# Is $f$ has a fixed point $l$ does that mean there exists an interval such that $f((l-\epsilon,l+ \epsilon)) \subset (l - \delta, l + \delta)?$

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?

• Use the $\epsilon\delta$ definition of continuity – Hagen von Eitzen Sep 30 '17 at 12:31
• 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 – Nathanael Skrepek Sep 30 '17 at 12:39
• @NathanaelSkrepek What do you mean by regardless of $\delta$. I can't choose whichever I want, can I? – John Mayne Sep 30 '17 at 12:57
• @JohnMayne regardless means that it doesn't matter which $\delta>0$ you choose. – Nathanael Skrepek Oct 3 '17 at 18:25

Yes, because, since $f$ is continuous at $l$, there is a $\delta>0$ such that$$f\bigl((l-\delta,l+\delta)\bigr)\subset(f(l)-\varepsilon,f(l)+\varepsilon)=(l-\varepsilon,l+\varepsilon).$$
• And is there a way to determine the $\delta$? Or can we just state that $\delta$ exists and nothing more? – John Mayne Sep 30 '17 at 12:38
• @JohnMayne Under the hypotesis that you stated, all we can say is that $\delta$ exists and nothing more. – José Carlos Santos Sep 30 '17 at 12:41