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I have a question for you.

Let $f\colon\mathbb{R}\rightarrow \mathbb{R}$ continuous. Assume that there exists $s,t\in\mathbb{R}$, with $t>s$, such that $f(s)=0$ and $f(t)>0$.

I want to prove that there exists an interval $I=[a,b]$ contained in $[s,t]$ such that $f(a)=0$ and $f(x)>0$ for each $x\in(a,b]$. To me it sounds correct, by I would like to have a formal proof.

Thanks!

francesco
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  • Hint: Use continuity to show that there is an interval $J = (t - \epsilon, t+ \epsilon)$ such that $f(x) > 0$ for all $x \in J$. Then let $I = J \cap [s,t]$. – Viktor Vaughn Jun 21 '15 at 16:45

3 Answers3

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Ussing the theorem of sing conservation, there is $\epsilon >0,$ such that for any $x\in [t-\epsilon,t+\epsilon]=I_\epsilon$ $f(x)>0,$ then $I_\epsilon\cap[s,t]=I\subset [s,t]$.

sti9111
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Edit: the question has changed -- this argument does not guarantee the $f(a)=0$ part.

By the Intermediate Value Theorem, there exists $r\in(s,t)$ such that $f(r)>0$. Take $\varepsilon\stackrel{\rm def}{=}\frac{f(r)}{2}$, and use the definition of continuity on $r$ with $\varepsilon$: there exists $\delta >0$ such that for all $x\in [r-\delta,r+\delta]$, $$\lvert f(x)-f(r) \rvert \leq \varepsilon.$$

Now, take $\delta^\prime \stackrel{\rm def}{=} \min(\delta, t-r,r-s)$, so that $I\stackrel{\rm def}{=}[r-\delta^\prime,r+\delta^\prime]\subseteq [s,t]$ (and the equation above still holds, clearly, for all $x\in I$) . This interval $I$ is a good candidate — can you see why?

Second edit: tackling the new question. Starting from above, for instance (with the $r$ guaranteed by the IVT) Let $$S \stackrel{\rm def}{=} \{ x \in [s,r) \mid f(x) \leq 0 \}$$ and $u\stackrel{\rm def}{=}\sup S$. Since $s\in S$, it is a non-empty bounded set, and thus has an supremum: $u$ is well-defined. By definition, $(u,r]\subseteq [s,t]$, and any $x\in (u,r]$ satisfies $f(x)>0$ as $x\neq S$. It remains to show $f(u)=0$. This can be shown in two steps:

  • First, $f(u)\geq 0$: indeed, suppose by contradiction that $f(u) < 0$. Then, by the IVT there is a $u^\prime \in (u,r)$ such that $f(u^\prime) = 0$, so that $u^\prime\in S$ and $u$ cannot be the supremum. Contradiction.

  • Second, $f(u) \leq 0$. This follows from the definition of $S$, the fact that $u=\sup S$ and continuity of $f$ (take any sequence $u_n\to u$ in $S$: $0 \geq f(u_n)\to f(u)$).

This being done, define $I=[u,r]\subseteq [s,t]$.

Clement C.
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  • Exactly. Sorry for changing the question. – francesco Jun 21 '15 at 16:49
  • @francesco see my edit. – Clement C. Jun 21 '15 at 17:01
  • Thanks! Very interesting. Otherwise, I came up with another proof. – francesco Jun 21 '15 at 17:15
  • Thanks! Very interesting. Otherwise, I came up with another proof. Let me know if it works for you. By continuity of $f$ and the fact that $f(t)>0$, there exist $\varepsilon>0$, such that $f(x)>0$ for each $x\in(t-\varepsilon, t]$. Now, from continuity, either $f(t-\varepsilon)=0$ or $f(t-\varepsilon)>0$. Repeating this construction, I believe that by induction the proof can be completed. Thank you again! – francesco Jun 21 '15 at 17:22
  • @francesco the issue here is that if you argue that way, you have no control on $\varepsilon$. The successive epsilons you get could get smaller and smaller (like, $1/2^k$ for the $k$-th) -- this still will be OK for the statement "there exists $\varepsilon > 0$," but the sequence of increasing intervals could possibly converge to an interval whose endpoint's value is not $0$ for $f$. – Clement C. Jun 21 '15 at 22:06
  • Thanks for replying. What you say is correct. I finally built another proof, which looks simpler. Anyway, I think that one can prove by contradiction that the sequence $\epsilon_k$ cannot converge to a point for which $f$ is strictly positive. – francesco Jun 22 '15 at 23:10
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Just use $b:=t$ and $a:=\sup\{u\in[s,t): f(u)=0\}$. By the continuity of $f$, $f(a)=0$. Also, $a<b$ and if $x\in(a,b]$ then $f(x)\not=0$. If $f(x)$ were $<0$ for some $x\in(a,b]$ then there would be $y\in(x,b)$ with $f(y)=0$ (Intermediate Value Theorem) in violation of the definition of $a$.

John Dawkins
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