Difference between revisions of "Jets7.3 Old Kiwi" - Rhea

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

@@ Line 1: / Line 1: @@
-Suppose we know the conclusion of problem 8,
+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
+We have
-Let <math>X</math> be a finite measure space. If <math>f</math>  is measurable, let
-<math>E_n = \{x \in X : n-1 \leq |f(x)| < n \}</math>. Then
+<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
++\int_{\mu(X)^{\frac{-1}{p}}}^\infty ry^{r-1}\mu\left\{|f|>y\right\} dy  </math>
-<math>f \in L^1(X)</math> if and only if <math>\sum_{n=1}^{\infty}nm(E_n) < \infty.</math>
+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>,
+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>
-.<math>(\Rightarrow)</math> First we apply Tchebyshev to <math>E_n</math> and find that
+Letting <math>c = 1 + \frac{rc_0}{p-r}</math> finishes the proof.
-<math> (n-1)\left|\{x \in X \mid |f(x)| \geq n-1 \}\right| \leq \int_{E_n}|f|</math>
+-pw
-or rather
-<math>(n-1) m(E_n) \leq \int_{E_n}|f|</math>
-Since we have that <math>m(E_n)</math> is finite we can move it to the other side of the inequality.
-<math>nm(E_n) \leq \int_{E_n}|f| + m(E_n)</math>
-Since this is true for all <math>n</math> we take sums on both sides and note that the <math>E_n</math> are disjoint.
-<math>\sum_{n=1}^{\infty}nm(E_n) \leq \sum_{n=1}^{\infty}\int_{E_n}|f| + \sum_{n=1}^{\infty}m(E_n)</math>
-or
-<math>\sum_{n=1}^{\infty}nm(E_n)\leq \int_{X}|f| + m(X)</math>
-And we are in a finite measure space so <math>m(X) < \infty</math> and since <math>f \in L^1</math> we have <math>\int_{X}|f| < \infty</math>.
-Thus we have that  <math>\sum_{n=1}^{\infty}nm(E_n) < \infty</math>.
-<math>(\Leftarrow)</math> Since <math>|f|< n</math> in each <math>E_n</math> we have that
-<math>\int_X|f| = \sum_{n=1}^{\infty}\left(\int_{E_n}|f|\right) \leq \sum_{n=1}^{\infty}\left( n m(E_n)\right)< \infty</math>
-In other words, <math> f \in L^1</math>.

Difference between revisions of "Jets7.3 Old Kiwi" - Rhea

Latest revision as of 11:33, 11 July 2008

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