Is f(a) < f(c) < f(b) true for a continuous and one-to-one function?

[xsw001]

Let f:[a,b]->R be continuous and one-to-one such that f(a)<f(b).
Let a<c<b. Prove that f(a)<f(c)<f(b)

My first instinct is to apply intermediate value theorem. Let me know whether my proof makes sense or not.

Proof:
Since f:[a,b]->R is continuous and one-to-one.
Therefore f is strictly increasing function.
Suppose a<c<b
According to Intermediate Value Theorem
There exists f(c) such that f(a)<f(c)<f(b)

 

[fzero]

The IVT only uses continuity and merely guarantees that there exists at least one c, such that a < c < b and f(a) < f(c) < f(b). This is not the same statement as what is to be proved, namely that f(a) < f(c) < f(b) for all c in [a,b]. So your proof is incomplete.

You will need to use that f is 1-1 and the fact that you say f is therefore strictly increasing. However it might be illuminating to study the proof of the IVT in order to adapt it to this situation.

 

[xsw001]

Okay, I'm still not completely getting the right approach to complete this proof. "fzero", you were saying that I need to use the fact of f is 1-1, along with f is continuous on [a, b] that only implies the function is NOT a constant function, so it is strictly monotone, isn't it?

The IVT states if f:[a,b]->R is continuous, and if f(a)<c<f(b) or f(b)<c<f(a), then there exists an x in (a,b) such that f(x)=c.

I'm a little unclear and confused as of how to approach it the right way, can you further clarify a little? Thanks.

 

[fzero]

Okay, I'm still not completely getting the right approach to complete this proof. "fzero", you were saying that I need to use the face of f is 1-1, along with f is continuous on [a, b] that only implies the function is NOT a constant function, so it is strictly monotone, isn't it?

The IVT states if f:[a,b]->R is continuous, and if f(a)<c<f(b) or f(b)<c<f(a), then there exists an x in (a,b) such that f(x)=c.

I'm a little unclear and confused as of how to approach it the right way, can you further clarify a little? Thanks.

It's clear that f is strictly increasing, but I'm not sure that we've proven that with the sufficient rigor that might be expected in your course.

The problem with direct application of the IVT is that it starts with a u such that f(a) < u < f(b) and tells you that there's an x in [a,b] such that f(x) = u. In your conjecture we start with c such that a < c < b. The IVT doesn't directly say anything about f(c).

You can use the IVT to prove this by contradiction. Assume that f(c) > f(b) and apply IVT to the intervals [a,c] and [c,b]. This will lead to a contradiction with the fact that f is 1-1.

 

[xsw001]

Oh, ic, thanks!

 

[fzero]

Oh, ic, thanks!

One more note on this. I forgot that you have to exclude f(c) < f(a), but I think that's straightforward too.

 

[xsw001]

Here is the sketch of the proof.
Assume by contradiction that a<c<b, and case 1) f(a)<f(b)<f(c) or case 2) f(c)<f(a)<f(b)
Since the function is one-to-one, therefore the graph of the continuous function can't oscillate, so it is strictly monotone, either strictly increasing or decresing.
case 1) if a<c<b but f(a)<f(b)<f(c)
then f(a)<f(b)<f(c) => a<b<c if it is strictly increasing which is a contradiction.
also f(a)<f(b)<f(c) => c<b<a if it is striclty decreasing which is also a contradiction.
case 2) if if a<c<b but f(c)<f(a)<f(b)
then f(c)<f(a)<f(b) => c<a<b if it is strictly increasing which is a contradiction.
also f(c)<f(a)<f(b) => b<a<c if it is striclty decreasing which is also a contradiction.
Hence if a<c<b, then f(a)<f(c)<f(b), and the function is strictly increasing.

Is it about right?

 

[fzero]

Well I wouldn't have used the monotonicity because, while it's clear that it's true, we haven't proved it. In essence this theorem is equivalent to monotonicity, so it seems like a case of assuming what is to be proven.

My idea was to take a<c<b, and f 1-1. Since f is 1-1, f(c) cannot be equal to f(a) or f(b). First we assume that f(c)>f(b). Then we consider u such that f(b)<u<f(c), so IVT tells us that there is an x in [c,b] such that f(x) = u. However f(a)<u<f(c), so there is an y in [a,c] such that f(y) =u. But f was 1-1 and x and y are in different sets, so we have a contradiction. We conclude that f(c)<f(b). By repeating the argument for f(c)<f(a), we show that f(a)<f(c).

 

[xsw001]

Thanks so much for the clarification fzero!