A function $f$ is increasing on $A$ if $f(x)\leq f(y)$ for all $x<y$ in $A$. Show that the Intermediate Value Theorem does have a converse if we assume $f$ is increasing on $[a,b]$.
The converse of IVT along with the additional hypothesis would look like this: Lef $f:[a,b] \rightarrow \mathbb{R}$ be an increasing function on $[a,b]$ which satisfies: if $L$ is a real number such that $f(a)<L<f(b)$ (or $f(a)>L>f(b)$), then there exists a point $c\in (a,b)$ where $f(c)=L.$ If the preceding conditions are met, then $f$ is continuous on $[a,b]$.
The proof (if it's correct) I'm going to show is only for the $c \in (a,b)$, but I think for end points ($a$ or $b$), the method should be similar.
We want to show that given $c \in (a,b)$, for all $\epsilon>0$, there exists a $\delta>0$ such that $|x-c|<\delta \implies |f(x)-f(c)|<\epsilon$.
Since $f$ is increasing, we know that $f(a)\leq f(c)$. If $f(c)-\epsilon<f(a)$, then set $x_1=a$, if $f(c)-\epsilon\geq f(a)$, then we know by hypothesis that a $x_1<c$ exists such that $f(x_1)=f(c)-\epsilon.$ In either case, we have for $x \in (x_1,c]$ $$ f(c)-\epsilon \leq f(x_1) \leq f(x) \leq f(c) .$$
In a similar way, since we know that $f(c)\leq f(b)$, we can deduce that a $x_2>c$ exists such that, for $x \in [c,x_2)$ $$f(c)\leq f(x)\leq f(x_2)\leq f(c)+\epsilon .$$
Pick $\delta = \min \{c-x_1,x_2-c \}$, we then have $$|x-c|<\delta \implies |f(x)-f(c)|<\epsilon .$$
Is this correct? In the proof that I saw, they originally used $\frac{\epsilon}{2}$, instead of just $\epsilon$, but I think this one works just as well, what do you guys think?
Best Answer
Your proof is correct.${}{}{}$