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So I was answering a problem in which:

  • $f$ is continuous on $[a;b]$;

  • there exists $\alpha$ and $\beta $ in $[a;b]$ such as $f(\alpha)f(\beta)<0$

and while trying to solve the problem I needed to prove the existence of two intervals $I$ and $J$ such as:

  • ($\forall x\in I$) $f(x)>0$
  • ($\forall x\in J$) $f(x)<0$

Can you help me prove the existence of both these intervals?

Thank you very much!

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    You can start by assuming $f(\alpha)>0$, for example. Because $f$ is continuous, there exists an open neighbourhood of $\alpha$ ($I$) on which $f$ is positive. – 2017-02-25

3 Answers 3

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Let's start off by using the fact that $f$ is continuous. Since $f$ is continuous on $[a,b]$ given $\epsilon>0$ $\exists \delta>0$ such that $$|p-p_0| < \delta => |f(p)-f(p_0)| < \epsilon$$

$\forall p \epsilon[a,b]$

now set $p_0= \alpha$ and assume $f(\alpha)<0$ Now for any neighbourhood around $f(p)$ I should be able to find a neighourhood around $p_0$ such that the condition above holds. Now suppose $f(\alpha)<0$ but the neghbourhood $B(\alpha,\delta)$ does not have any element
p: $f(p)<0$ other than $\alpha$. Then for an $\epsilon < |f(\alpha)|$ I wont be able to find a corresponding $\delta$ Contradiction.

So there should exist $p \epsilon B(\alpha,\delta)$ such that $f(p)<0$ And this proves the existence of a Ball or an interval in that case where all elements are zero, since apply the following statement to all element between $p$ and $\alpha$ and you would end up with an interval consisting of negatives.

Consider $B$ as an interval. Apply the same for $f(\beta)>0$

What we have also proved is that somewhere the function is equal to $0$

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By our hypothesis we could assume that $f(\alpha)>0$, while $f(\beta)<0$. Now try to prove the lemma stating that if $f(x_0)>0$ for some interior point $x_0$, then $f(x)>0$ on some neighbourhood of $x_0$. It is enough. Could you continue from this place?

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    Yes! I believe The epsilon-delta definition of continuity would work amazingly. Am i right ? – 2017-02-25
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    @IsmailBourhaeil, yes, of course. – 2017-02-25
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    @IsmailBourhaeil Added an illustration for how the $\epsilon-\delta$ definition of continuity applies to this problem. – 2017-02-25
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WLOG, assume $f(\alpha) < 0$ and $f(\beta) > 0$. Recall this fact for continuous functions.

If $f:[a,b] \to \Bbb R$ is continuous at $x_0 \in [a,b]$ and $f(x_0)>0$, then these exists $\delta > 0$ such that $f(x) > 0 \,\forall x \in (x_0-\delta,x_0+\delta)$.

image

Picture from Wikipedia. View "$2$" as $x_0$ and take $\epsilon = f(x_0) / 2$. Then it's evident from the graph that there's an interval $I:=(x_0-\delta,x_0+\delta)$ such that $f(x) > 0 \,\forall x \in I$.

Apply this fact with $x_0 = \beta$ to get $\delta'>0$ such that $f(x) > 0 \,\forall x \in I := (\beta-\delta',\beta+\delta')$.

Similarly, by considering the function $-f$ and applying this fact, we get the desired interval $J: = (\alpha-\delta,\alpha+\delta)$ for a certain $\delta >0$.

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    This is precisely my answer with a detailed proof. Why all of us assume $f(\alpha)>0$, not $f(\beta)>0)$? :-) – 2017-02-25
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    I edited my answer and swapped $I$ and $J$ to match the question body. My daily vote limit reached. Upvote your ans tmr. I think it won't harm if there's a more detailed explanation, so that users at different level can benefit from them. – 2017-02-25
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    This was of course a joke with $\alpha,\beta$. – 2017-02-25
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    @GNU Supporter That's a nice graph! What software did you used? – 2017-02-25
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    @ΘΣΦGenSan As written in my post: "from Wiki": https://commons.wikimedia.org/wiki/File:Example_of_continuous_function.png – 2017-02-25