Proof that if f(a)<0 and f(b)>0 f and is continuous on [a,b] then f changes sign at some c in (a,b)

Question

In Calculus, people argue that if $f(a)<0$ and $f(b)>0$, then IVT guarantees that $f(x)$ changes sign at some point in $(a,b)$. I understand that IVT guarentees that there is a $c$ such that $f(c)=0$. I was able to prove using the bisection method that there is a $c\in (a,b)$ such that $f$ changes sign at $c$ in the sense that $\forall \delta>0$, $\exists x\in (c,c+\delta)$, $f(x)<0$ and $\exists x\in (c,c+\delta)$, $f(x)>0$.

Assuming that $f(a)<0$ and $f(b)>0$ and $f$ is continuous on $[a,b]$, I would like to show that there is some point $c\in(a,b)$ such that $f$ changes sign from negative to positive in the sense that $\exists \delta_1>0, \forall x\in(c-\delta_1,c),\ f(x)<0$ and $\exists \delta_2>0$, $\forall x\in(c,c+\delta_2), \ f(x)>0$. I am surprised that no textbook has proven this even though it is a crucial result in Calculus.

I know that the sense of changing sign in paragraph 1 is not equivalent to the sense in paragraph 2, as

$f(x)=\begin{cases} xsin\left(\frac{1}{x}\right), & \text{if $x\ne0$} \\ 0, & \text{if $x=0$} \end{cases}$

satisfies the definition of changing sign in the first paragraph but not the definition of changing sign in the second paragraph.

I tried arguing by contradiction, assuming that $f$ does not change sign from positive to negative at any point in $(a,b)$. Then $\forall c\in (a,b)$, either $\forall \delta_1>0, \exists x\in(c-\delta_1,c),\ f(x)\ge 0$ or $\exists \delta_2>0$, $\exists x\in(c,c+\delta_2), \ f(x)\le0$.

The problem is that I don't know how to deal with the or in bold. I have been able to prove that assuming $\forall c\in (a,b)$ $\forall \delta_1>0, \exists x\in(c-\delta_1,c),\ f(x)\ge 0$ results in a contradiction, but this is not what I was trying to prove was false.

Answer 1

We have the following stronger result:

Theorem (Strong IVT): Let $f:[a, b] \to\mathbb {R} $ be a function which is continuous on $[a, b] $ such that $f(a) f(b) <0$. Then there exist two points $c, d$ in $(a, b)$ such that $c\leq d, f(c) =f(d) =0$ and $f(x) $ maintains a constant sign in $[a, c) $ and in $(d, b] $. In other words there is a first and a last value of $x$ in $[a, b] $ for which $f(x) =0$.

To prove it consider set $$A=\{x\in[a, b] \mid f\text{ maintains a constant sign in }[a, x] \}$$ The set $A$ is bounded above by $b$ and includes all points of interval $[a, a+h] $ for some $h>0$. It follows that $c=\sup A$ exists and $a<c\leq b$. Similarly for set $B$ defined by $$B=\{x\in[a, b] \mid f\text{ maintains a constant sign in }[x, b] \} $$ $d=\inf B$ exists and $a\leq d<b$. It can further be proved that $c, d$ satisfy all the properties in conclusion of the theorem (prove this!).

If $c=d$ then you can see the sign change happening at $c$ in exactly the way you want. But this is not guaranteed all the time. For example if $f(x) =1-x$ for $0\leq x\leq 1$,$f(x)=0$ for $1<x<2$ and $f(x) =2-x$ for $2\leq x\leq 3$ then $f(0)f(3)<0$ and yet no sign change occurs at a single point.