Intersection of concave functions

[hermanni]

Hi all,
I have a question. Suppose f : [ 0, l) \rightarrow ℝ is concave , increasing and continuous where l < ∞ and g : [ 0, l) \rightarrow ℝ is also concave, nondecreasing and continuous on the same interval. Can we claim that f and g intersect finitely many times in this interval (possibly 0) ? What if number l replaces with infinity?
Thanx in advance, H.

 

[AlephZero]

This looks a bit like a homework question, so I'm not going to just give you the answer!

The important thing here is that the functions are defined on [0,1), not on [0,1]. In other words they can tend to infinity as $x \rightarrow 1$.

EDIT: Sorry, I misread "concave" as "convex" here.

 

[Bacle2]

Maybe it is helpful to say that if f=g infinitely-often, then h=f-g has infinitely-many

zeros in [0,l). Only continuous ,monotone function with infinitely-many zeros I can

think off is a function of the type d(x,S) , i.e., the distance function between a

point and a set. Edit: there is a result that every closed set is the zero set of a smooth function --

more so a continuous one, so there are a lot of options for h=f-g.

Also, if the functions are monotone, then they are a.e. differentiable,

so it may make sense to assume differentiability to see what happens.

Let me think it through some more, tho.

 

[hermanni]

This looks a bit like a homework question, so I'm not going to just give you the answer!

The important thing here is that the functions are defined on [0,1), not on [0,1]. In other words they can tend to infinity as $x \rightarrow 1$.

Alephero, can you explain that last line (maybe give reference )? Also I think here being increasing is also important , I think only concaveness is not enough here.

 

[hermanni]

and it's not a homework question but I need it as a part of an bigger argument . AlephZero, your last line made me wonder : As far as I get, the domain is also important here. What if f, g : [0, ∞ ) \rightarrow ℝ, can we claim they intersect finitely many times in [0,1) ? Here I guess there's no problem with right end point 1.

 

[AlephZero]

Alephero, can you explain that last line (maybe give reference )? Also I think here being increasing is also important , I think only concaveness is not enough here.

For example take f(x) = 1/(1-x).

Define g(x) by an infinite set of straight line segments joined end to end on the intervals [0, 1/2], [1/2, 3/4], [3/4, 7/8], ...
You can make f(x) and g(x) intersect twice within each interval.

That type of example doesn't work if f(x) is defined on the closed interval [0,1] , because you can't define "$f(1) = \infty$".

You can make the same idea work for the interval [0, ∞ ). For example take. with $f(x) = x^2$. If f(0) and f(1) ar both finite, this idea for getting an infinite number of intersections doesn't work. The theorem that a continuous function on a closed interval is uniformly continuous will probably come into a proof that there are only a finite number of intersections if f(0) and f(1) are both finite.

EDIT: Again, sorry, this is about convex functions not concave.

 

[hermanni]

For example take f(x) = 1/(1-x).

Define g(x) by an infinite set of straight line segments joined end to end on the intervals [0, 1/2], [1/2, 3/4], [3/4, 7/8], ...
You can make f(x) and g(x) intersect twice within each interval.

That type of example doesn't work if f(x) is defined on the closed interval [0,1] , because you can't define "$f(1) = \infty$".

You can make the same idea work for the interval [0, ∞ ). For example take. with $f(x) = x^2$. If f(0) and f(1) ar both finite, this idea for getting an infinite number of intersections doesn't work. The theorem that a continuous function on a closed interval is uniformly continuous will probably come into a proof that there are only a finite number of intersections if f(0) and f(1) are both finite.

OK, my functions f and g are defined on [0, ∞) and I needed finiteness of number of intersections in [0, 1). They're both bounded, concave, increasing and BOUNDED on [0, ∞). I have now boundedness, can I claim finite # of intersections on [0,1)? I think I can do it for [0,1] based on what you said, here does endpoint 1 matter now?

 

[haruspex]

Are we all using the same definition of concave? The definition I'm used to is the one at http://en.wikipedia.org/wiki/Concave_function. According to that, 1/(1-x) (on [0,1)) would be convex.
hermanni, please clarify.

 

[hermanni]

Are we all using the same definition of concave? The definition I'm used to is the one at http://en.wikipedia.org/wiki/Concave_function. According to that, 1/(1-x) (on [0,1)) would be convex.
hermanni, please clarify.

Well, I'm using the same definition and you're right, I didn't read carefully and 1/1-x is convex.

 

[AlephZero]

Are we all using the same definition of concave? The definition I'm used to is the one at http://en.wikipedia.org/wiki/Concave_function. According to that, 1/(1-x) (on [0,1)) would be convex.

Oops. So ignore my comments about the open and closed intervals!

Bu I don't think this changes my assertioon that there can be an infinite number of intersections, with the conditions in the OP. Given a smooth enough function f(x) you can still construct a function g(x) from an infiite number of line segments, with an infinite number of intersections.

Of crosue the OP's functions may have some more properties we don't know about, which prevent this.

 

[hermanni]

Oops. So ignore my comments about the open and closed intervals!

Bu I don't think this changes my assertioon that there can be an infinite number of intersections, with the conditions in the OP. Given a smooth enough function f(x) you can still construct a function g(x) from an infiite number of line segments, with an infinite number of intersections.

(*)Of crosue the OP's functions may have some more properties we don't know about, which prevent this.

OK, I got the idea of construction, thank you very much.
(*) Do you think boundedness is one of these properties? If so, how can I show?

 

[haruspex]

Do you think boundedness is one of these properties?

I don't think boundedness gets you far. Here's another construction:
Consider the points (1-1/n, 1-1/n²), n = 1, 3, 5... Join the dots with straight lines. Now do the same with n = 2, 4, 6... And smoothness is not going to help - it wouldn't be hard to smooth out the corners.
Conditions on higher derivatives might do it, but I'd be surprised.

 

[Bacle2]

For example take f(x) = 1/(1-x).

Define g(x) by an infinite set of straight line segments joined end to end on the intervals [0, 1/2], [1/2, 3/4], [3/4, 7/8], ...
You can make f(x) and g(x) intersect twice within each interval.

That type of example doesn't work if f(x) is defined on the closed interval [0,1] , because you can't define "$f(1) = \infty$".

You can make the same idea work for the interval [0, ∞ ). For example take. with $f(x) = x^2$. If f(0) and f(1) ar both finite, this idea for getting an infinite number of intersections doesn't work. The theorem that a continuous function on a closed interval is uniformly continuous will probably come into a proof that there are only a finite number of intersections if f(0) and f(1) are both finite.

EDIT: Again, sorry, this is about convex functions not concave.

I see, so the idea is that the function/curve is rectifiable, right?

How is f(x)=1/(1-x) non-decreasing?