Functions in L1 that are not in L2

[J$C]

It is clear that there are functions in L2 that are not in L1, but what about the other way? And what effect does considering L2(R) versus L2([a,b]) have?

Thanks.

 

[HallsofIvy]

To simplify this, let's make it [1,\infty]. L2 contains functions f(x) such that \int_1^\infty f^2(x)dx[/itex] is finite. L1 contains functions f(x) such that \int_1^\infty f(x)dx[/itex] is finite. <br /> <br /> Try f(x)= 1/x. The anti-derivative of 1/x is ln(x) which is not defined at \infty but the anti-derivative of 1/x^2 is -1/x which is defined at both 1 and infinity.

 

[J$C]

So f(x)=1/x would fall into the L2 but not L1 category. I'm interested to see if there are functions in L1 but not in L2, or if L1 is a subset of L2.

 

[D H]

The answer is yes. I don't know if this is homework, so I won't tell you a whole lot yet.

Suppose the domain is infinite. The function is certainly not in either L1 or L2 if |f(x)| does not asymptote to zero as x\to\infty. That |f(x)|^2 < |f(x)|\;\forall x:|f(x)|< 1 should tell you that the behavior as x\to\infty is not a factor in finding a function that is L1 but not L2.

You need to look for a non-negative function that is integrable across (or starting at) a singularity but whose square is not.

 

[maze]

Squaring a number greater than 1 makes it larger, whereas squaring a number less than 1 makes it smaller. Thus squaring a function makes the tail smaller, but makes singularities larger.

The basic idea is to construct a function with a singularity that is just barely integrable, but is no longer integrable when you square it.

For example, integrate 1/sqrt(x) from 0 to 1, and you get 2, whereas integrate 1/x from 0 to 1 and you get infinity.

Edit: Here is a picture:
http://img54.imageshack.us/img54/3531/integrablesingularityaf6.png

 

[maze]

D H's point about behavior at infinity actually got me thinking a little bit, and I think you actually can find an example that depends on the behavior at infinity. Construct a series of boxes that get thinner and taller so that the L2 norm is the harmonic series and the L1 norm is the Basel series:

Here is an image:
http://img410.imageshack.us/img410/1378/basell2l1db3.png

 

[D H]

Squaring a number greater than 1 makes it larger, whereas squaring a number less than 1 makes it smaller. Thus squaring a function makes the tail smaller, but makes singularities larger.

That's the key.

However, you shouldn't give out complete answers, at least not until the J$C says the query asked in the original post was not homework.

D H's point about behavior at infinity actually got me thinking a little bit, and I think you actually can find an example that depends on the behavior at infinity.

Very nice, but once again don't give out complete answers.

 

[maze]

Well its nearly impossible to have a worthwhile discussion if you can't actually say what you're talking about. Taking the 2nd example of the boxes, I guarantee you the conversation would go like this:

D H: The behavior at infinity doesn't matter.

Maze: I have an example where the behavior at infinity does matter; it has to do with boxes.

D H: No, I'm pretty sure that won't work.

Maze: No, I'm pretty sure it does work.

...

Sort of a pointless conversation, wouldn't you say?

 

[J$C]

Very interesting discussion.

I assure you it is not a homework question, although it did stem from a lecture. The professor seemed to imply it was a simple concept, which kept me from asking a question (It was also a bit off topic for that particular lecture). Of course with the hypothetical anonymity of the internet I am free to ask all types of questions, whether they be stupid, redundant, or maybe even insightful.

He also claimed that with a finite interval they were equivalent. Now I know norms on finite dimensional spaces are equivalent, but even with a finite interval L1 and L2 are certainly still infinite dimensional. Now, the discussion here suggests that the most accessible example of an L1 function that is not in L2 would involve a singularity, not behavior as x goes off to infinity. This thought seems to contradict my professors claim that L1 and L2 are equivalent on a finite interval, as this is where the singularity I search for in a counter example (L1 but not L2 function) would lie.

Thoughts?

 

[J$C]

Maze, I just looked at your example 1/sqrt(x) from 0 to 1. This of course makes sense.

So to clarify my last post then, here we have an example on a finite interval in L1 but not in L2. I need to add the condition that my professor was referring to a closed interval i.e. L1[a,b]. Perhaps the closed condition would preclude all of the singularity counter examples and in fact L1[a,b]=L2[a,b].

Thanks for this example.

 

[D H]

Perhaps the closed condition would preclude all of the singularity counter examples and in fact L1[a,b]=L2[a,b].

No. Maze's first example is exactly what I had in mind in post #4. In fact, any function of the form f(x) = c/x^a,\, a\in\,[1/2,1) will be a member of L1 but not L2 for any closed domain that includes 0.


Maze's example for an infinite domain is a nice one. One way to represent that function is

\aligned<br /> f(x) &amp;= \sum_{n=1}^{\infty} f_n(x) \\<br /> f_n(x) &amp;= n\,\text{if}\,x\in(n-1/n^3,n),\,0\,\text{otherwise}

which again takes advantage of the fact that if x>1, x²>x.

 

[J$C]

So a function such as 1/sqrt(x) is considered a member of L1[0,1] even though the function is not defined at 0?

And yes, Maze's infinite domain solution is quite slick.

 

[D H]

So a function such as 1/sqrt(x) is considered a member of L1[0,1] even though the function is not defined at 0?

Sure.

L1, L2, etc are defined in terms of integrability, so the function needs to be integrable over the domain -- which means that it must be defined almost everywhere over the domain.

Boundedness is not required (except for L-infinity, of course).

 

[Office_Shredder]

So a function such as 1/sqrt(x) is considered a member of L1[0,1] even though the function is not defined at 0?

And yes, Maze's infinite domain solution is quite slick.

You could just define the function to be 0 at x=0 if you're worried about that.

 

[J$C]

You're right, I was stuck in a continuous mindset I suppose.

Thanks, good discussion.

 

[kilimanjaro]

If a function f is in L^2([a,b]), then it is in L^1([a,b]) by the Cauchy-Schwarz inequality (just consider the L^1 norm of f as being the inner product of |f| and 1). They key point here is that [a,b] has finite measure; the argument wouldn't work on something like the real line, and specific counterexamples have been mentioned.

And if you try to go the other way around, i.e. figure out when L^1 is a subset of L^2, you hit the usual issues when working with functions on R or on [a,b]. Basically, the problem with these spaces is that there are plenty of disjoint sets with arbitrarily small measure. If you were working with L^1(Z) and L^2(Z) (i.e. summable and square-summable series) these counterexamples don't work, and you do get the proper inclusion.

 

[The Genius]

Sure.

L1, L2, etc are defined in terms of integrability, so the function needs to be integrable over the domain -- which means that it must be defined almost everywhere over the domain.

Boundedness is not required (except for L-infinity, of course).

The example of f(x) = x^(-0.5) is not continuous in [0,1] (note that I've included 0)...So, is the function still a part of L1?

I mean is it not required that a function be continuous to be a part of L1?