begingroup$ Use the $\epsilon\delta$ definition of continuity $\endgroup$ – Hagen von Eitzen Sep 30 '17 at 12:31
  • $\begingroup$ Let $f$ be the constant function $l$ then clearly $f(l) = l$ and $f\big(( l-\delta,l+\delta) \big) = \{l\} \subseteq (l-\epsilon,l+\epsilon)$ for all $\epsilon>0$ regardless of $\delta>0$. So the relation of $\delta$ and $\epsilon$ can be anything $\endgroup$ – Nathanael Skrepek Sep 30 '17 at 12:39
  • $\begingroup$ @NathanaelSkrepek What do you mean by regardless of $\delta$. I can't choose whichever I want, can I? $\endgroup$ – John Mayne Sep 30 '17 at 12:57
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    $\begingroup$ @JohnMayne regardless means that it doesn't matter which $\delta>0$ you choose. $\endgroup$ – Nathanael Skrepek Oct 3 '17 at 18:25

Yes, because, since $f$ is continuous at $l$, there is a $\delta>0$ such that$$f\bigl((l-\delta,l+\delta)\bigr)\subset(f(l)-\varepsilon,f(l)+\varepsilon)=(l-\varepsilon,l+\varepsilon).$$

  • $\begingroup$ And is there a way to determine the $\delta$? Or can we just state that $\delta $ exists and nothing more? $\endgroup$ – John Mayne Sep 30 '17 at 12:38
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    $\begingroup$ @JohnMayne Under the hypotesis that you stated, all we can say is that $\delta$ exists and nothing more. $\endgroup$ – José Carlos Santos Sep 30 '17 at 12:41

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