Finite measure and Lebesgue spaces

[salvozungri]

Let $mu$ be a finite measure on $(X, A)$, where $X$ is a non-empty set, $A$ a $sigma$-algebra of subset of $X$.
Prove that if $1<= r $||f||_{r}<= ||f||_s \mu(X)^(1/r-1/s)$

Hope you enjoy it. .

[size=75]ps: this is my first time in this section, My knowledge of English is really poor, that is why I want and i must improve myself [/size]

[fu^2]

$|E|_{mu}$ maybe is only my notation because normally i use $|E|$ to write the lebesgue's misure of $E$. Therefore if $mu$ is another misure, $|E|_{mu}=mu(E)$ (standard notation i think), i.e. the misure of $E$ under $mu$.

For example if $mu$ is a probability misure and $X=RR$, $A=B(RR)$, we have that $mu(RR)=1$ or $|RR|_{mu}=1$.

in your case you have a finite misure on $X$ (i.e. $mu(X)<+oo$), then the proof is general! if the proof is correct.

[salvozungri]

What do you mean with $|E|_\mu$? I never see that notation... Sorry

[fu^2]

"Mathematico":Let $mu$ be a finite measure on $(X, A)$, where $X$ is a non-empty set, $A$ a $sigma$-algebra of subset of $X$.
Prove that if $1<= p_1

i want write another proof of this theorem that not use the Holder's inequality, but i need that $|E|_{mu}<+oo$, where $E\sub X$. (i'll use $p_1,p_2$ instead of $s,r$ because is more simple for my mind remember the larger number in the first inequality).
I show that if $p_2>p_1$, $L_{mu}^{p_2}(E)\sub L_{mu}^{p_1}(E)$, where $E\sub X$, $|E|=|E|_{mu}<+oo$.

for all $f\in L_{mu}(E)$ we can write

$int_E f=int_{E_1}f+int_{E_2}f$, where $E_1={|f|<=1}$,$E_2={|f|>1}$.

Then $int_E f<=|E_1|+int_{E_2}f$.

we use this inequality proving the sentence, in fact if $f>1=>f^{p_2}>f^{p_1}$, therefore $int_E f^{p_1}<=|E_1|+int_{E_2} f^{p_1}<=|E_1|+int_{E_2} f^{p_2}$, so if $f\in L^{p_2}$ we have that $|E_1|+int_{E_2} f^{p_2}<+oo $, that is $f\in L^{p_1}$ and the theorem follows.

is ok?

[salvozungri]

Sorry ViciousGoblin , i hope you can forgive me for this . By the way, you are right .

Here my attempt.

Let $\alpha= r/s \in (0,1) => r= \alpha s$. Now let $\alpha$ be $\alpha=1/p$ while $1-\alpha= 1/q$ so we have that $1/p+1/q=1$, with $p>=1$.
Since $f\in L_s(\mu)$ by hp, then $|f|^r\in L_p(\mu)$ in fact:
$|f|^(rp)= |f|^(\alpha s p)= |f|^s\in L_1(\mu)$. $1\in L_q(\mu)$ because $\mu(X)<+\infty$, by Holder inequality, we have that $|f|^r*1\in L_1(\mu)$ and:
$|| |f|^r*1||_1<=|| |f|^r||_p*||1||_q=>$
$=> ||f||_r^r <= (\int_X |f|^(rp) d\mu)^(1/p) \mu(X)^(1/q)$, but $rp= s$ so:
$||f||_r^r <= [(\int_X |f|^(s) d\mu)^(1/s)]^(s/p) \mu(X)^(1/q)= ||f||_s^(s/p)\mu(X)^(1/q)= ||f||_s^r\mu(X)^(1-\alpha)$, taking the rth-root on both sides:
$||f||_r<= ||f||_s \mu(X)^(1/r-1/s)$. QED

[ViciousGoblin]

"Mathematico":[quote="ViciousGoblin"][...]Of course you can repeat the argument with $r So, whenerver $r
That's it.

congratulations, it's a perfect analysis. Now it remains to show the inequality .[/quote]

How boring ...

Let $1\leq r
$\int_0^1|f(x)|^r dx=\int_0^1|f(x)|^r\cdot 1dx\leq(|f(x)|^{rp}dx)^{1/p}(\int_0^1 1^q dx)^{1/q}=(\int_0^1 |f(x)|^{s} dx)^{r/s}$

Taking the power $1/r$ of both sides

$||f||_{L^r}\leq ||f||_{l^s}$.

Is there anything missing ?

[salvozungri]

"ViciousGoblin":[...]Of course you can repeat the argument with $r So, whenerver $r
That's it.

congratulations, it's a perfect analysis. Now it remains to show the inequality .

[fu^2]

"Mathematico":[quote="fu^2"]sorry, but in my opinion $L^r(\mu)\subset L^s(\mu)$, if $s>r>=1$, by Holder's inequality.

Can you prove this statement? I would like to see your attempt. .

Edit 1: What do you think about the function $f(x)= 1/\sqrt(x)$? $f\in L^1(0,1)$ but $f\notin L^2(0,1)$[/quote]

sorry... this is the result when i write before thinking.

[ViciousGoblin]

"Mathematico":[quote="fu^2"]sorry, but in my opinion $L^r(\mu)\subset L^s(\mu)$, if $s>r>=1$, by Holder's inequality.

Can you prove this statement? I would like to see your attempt. .

Edit 1: What do you think about the function $f(x)= 1/\sqrt(x)$? $f\in L^1(0,1)$ but $f\notin L^2(0,1)$[/quote]

The example just confirms that $L^1(0,1)$ is bigger than $L^2(0,1)$ - to prove the inclusion notice that

$\int_0^1|f(x)| dx=\int_0^1|f(x)|\cdot 1 dx\leq(|f(x)|^2 dx)^{1/2}(\int_0^1 1^2 dx)^{1/2}=(\int_0^1 |f(x)|^2 dx)^{1/2}$
(by Hoelder's inequality). Of course you can repeat the argument with $r So, whenerver $r
That's it.

[salvozungri]

"fu^2":sorry, but in my opinion $L^r(\mu)\subset L^s(\mu)$, if $s>r>=1$, by Holder's inequality.

Can you prove this statement? I would like to see your attempt. .

Edit 1: What do you think about the function $f(x)= 1/\sqrt(x)$? $f\in L^1(0,1)$ but $f\notin L^2(0,1)$

[fu^2]

sorry, but in my opinion $L^r(\mu)\subset L^s(\mu)$, if $s>r>=1$, by Holder's inequality.