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

In summary, the conversation was about finding a continuous function in L1[0,infinity) that does not converge to 0 as the function goes out to infinity. The person seeking help had tried using decreasing oscillating functions and rapid oscillating functions, but had not found success. They were looking for further ideas and clarification on how to prove that the function does not converge to 0. Other commenters suggested using a function that is continuous but behaves more like a discontinuous function at high values, or using a piecewise function with infinitely decreasing bumps. The conversation ended with a suggestion to use a saw-tooth function that is nonzero on a sequence of decreasing intervals between each natural number.
  • #1
eeMath
10
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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?
 
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  • #2
I don't think you can do it with smooth functions. Perhaps something discontinuous might do the trick!
 
  • #3
clamtrox said:
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?
 
  • #4
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
 
  • #5
clamtrox said:
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?
 
  • #6
eeMath said:
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.
 
  • #7
Dick said:
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?
 
  • #8
eeMath said:
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.
 
  • #9
Dick said:
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?
 
  • #10
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.
 
  • #11
eeMath said:
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 [itex] \lbrace [0,1] \cup [1,1+1/2] \cup [2,2+1/4] \cup [3, 3+1/8] \cup ... \rbrace [/itex]
 
  • #12
Dick said:
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?
 
  • #13
clamtrox said:
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 [itex] \lbrace [0,1] \cup [1,1+1/2] \cup [2,2+1/4] \cup [3, 3+1/8] \cup ... \rbrace [/itex]

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?
 
  • #14
Dick said:
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)?
 
  • #15
eeMath said:
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.
 
  • #16
Dick said:
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) = [itex]\sum^{\infty}_{k=0} [/itex] abs{cos([itex]2^{k}[/itex](t - [itex]\frac{3(2^{k-1})\pi }{2^{k+1}}[/itex]} where [itex]\frac{3(2^k)\pi - 4 \pi}{2^{k+1}}[/itex] [itex]\leq[/itex] t [itex]\leq[/itex] [itex]\frac{3(2^k)\pi - 2 \pi}{2^{k+1}}[/itex]

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).
 
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  • #17
eeMath said:
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) = [itex]\sum^{\infty}_{k=0} [/itex] abs{cos([itex]2^{k}[/itex](t - [itex]\frac{3(2^{k-1})\pi }{2^{k+1}}[/itex]} where [itex]\frac{3(2^k)\pi - 4 \pi}{2^{k+1}}[/itex] [itex]\leq[/itex] t [itex]\leq[/itex] [itex]\frac{3(2^k)\pi - 2 \pi}{2^{k+1}}[/itex]

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.
 
  • #18
Dick said:
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...
 
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  • #19
eeMath said:
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).
 
  • #20
Dick said:
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).
 
  • #21
eeMath said:
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.
 

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

What is a continuous function?

A continuous function is a mathematical function in which small changes in the input result in small changes in the output. This means that the function is smooth and does not have any abrupt changes or discontinuities.

What is L1 space?

L1 space, also known as the Lebesgue space, is a mathematical space that consists of functions that are Lebesgue measurable and have finite Lebesgue integral.

What does it mean for a function to have no limit at infinity?

A function has no limit at infinity if the values of the function do not approach a specific value as the input approaches infinity. This means that the function does not have a well-defined behavior as the input gets larger and larger.

Can a continuous function in L1 space have no limit at infinity?

Yes, it is possible for a continuous function in L1 space to have no limit at infinity. This means that the function is continuous and measurable, but its values do not approach a specific value as the input gets larger.

Can you provide an example of a continuous function in L1 space with no limit at infinity?

One example is the function f(x) = sin(x)/x, which is continuous and measurable in the L1 space but does not have a limit as x approaches infinity.

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