(New page: Suppose we know the conclusion of problem 8, Problem 8 Let <math>(X,A,mu)</math> be a finite measure space. If f is <math> A </math>measurable, let En = fx 2 X : n 􀀀 1 � jf(x)j < ng:...)
 
 
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Suppose we know the conclusion of problem 8,
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We know that <math>\int_X |f|^r = \int_0^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy</math>.  If <math>\mu(X) = \infty</math>, the statement is trivial, so assume it's finite.
  
Problem 8
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We have
Let <math>(X,A,mu)</math> be a finite measure space. If f is <math> A </math>measurable, let
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En = fx 2 X : n 􀀀 1 � jf(x)j < ng:
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<math>\int_0^\infty ry^{r-1}\mu(|f|>y) dy = \int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1}\mu\left\{|f|>y\right\} dy
Show that f 2 L1 if and only if P1 n=1 n�(En) < 1.
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+\int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy  </math>
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Now <math>\mu(|f|>y)\leq\mu(X) \Rightarrow \int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1}\mu(|f|>y) dy \leq \mu(X)\int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1} dy = \mu(X)^{1-\frac{r}{p}}</math>,
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and by hypothesis, <math>\mu(|f|>y) \leq c_0y^{-p} \Rightarrow \int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy \leq \int_{\mu(X)^{\frac{-1}{p}}}^\infty rc_0y^{-1-p+r} dy = \frac{rc_0}{p-r}\mu(X)^{1-\frac{r}{p}}</math>
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Letting <math>c = 1 + \frac{rc_0}{p-r}</math> finishes the proof.
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-pw

Latest revision as of 11:33, 11 July 2008

We know that $ \int_X |f|^r = \int_0^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy $. If $ \mu(X) = \infty $, the statement is trivial, so assume it's finite.

We have

$ \int_0^\infty ry^{r-1}\mu(|f|>y) dy = \int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1}\mu\left\{|f|>y\right\} dy +\int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy $

Now $ \mu(|f|>y)\leq\mu(X) \Rightarrow \int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1}\mu(|f|>y) dy \leq \mu(X)\int_0^{\mu(X)^{\frac{-1}{p}}} ry^{r-1} dy = \mu(X)^{1-\frac{r}{p}} $,

and by hypothesis, $ \mu(|f|>y) \leq c_0y^{-p} \Rightarrow \int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy \leq \int_{\mu(X)^{\frac{-1}{p}}}^\infty rc_0y^{-1-p+r} dy = \frac{rc_0}{p-r}\mu(X)^{1-\frac{r}{p}} $

Letting $ c = 1 + \frac{rc_0}{p-r} $ finishes the proof.

-pw

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood