Analysis Question on Continuity

[dhong]

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 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!

I would start with ##S=\{x\in [0,1]: f(x) = 0\}##.

 

[pasmith]

I would start with ##S=\{x\in [0,1]: f(x) = 0\}##.

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

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

 

[LCKurtz]

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

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

 

[exclamationmarkX10]

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

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

Don't you mean [itex]f(\sup P) > 0[/itex]?

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

 

[pasmith]

Don't you mean [itex]f(\sup P) > 0[/itex]?

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

 

[exclamationmarkX10]

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

My mistake, I thought you were saying [itex]\sup P=0[/itex]