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Interpolation of L^p Spaces

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Homework Statement



If [itex]0 < \alpha < n[/itex], define an operator [itex]T_{\alpha}[/itex] on function on [itex]\mathbb{R}^n[/itex] by

[tex]T_{\alpha}f(x) = \int |x-y|^{-\alpha}f(y)dy[/tex]

Then prove that [itex]T_{\alpha}[/itex] is weak type [itex](1,(n-\alpha )^{-1})[/itex] and strong type (p,r) with respect to Lebesgue measure on [itex]\mathbb{R}^n[/itex], where [itex]1 < p < n\alpha ^{-1}[/itex] and [itex]r^{-1} = p^{-1} -\alpha n^{-1}[/itex].

Homework Equations



Let X be a set, [itex]\mu[/itex] a measure on this set. For [itex]0 < q < \infty[/itex] define the Lq norm of a function [itex]g : X \to \mathbb{C}[/itex] with respect to [itex]\mu[/itex] to be:

[tex]||g||_q = \left (\int _X |g|^qd\mu \right )^{1/q}[/tex]

Define the weak Lq norm of such a function to be:

[tex][g]_q = \left [\mbox{sup} _{\beta > 0}(\beta ^q\mu \{ x : |g(x)| > \beta \})\right ]^{1/q}[/tex]

Define the space of functions [itex]\mathbf{L^q(\mu )}[/itex] to be the set of function with finite Lq norm. Define the space of functions [itex]\mathbf{weak\ L^q(\mu )}[/itex] to be those function with finite weak Lq norm.

An operator T is sublinear if |T(f + g)| < |Tf| + |Tg| and |T(cf)| = c|Tf| for every function in the domain of T (which is some vector space of functions). A sublinear operator T is strong type (a,b) if [itex]L^a(\mu )[/itex] is contained in its domain, T maps [itex]L^a(\mu )[/itex] into [itex]L^b(\mu )[/itex], and there exists C > 0 such that [itex]||Tf||_b \leq C||f||_a[/itex] for all f in [itex]L^a(\mu )[/itex]. A sublinear operator T is weak type (a,b) if [itex]L^a(\mu )[/itex] is contained in its domain, T maps [itex]L^a(\mu )[/itex] into [itex]weak\ L^b(\mu )[/itex], and there exists C > 0 such that [itex][Tf]_b \leq C||f||_a[/itex] for all f in [itex]L^a(\mu )[/itex].

As this is real analysis, there are few relevant equations, instead there are inequalities. They include:

Holder's inequality
Minkowski's inequality
Chebyshev's inequality
Minkowski's inequality for integrals
The Riesz-Thorin Interpolation Theorem
The Marcinkiewicz Interpolation Theorem
and a few other propositions and lemmas that I would take too long to write out.

The Attempt at a Solution



I've only started on the "weak type" part of the problem, and I've only gotten as far as writing out what I need to prove in terms of the definitions. Then I guess I have to find one of the inequalities in my book and find some non-obvious way to apply it which ends up giving the right answer, but I have no clue really of what to do. So this is all I have:

I need to find C > 0 such that

[tex]\left [\mbox{sup} _{\beta > 0} \left (\beta ^{(n-\alpha )^{-1}}m\{ x : |\int f(y)|x-y|^{-\alpha }dy| > \beta \}\right ) \right ]^{n-\alpha } \leq C\int |f|[/tex]

I've determined that this is equivalent to proving:

[tex]\left [\mbox{sup} _{\beta > 0} \left (\beta ^{(n-\alpha )^{-1}}m\{ x : \int |f(y)||x-y|^{-\alpha }dy > \beta \}\right ) \right ]^{n-\alpha } \leq C\int |f|[/tex]

but I don't know if that's any use. Help!
 
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Answers and Replies

  • #2
matt grime
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I'm sorry, was there a question there?
 
  • #3
AKG
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Sorry, that's how my book does things. The exercises are just statements, and it's assumed you have to prove them, it never says things like "Prove that ...". I added it in now though.
 
  • #4
matt grime
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We don't have your book, so we have no idea where what the book proves diverges from what the book 'states' in the problems.

Of course, I have no idea what any of the things you're asking about are, so I can't help. (e.g. weak type (X,Y)... not a clue, sorry.)
 
  • #5
AKG
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I want to prove that there exists a constant C > 0 such that if f is a function on Rn such that:

[tex]\int _{\mathbb{R}^n}|f(z)|dz < \infty[/tex]

then

[tex]\left (\mbox{sup} _{\beta > 0} \left [\beta ^{\frac{1}{n-\alpha }}\, m\left ( \left \{ x : \int _{\mathbb{R}^n} |f(y)||x-y|^{-\alpha }dy > \beta \right \}\right )\right ] \right )^{n-\alpha } \leq C\int _{\mathbb{R}^n} |f(z)|dz[/tex]

or, equivalently,

[tex]\left (\mbox{sup} _{\beta > 0} \left [\beta ^{\frac{1}{n-\alpha }}\, m\left ( \left \{ x : \left | \int _{\mathbb{R}^n} f(y)|x-y|^{-\alpha }dy\right | > \beta \right \} \right ) \right ] \right )^{n-\alpha } \leq C\int _{\mathbb{R}^n} |f(z)|dz[/tex]

where m is the Lebesgue measure, and all integrals are Lebesgue integrals.
 
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  • #6
matt grime
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That still doesn't make sense to me.

You have a set {x: ... some integral >b} then that's inside ( ) and multiplied be a some thing, then supped, then raised to a power.
 
  • #7
AKG
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You have m({x:... > b}), which stands for the Lebesgue measure of {x:... > b}. You have the measure of that set, multiplied by b^(1/(n-a)). This product is supped. Finally, the supremum is rasied to n-a.
 
  • #8
StatusX
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The sup of a set being less than or equal to some value is equivalent to every element in the set being less than or equal to that value. So you can remove all the sups as long as its implicit that beta can take on any positive value. This isn't going to solve the problem, but at least it makes it a little neater.

Also, what you have seems to basically be a convolution. Do you have any convolution inequalities that might be helpful?
 
  • #9
AKG
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Let me remove the "sup" StatusX. We had:

[tex]\left (\mbox{sup} _{\beta > 0} \left [\beta ^{\frac{1}{n-\alpha }}\, m\left ( \left \{ x : \int _{\mathbb{R}^n} |f(y)||x-y|^{-\alpha }dy > \beta \right \}\right )\right ] \right )^{n-\alpha } \leq C\int _{\mathbb{R}^n} |f(z)|dz[/tex]

This is:

[tex]\mbox{sup} _{\beta > 0} \left [\beta ^{\frac{1}{n-\alpha }}\, m\left ( \left \{ x : \int _{\mathbb{R}^n} |f(y)||x-y|^{-\alpha }dy > \beta \right \}\right )\right ] \leq \left (C\int _{\mathbb{R}^n} |f(z)|dz\right )^{\frac{1}{n-\alpha }}[/tex]

which means for all positive [itex]\beta[/itex]:

[tex]\beta ^{\frac{1}{n-\alpha }}\, m\left ( \left \{ x : \int _{\mathbb{R}^n} |f(y)||x-y|^{-\alpha }dy > \beta \right \}\right ) \leq \left (C\int _{\mathbb{R}^n} |f(z)|dz\right )^{\frac{1}{n-\alpha }}[/tex]

[tex]\beta m\left ( \left \{ x : \int _{\mathbb{R}^n} |f(y)||x-y|^{-\alpha }dy > \beta \right \}\right )^{n-\alpha } \leq C\int _{\mathbb{R}^n} |f(z)|dz[/tex]

[tex]m\left ( \left \{ x : \int _{\mathbb{R}^n} \frac{|f(y)|}{\beta }|x-y|^{-\alpha }dy > 1 \right \}\right )^{n-\alpha } \leq C\int _{\mathbb{R}^n} \frac{|f(z)|}{\beta }dz[/tex]

Let me post this to see what it looks like.
 
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  • #10
matt grime
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You have m({x:... > b}), which stands for the Lebesgue measure of {x:... > b}.

That was really not very clear from what you wrote. m appeared to just be a constant. You should try to write things more clearly in future.
 
  • #11
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Try Folland's "Real Analysis: Modern Techniques and Their Applications"
 
  • #12
AKG
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That still doesn't make sense to me.

You have a set {x: ... some integral >b} then that's inside ( ) and multiplied be a some thing, then supped, then raised to a power.
The post says quite clearly:

"where m is the Lebesgue measure, and all integrals are Lebesgue integrals."
gammamcc said:
Try Folland's "Real Analysis: Modern Techniques and Their Applications"
Try what exactly? The question I've asked is problem 45 from chapter 6 in Folland's "Real Analysis: Modern Techniques and Their Applications".
 
  • #13
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OK. I get that T_{\alpha} is weak type (1,n(n-\alpha)^{-1}) using the last theorem in the section.
 
  • #14
AKG
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Really? If we let K(x,y) = |x-y|-a, then the only q for which [K(x,.)]q and [K(.,y)]q are bounded for a.e. x and y is q = n/a. So that theorem gives that T is weak type (1, n/a). I don't know how you're getting that it's weak type (1, n/(n-a)) from that theorem. Moreover, I need to prove that T is weak type (1, 1/(n-a)), not (1, n/a) nor (1, n/(n-a)).
 
  • #15
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oh well, what can I say. I'm not wrong.*** for \alpha less than or equal to n/2.
 
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  • #16
matt grime
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The post says quite clearly:

"where m is the Lebesgue measure, and all integrals are Lebesgue integrals.

Sorry. Going blind, perhaps.
 
  • #17
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Moreover, I got the same weak estimate as you did for K, but that is a stronger estimate than is needed to satisfy the theorem. Good Luck.

Hint: How does x/(x-1) compare to x when x>1?

BTW. It's OK Matt, I hope this forum is not always so uptight.
 
  • #18
AKG
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Moreover, I got the same weak estimate as you did for K, but that is a stronger estimate than is needed to satisfy the theorem. Good Luck.
Huh?
Hint: How does x/(x-1) compare to x when x>1?
It depends on whether x < 2, x = 2, or x > 2.
 
  • #19
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yep. say x=n/alpha. when x<2 compare K to function more convenient.
This applies, but I won't give it all away. This was a fun problem, thanks.
 
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  • #20
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I wonder if there is an alternate proof using Calderon-Zygmund decomposition combined with L^p inequalities.
 
  • #22
AKG
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Oh geez, thanks a lot gammamcc.
 

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