Cfrgtkky

In a space with measure $1$, $||f||_p$ is an increasing function with respect of $p$. To show that $lim_{p rightarrow infty} ||f||_p=||f||_{infty}$ we have to show that $||f||_{infty}$ is the supremum, right??

To show that, we assume that $||f||_{infty}-epsilon$ is the supremum.

From the essential supremum we have that $m({|f|>||f||_{infty}-epsilon})=0$.

So, we have to show that $m({|f|>||f||_{infty}-epsilon})>0$.

Let $A={|f|>||f||_{infty}-epsilon}$.

We have that $int_A |f|^p leq int |f|^p leq ||f||_{infty}^p$.

$int_A |f|^p >int_A (||f||_{infty}-epsilon)^p=(||f||_{infty}-epsilon)^p m(A)$

So, $m(A)^{1/p} (||f||_{infty}-epsilon)<||f||_p leq ||f||_{infty}$

How could we continue to show that $m(A)>0$??

EDIT:

Is it as followed??

We have that $0<||f||_{infty}-epsilon<||f||_{infty}$ for some $epsilon>0$.

$||f||_{infty}$ is the essential supremum. So, from the definition we have that $mleft ( {|f(x)>||f||_{infty}-epsilon} right )>0$.

Let $A={|f|>||f||_{infty}-epsilon}$.

We have that $int_A |f|^p leq int |f|^p leq ||f||_{infty}^p$.

$int_A |f|^p >int_A (||f||_{infty}-epsilon)^p=(||f||_{infty}-epsilon)^p m(A)$

So, $m(A)^{1/p} (||f||_{infty}-epsilon)<||f||_p leq ||f||_{infty}$

Taking the limit $p rightarrow +infty$ we have the following:

$$lim_{p rightarrow +infty}m(A)^{1/p} (||f||_{infty}-epsilon)<lim_{p rightarrow +infty} ||f||_p overset{ m(A)>0 Rightarrow lim_{p rightarrow +infty}m(A)^{1/p}=1}{Longrightarrow} \ ||f||_{infty}-epsilon<lim_{p rightarrow +infty} ||f||_p$$

Is this correct?? How do we conclude that $lim_{p rightarrow +infty} ||f||_p=||f||_{infty}$ ??

First of all you have to show that $||f||_infty$ is a superior bound. But this is immediate by the definition of $| cdot|_infty$.

Now you want to show that, for all $epsilon$ we have $$lim_{p rightarrow infty} |f|_p > ||f||_infty - epsilon$$

As already observed and using your notation, $m(A) = m(|f| > ||f||_infty -epsilon)>0$. Now putting the things together you got

$$|f|_p > (||f||_infty- epsilon)m(A)^{1/p}$$ for all values of $p$. Now just take the limit, which exists since the sequence is increasing and bounded, to obtain what you desire.

The final step follows from the definition of the essential supremum.

Proposition

If $m({|f| > C}) = 0$ then necessarily $C ge |f|_infty$.

Proof

Let $N := {|f|>C}$. By assumption $m(N) = 0$. This means
$$|f|_infty = inf_{m(N') = 0} sup_{xnotin N'} |f(x)| le sup_{xnotin N} |f(x)| = sup_{{|f| > C}} |f(x)| le C$$
as was to be shown.

Now use this proposition together with your initial steps to conclude a contradiction for $C = |f|_infty - epsilon < |f|_infty$.