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Prove sinx/x in L2 but not in L1

  • Thread starter Yaelcita
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1. Homework Statement
Prove that [tex]f(x)=\sin(\pi x)/(\pi x)[/tex] is in [tex]L^2(R)[/tex] but not in [tex]L^1(R)[/tex]
This is in a chapter of the book dealing with Inverse Fourier Transform

f is in [tex]L^1[/tex] if [tex]\int|f|<\infty[/tex]
f is in [tex]L^2[/tex] if [tex]\sqrt{\int|f|^2}<\infty[/tex]

2. Homework Equations

I just have no idea how to do it

3. The Attempt at a Solution

Because the question was in a chapter on Fourier Transform, and many of the theorems there only work for L1 functions, I thought maybe I could prove the conclusion doesn't hold for [tex]\sin(\pi x) /(\pi x)[/tex], so it's not in L1. But so far I haven't been very successful, because every integral I come across is extremely complicated and I have no idea how to solve it. Also, I have nothing to prove it is in L2.

So now I've turned to trying to prove it directly, from the definition of L1 and L2, but I'm still getting nowhere. I just don't know how to solve the integrals.
 

Dick

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It's easier just to show the integrals exist or don't exist by a comparison test with something you can integrate. For the first one I suggest an infinite step function.
 
By the first one you mean the L1 one or the L2? And what do you mean by infinite step function? f=1 for all x in R?

I'm not sure how this would work, because that integral would be infinite, but |sin x/x| is less than 1, so I can't conclude its integral is infinite too. And if you mean to use it to prove the L2 integral is finite, well, I need something finite to compare it to.

I've been thinking of using that method, but I just can't seem to come up with anything useful to compare it to. Any suggestions?
 
Consider the integral of |sin(pi x)|/x from some large N to infinity. You want to prove that this "tail" will diverge. You can write this as a summation over integers n from N to infinity of f(n), where:

f(n) = Integral of |sin(pi x)|/x dx from x = n to n+1

You can find a lower bound for f(n) by replacing 1/x in this integral by 1/(n+1). The integral is easily computed. So, you then find that the tail of the summation is larger than some constant times the tail of the Harmonic series, and the result then follows.

The L^2 problem is easier, as you can then replace sin^2 in the integral by 1 to obtain an upper bound for the tail which converges.
 

Dick

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Compare |f(x)|^2 with 1/x^2 for large x. For small x, you know it's bounded. For |sin(pi*x)/(pi*x)|, you know that is the numerator is equal to 1 for x=pi(n+1)/2, right? That means you should be able to show that |sin(pi*x)| is greater than 1/2 for a small interval of fixed size d around every point x=pi(n+1)/2. Use that to define an infinite number of step functions centered on each point x=pi(n+1)/2, the sum of whose areas diverges as you sum n from 0 to infinity. Since the contribution of each is approximately proportional to 1/n for large n.
 
Ok, I'll try that for L1, but now my problem is that Mathematica doesn't agree with me! If I do the sin^2 integral in Mathematica, it says the answer is 1, but if sin^2 is less than 1, then the integral is less than the integral of 1/x^2, which is zero between - and + infinity. So my whole integral is zero (cause the lower bound is zero too, since sin^2>0). What am I dong wrong?
 

Dick

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The integral of 1/x^2 between -infinity and +infinity is divergent. It diverges near x=0. If Mathematica is telling you it's 0, then it's lying. You have to apply another approximation near x=0.
 
Ok, there's somethign I'm missing here. You say construct little step functions that are 1/2 where sin x is greater than 1/2 and zero where it's not, right? Or in a delta neighborhood around the point where it's 1, which is the same. So the area under the step functions is infinity because each one is delta/2 and there are infinitely many of them (please tell me if I understood this correctly)

Now my problem is that I also need to do something about x, because it's not obvious (to me, at least) that a sum of little bits of 1/2x is divergent as well. If x were smaller than the 1/2, then I'd be set, but x is in a neighborhood of (n+1)/2>1/2, so this is not the case.

I'm sorry if it seems I'm asking every little step, but I've been working on this for the last 8 hours and my brain is quite tired :)
 

Dick

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The sum of 'little bits' of 1/x is divergent. Because 1/x is divergent. Without trying to finesse this at all, I know if |x-pi*(n+1)/2|<1/100, that |sin(x)|>1/2. That means that the integral of |f(x)| is greater than the sum over all n of (2/100)*(1/2)/(pi*(n+1)/2+d). You don't even have to make an accurate estimate to show it diverges.
 
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