An example of a continuous function in L1 space with no limit at infinity

[eeMath]

Homework Statement

I am trying to come up with a continuous function in L1[0,infinity) that doesn't converge to 0 as the function goes out to infinity.

Homework Equations

I am trying to show an example of an f in L1[0,infinity) (i.e. ∫abs(f) < infinity) where the limit as the function goes to infinity does not exist.

The Attempt at a Solution

My professor said it could be done in the lecture, but I have had no success... I have been experimenting with decreasing oscillating functions (e.g sinx/x - where I can't stop it converging to 0) and rapid oscillating functions (e.g. sin(e^x) - can't seem to get an integral on the absolute value) and haven't figured it out yet. Anyone have any ideas about a function that would work?

 

[clamtrox]

I don't think you can do it with smooth functions. Perhaps something discontinuous might do the trick!

 

[eeMath]

I don't think you can do it with smooth functions. Perhaps something discontinuous might do the trick!

My professor said clearly that you can do it with a continuous function. With a discontinuous function, you could put in pieces with measure zero that would mess up the limit, while still being in L1...

Any more ideas?

 

[clamtrox]

Yeah sorry; I typed my answer horribly wrong :D What I meant is that you might want to try to find a function which is continuous, but behaves more and more like a discontinuous function when you go to high x

 

[eeMath]

Yeah sorry; I typed my answer horribly wrong :D What I meant is that you might want to try to find a function which is continuous, but behaves more and more like a discontinuous function when you go to high x

Thanks, that sounds more like it.
I have been trying some functions like sin(e^x) which is continuous and has no limit, but I can't prove it is in L1. Is it in L1? Do I need to change the function in some way to get it into L1?

 

[Dick]

My professor said clearly that you can do it with a continuous function. With a discontinuous function, you could put in pieces with measure zero that would mess up the limit, while still being in L1...

Any more ideas?

Sum a series of bumps of height 1 but smaller and smaller area centered on points running off to infinity.

 

[eeMath]

Sum a series of bumps of height 1 but smaller and smaller area centered on points running off to infinity.

Thanks for the reply...
With abs(sin(e^x)) I have a function that does that, but I am getting a limit of 0 according to wolfram alpha... am I doing something wrong?

 

[Dick]

Thanks for the reply...
With abs(sin(e^x)) I have a function that does that, but I am getting a limit of 0 according to wolfram alpha... am I doing something wrong?

It doesn't have limit 0, but it's also not in L1. Define the function piecewise. Can you define a function f(x) such that f(0)=1 and f(x)=0 for x outside of the interval [-1,1]? That's what I mean by a 'bump'. What's the area under your f(x)? What would be the area under f(cx) for some constant c? Now think about making an infinite sum of them.

 

[eeMath]

It doesn't have limit 0, but it's also not in L1. Define the function piecewise. Can you define a function f(x) such that f(0)=1 and f(x)=0 for x outside of the interval [-1,1]? That's what I mean by a 'bump'. What's the area under your f(x)? What would be the area under f(cx) for some constant c? Now think about making an infinite sum of them.

The function f=abs(cos x) satisfies f(0)=1 and f(pi/2)=0. The area under the curve is 2 for every bump (-pi/2 to pi/2). If you take f(2x), the area under the curve is 1 for the bump (-pi/4 to pi/4). I see how the area and interval are cut by the value of c, so if the c increases each bump has progressively less area under the curve over a progressively smaller interval.
But, what is the proper notation for setting up an infinite sum of these bumps and how can I prove this doesn't converge to 0?

 

[Dick]

f(k(x-c)) is going to be a bump centered at c. Pick a sequence of points c going to infinity. Arrange the values of the k so the infinite series that sums the areas converges. You could express the final result using sigma notation, but describing it in words should be fine too. All you have to do to show the sum doesn't converge to 0 as x->infinity is notice that your sum assumes the value 1 for arbitrarily large values of x.

 

[clamtrox]

The function f=abs(cos x) satisfies f(0)=1 and f(pi/2)=0. The area under the curve is 2 for every bump (-pi/2 to pi/2). If you take f(2x), the area under the curve is 1 for the bump (-pi/4 to pi/4). I see how the area and interval are cut by the value of c, so if the c increases each bump has progressively less area under the curve over a progressively smaller interval.
But, what is the proper notation for setting up an infinite sum of these bumps and how can I prove this doesn't converge to 0?

As I said earlier, you can't really use a smooth function like that (or you have to choose a very weird smooth function..) Easiest thing to do would be just to have a saw-tooth function which is nonzero on a sequence of decreasing intervals between each natural number. So for example it's nonzero in \lbrace [0,1] \cup [1,1+1/2] \cup [2,2+1/4] \cup [3, 3+1/8] \cup ... \rbrace

 

[eeMath]

f(k(x-c)) is going to be a bump centered at c. Pick a sequence of points c going to infinity. Arrange the values of the k so the infinite series that sums the areas converges. You could express the final result using sigma notation, but describing it in words should be fine too. All you have to do to show the sum doesn't converge to 0 as x->infinity is notice that your sum assumes the value 1 for arbitrarily large values of x.

Thanks, this should help, but I meant how do I show that that the limit as x-> doesn't exist?

 

[Dick]

As I said earlier, you can't really use a smooth function like that (or you have to choose a very weird smooth function..) Easiest thing to do would be just to have a saw-tooth function which is nonzero on a sequence of decreasing intervals between each natural number. So for example it's nonzero in \lbrace [0,1] \cup [1,1+1/2] \cup [2,2+1/4] \cup [3, 3+1/8] \cup ... \rbrace

Probably should have been more careful to say explicitly, "define the function f(x)=cos(x) for x in [-pi/2,pi/2] and f(x)=0 otherwise". That's what I was thinking was intended, but perhaps not. Does that make it clearer why a function defined as an infinite sum of bumps in the right way doesn't approach 0 as x->infinity?

 

[eeMath]

Probably should have been more careful to say explicitly, "define the function f(x)=cos(x) for x in [-pi/2,pi/2] and f(x)=0 otherwise". That's what I was thinking was intended, but perhaps not. Does that make it clearer why a function defined as an infinite sum of bumps in the right way doesn't approach 0 as x->infinity?

Totally; I think I was momentarily confusing the value of the function (will always fluctuate between 1 and 0) and the area under the curve. So I guess the final thing I don't know how to prove is that the integral for the entire absolute value of the function is less than infinity so that the function remains in L1[0,infinity). Is it enough to say that the bumps get infinitely thin (area under each bump approaches 0)?

 

[Dick]

Totally; I think I was momentarily confusing the value of the function (will always fluctuate between 1 and 0) and the area under the curve. So I guess the final thing I don't know how to prove is that the integral for the entire absolute value of the function is less than infinity so that the function remains in L1[0,infinity). Is it enough to say that the bumps get infinitely thin (area under each bump approaches 0)?

Yes. You have to specify how the bumps get thin and where they are centered. Then make sure the sum of the area under all of the bumps is a convergent series.

 

[eeMath]

Yes. You have to specify how the bumps get thin and where they are centered. Then make sure the sum of the area under all of the bumps is a convergent series.

So I figured out a function that starts with an area of 2 under the first bump and each successive area is cut in half...

f(t) = \sum^{\infty}_{k=0} abs{cos(2^{k}(t - \frac{3(2^{k-1})\pi }{2^{k+1}}} where \frac{3(2^k)\pi - 4 \pi}{2^{k+1}} \leq t \leq \frac{3(2^k)\pi - 2 \pi}{2^{k+1}}

The widths diminish: pi,pi/2,pi/4,pi/8... and integrals diminish: 2,1,1/2,1/4...
How does it look to you?
Can I just use the standard proof that Ʃ 1/(2^n) converges to 1? It would give a total area under the curve as 3 (from 0 to infinity).

 

[Dick]

So I figured out a function that starts with an area of 2 under the first bump and each successive area is cut in half...

f(t) = \sum^{\infty}_{k=0} abs{cos(2^{k}(t - \frac{3(2^{k-1})\pi }{2^{k+1}}} where \frac{3(2^k)\pi - 4 \pi}{2^{k+1}} \leq t \leq \frac{3(2^k)\pi - 2 \pi}{2^{k+1}}

The widths diminish: pi,pi/2,pi/4,pi/8... and integrals diminish: 2,1,1/2,1/4...
How does it look to you?
Can I just use the standard proof that Ʃ 1/(2^n) converges to 1? It would give a total area under the curve as 3 (from 0 to infinity).

It's a good try. But the center point of your bumps isn't going to infinity. They are piling up at a single point, the limit of 3*2^(k-1)pi/2^(k+1) as k->infinity. I think you should just define it in terms of of the function I defined in post 13. You could write it as the sum of f(2^k*(x-ck)). Pick a sequence for ck that goes to infinity.

 

[eeMath]

It's a good try. But the center point of your bumps isn't going to infinity. They are piling up at a single point, the limit of 3*2^(k-1)pi/2^(k+1) as k->infinity. I think you should just define it in terms of of the function I defined in post 13. You could write it as the sum of f(2^k*(x-ck)). Pick a sequence for ck that goes to infinity.

But if I pick some sequence for ck that goes to infinity, how will the bumps line up to create a continuous function? I may be totally missing something here - could you please give me an example of a function for ck that goes to infinity and how the whole function would look? At this point I am lost...

 

[Dick]

But if I pick some sequence for ck that goes to infinity, how will the bumps line up to create a continuous function? I may be totally missing something here - could you please give me an example of a function for ck that goes to infinity and how the whole function would look? At this point I am lost...

I'm not sure why you think the bumps should 'line up'. Each function in your sum is continuous all by itself. How about just ck=k? Then you have an infinite sequence of bumps of height 1 getting narrower and narrower as the centers go to infinity (well except maybe at the beginning, the first few might overlap).

 

[eeMath]

I'm not sure why you think the bumps should 'line up'. Each function in your sum is continuous all by itself. How about just ck=k? Then you have an infinite sequence of bumps of height 1 getting narrower and narrower as the centers go to infinity (well except maybe at the beginning, the first few might overlap).

I see... I was confusing continuity as lining up everything perfectly. Even if the terms overlap, the bumps will just look a little wobbly over the overlapping interval. One last thing, for L1[0,infinity) do the centers always need same distance from each other, or can the distance get larger? (From the earlier example, the sum converges to quickly if the distance gets smaller).

 

[Dick]

I see... I was confusing continuity as lining up everything perfectly. Even if the terms overlap, the bumps will just look a little wobbly over the overlapping interval. One last thing, for L1[0,infinity) do the centers always need same distance from each other, or can the distance get larger? (From the earlier example, the sum converges to quickly if the distance gets smaller).

No, no reason the centers have to be the same distance from each other. ck=k^2 works too. But you do want them to approach infinity. Otherwise you won't be able to say the limit x->infinity is not zero.