# Analysis Question on Continuity

## Homework Statement

Suppose the function ##f:[0,1] \to \mathbb{R}## is continuous, ##f(0) > 0## and ##f(1)=0##. Prove that there is a number ## x_0 \in (0,1] : f(x_0) = 0## and ##f(x) > 0## for ##0 \leq x \leq x_0##.

## Homework Equations

We can't use the IVT. Additionally, the definition of continuity we have been given is, a function ##f: D \to \mathbb{R}## is continuous at ##x_0## in ##D## if whenever ##\{x_n \}## is a sequence in ##D## that converges to ##x_0## the image sequence ##\{ f(x_n) \}## converges to ##f(x_0)##.

## The Attempt at a Solution

I was thinking about considering the set ##S=\{f([0,1]) \}## and using the Completeness Axiom to label the ##\inf S## then using the Bolzano-Weierstrass Theorem as part of a constructive existence proof.

Can you tell me if I'm on the right track, or is there a better way to start?

Thanks a ton!

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Suppose the function ##f:[0,1] \to \mathbb{R}## is continuous, ##f(0) > 0## and ##f(1)=0##. Prove that there is a number ## x_0 \in (0,1] : f(x_0) = 0## and ##f(x) > 0## for ##0 \leq x \leq x_0##.

## Homework Equations

We can't use the IVT. Additionally, the definition of continuity we have been given is, a function ##f: D \to \mathbb{R}## is continuous at ##x_0## in ##D## if whenever ##\{x_n \}## is a sequence in ##D## that converges to ##x_0## the image sequence ##\{ f(x_n) \}## converges to ##f(x_0)##.

## The Attempt at a Solution

I was thinking about considering the set ##S=\{f([0,1]) \}## and using the Completeness Axiom to label the ##\inf S## then using the Bolzano-Weierstrass Theorem as part of a constructive existence proof.

Can you tell me if I'm on the right track, or is there a better way to start?

Thanks a ton!

Last edited:
pasmith
Homework Helper

The only drawback there is that resort to the IVT seems necessary to show that $f(x) > 0$ when $x < \inf S$.

I would suggest instead $P = \{ x \in [0,1] : \mbox{f is strictly positive on [0,x)} \}$, and all that is necessary is to show that $f(\sup P) = 0$.

LCKurtz
Homework Helper
Gold Member
The only drawback there is that resort to the IVT seems necessary to show that $f(x) > 0$ when $x < \inf S$.

True enough. I didn't notice until after I read your post that he can't use the IVT.

The only drawback there is that resort to the IVT seems necessary to show that $f(x) > 0$ when $x < \inf S$.

I would suggest instead $P = \{ x \in [0,1] : \mbox{f is strictly positive on [0,x)} \}$, and all that is necessary is to show that $f(\sup P) = 0$.

Don't you mean $f(\sup P) > 0$?

Another route you could take to prove the result is by contradiction: Suppose that there does not exist $x_0 \in (0,1]$ such that $f(x_0) = 0$ and $f(x) > 0\ \forall\ x \in\ [0, x_0]$. That would mean that for all $x_0 \in (0,1], f(x_0) \neq\ 0$ or $f(x) \leq\ 0$ for some $x \in\ [0, x_0]$. Construct a sequence: $x_1, x_2, \ldots$ such that it converges to 0 but the sequence $f(x_1), f(x_2), \dots$ does not converge to a positive number, contradicting the fact that $f$ is continuous.

Last edited:
pasmith
Homework Helper
Don't you mean $f(\sup P) > 0$?

No, $f(\sup P) = 0$. I'd explain why $f(\sup P) > 0$ is impossible, but that would be doing half of the OP's work for him.

No, $f(\sup P) = 0$. I'd explain why $f(\sup P) > 0$ is impossible, but that would be doing half of the OP's work for him.
My mistake, I thought you were saying $\sup P=0$