Does the limit of $||f||_p$ equal $||f||_{\infty}$ as $p$ approaches infinity?

[mathmari]

Hey!

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

Taking the limit $p \rightarrow +\infty$ we have the following:$$\lim_{p \rightarrow +\infty}m(A)^{1/p} (||f||_{\infty}-\epsilon)<\lim_{p \rightarrow +\infty} ||f||_p $$

Since we have supposed that $||f||_{\infty}-\epsilon$ is the supremum, is the limit of $||f||_p$ equal to $||f||_{\infty}-\epsilon$?? (Wondering)

So, is it as followed??

$$\lim_{p \rightarrow +\infty}m(A)^{1/p} (||f||_{\infty}-\epsilon)<||f||_{\infty}-\epsilon $$

(Wondering)

How can we get a contradiction?? (Wondering)

 

[Opalg]

Hey!

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

Taking the limit $p \rightarrow +\infty$ we have the following:$$\lim_{p \rightarrow +\infty}m(A)^{1/p} (||f||_{\infty}-\epsilon)<\lim_{p \rightarrow +\infty} ||f||_p $$

Since we have supposed that $||f||_{\infty}-\epsilon$ is the supremum, is the limit of $||f||_p$ equal to $||f||_{\infty}-\epsilon$?? (Wondering)

So, is it as followed??

$$\lim_{p \rightarrow +\infty}m(A)^{1/p} (||f||_{\infty}-\epsilon)<||f||_{\infty}-\epsilon $$

(Wondering)

How can we get a contradiction?? (Wondering)

Your proof is going in the right direction. But instead of treating it as a proof by contradiction, just choose $\epsilon>0$. It follows from the definiton of $\|f\|_\infty$ as an essential supremum that your set $A$ must have positive measure: $m(A) > 0.$

You have shown that $\|f\|_p^p = \int_A |f|^p \geqslant\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$. When you take the $p$th root of that, you get $\|f\|_p \geqslant m(A)^{1/p}(||f||_{\infty}-\epsilon).$ By choosing $p$ large enough, you can get $m(A)^{1/p}$ as close as you like to $1$. Since $\epsilon$ could be made arbitrarily small it follows that by taking $p$ large enough you can get $\|f\|_p$ as close as you like to $\|f\|_\infty.$

 

[mathmari]

Your proof is going in the right direction. But instead of treating it as a proof by contradiction, just choose $\epsilon>0$. It follows from the definiton of $\|f\|_\infty$ as an essential supremum that your set $A$ must have positive measure: $m(A) > 0.$

Haven't we supposed that $||f||_{\infty}-\epsilon$ is the supremum?? (Wondering)

Why does it follow from the definiton of $\|f\|_\infty$ as an essential supremum that the set $A$ must have positive measure: $m(A) > 0.$ ?? (Wondering)

 

[Opalg]

Haven't we supposed that $||f||_{\infty}-\epsilon$ is the supremum?? (Wondering)

Why does it follow from the definiton of $\|f\|_\infty$ as an essential supremum that the set $A$ must have positive measure: $m(A) > 0.$ ?? (Wondering)

I think you need to look a bit more closely at the definition of $\|\,f\|_\infty$. Remember that when dealing with measurable functions we ignore anything that happens on a null set. So $\|\,f\|_\infty$ is not the least upper bound of $|\,f|$, but the least essential upper bound. This means that $|\,f|$ can take values larger than $\|\,f\|_\infty$, but only on a null set. However, if you take any number less than $\|\,f\|_\infty$ (for example $\|\,f\|_\infty - \epsilon$) then it is not an essential upper bound. This means that the set of points where $|\,f|$ takes values greater than $\|\,f\|_\infty - \epsilon$ is not a null set and must therefore have measure greater than $0$. That is the reason why $m(A)>0$.

 

[mathmari]

So, 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 $m\left ( \{|f(x)>||f||_{\infty}-\epsilon\} \right )>0$.

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:

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

So, we have that $$||f||_{\infty}-\epsilon<\lim_{p \rightarrow +\infty} ||f||_p \leq ||f||_{\infty}$$

Since $\epsilon$ is arbitrarily, it follows that $$\lim_{p \rightarrow +\infty} ||f||_p = ||f||_{\infty}$$

Is this correct?? (Wondering)

 

[Opalg]

Yes, that looks good. (Sun)

 

[mathmari]

Yes, that looks good. (Sun)

Great! Thank you very much! (Yes)