Analysis Question on Continuity

In summary, the function ##f:[0,1] \to \mathbb{R}## is continuous if and only if there exists a number ##x_0 \in (0,1]## such that f(x_0) = 0 and f(x) > 0 for all x in [0, x_0].
  • #1
dhong
2
0

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!
 
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  • #2
dhong said:

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\}##.
 
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  • #3
LCKurtz said:
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].
 
  • #4
pasmith said:
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.
 
  • #5
pasmith said:
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.
 
Last edited:
  • #6
! said:
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.
 
  • #7
pasmith said:
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]
 

Related to Analysis Question on Continuity

1. What is continuity and why is it important in analysis?

Continuity is a mathematical concept that describes the smoothness and connectedness of a function. In analysis, it is important because it allows us to make meaningful statements about the behavior of a function at a particular point or interval.

2. How is continuity different from differentiability?

While continuity describes the connectedness of a function, differentiability refers to the smoothness and rate of change of a function. A function can be continuous but not differentiable, meaning it has no abrupt jumps or holes but the rate of change is undefined at certain points.

3. How do you determine if a function is continuous at a specific point?

A function is continuous at a point if the limit of the function at that point exists and is equal to the value of the function at that point. In other words, the left-hand and right-hand limits of the function approach the same value.

4. Can a function be continuous but not continuous at a specific point?

Yes, a function can be continuous on an interval but not at a specific point. This occurs when the limit at that point does not exist or is not equal to the value of the function at that point.

5. Why is the concept of continuity important in real-life applications?

In real-life applications, continuity allows us to make predictions and analyze the behavior of systems. For example, in physics, we use continuity to study the motion of objects and in economics, we use it to model the changes in supply and demand.

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