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I'm reading the following proof:

Suppose $\mu(X) < \infty$ and $0 < p< q \leq \infty$. If $q=\infty$ then $L^{q}(\mu) \subseteq L^{p}(\mu)$.

Proof: If $q=\infty$ then $\int_X |f|^{p}d\mu \leq (||f||_{\infty})^{p} \int_X 1d\mu$.

Why we have the above inequality??

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    What ways do you know of bounding an integral such as $\int|f|^p$?2010-11-10
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    Certainly we always have $ |f| \leq ||f||_{\sup}$. But here the $||.||_{\infty}$ denotes the seminorm of essentially bounded functions not the usual supremum norm. So that's why I'm confused.2010-11-10
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    @user: The first thing we prove after define the norm $\| \cdot \|\_{\infty}$ is that $|f(x)| \le \| f \|_{\infty}$ almost everywhere. With that at hand we prove that $\| f \|\_{\infty} = 0 $ if and only if $f(x) = 0$ almost everywhere.2010-11-10
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    Isn't this inequality true for all p and q?2010-11-10
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    @user1736: consider X = (0,1) with \mu a standard Lebesgue measure on X. Then x^{-1/2} is certainly in L^1 but not in L^2.2010-11-10
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    @user1736: I just realized that I provided counter-example for the reversed relation. If I am reading it correctly this time, then yes: the relation holds, assuming 1 \leq p < q.2010-11-10
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    @user1736: Do you mean the inclusion $L^q(\mu) \subseteq L^p(\mu)$ if $1 \le p < q \le \infty $? If $\mu$ is a finite measure then it works for all p and q, but to prove the inclusion with $q \ne \infty$ we need Hölder's inequality. If $\mu$ isn't a finite measure you can find functions $f \in L^q(\mu)$ and $f \not \in L^p(\mu)$. You can probably find an example by yourself considere $\mu$ the Lebesgue measure.2010-11-10

2 Answers 2

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HINT: Note that $-||f|\|_{\infty} \leq f(x)\leq ||f||_{\infty}$ almost everywhere. Hence...

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Let's just write $\|f\|$ for the infinity norm. We have that $\|f\|$ is defined as the essential sup of the values that $f$ takes. But then $|f(x)|^p\le\|f\| ^p$ for any $x\in X$ except in a set of measure 0.

Hence, what you have is $\int|f|^p\le \int\|f\| ^p$, because the set where the inequality between $|f|$ and $\|f\|$ is reversed is negligible. And, since $\|f\| ^p$ is a constant, we can move it out of the integral.

(Note $\|f\|$ could very well be infinite, but in that case the inequality you need is obvious, because anything is $\le\infty$. So, only the case where $\|f\| <\infty$ needs arguing.)


Ok, I was using the wrong norm. You have $\|f\|$ defined as the infimum of $\|h\|_ {\rm sup}$ where $h$ ranges over all (measurable) functions that agree with $f$ a.e.

Fix any such $h$. Then $|f|=|h|$ a.e., so $|f|^p=|h|^p\le \|h\|^ p_ {\rm sup}$ a.e. So $\int |f|^p=\int |h|^p\le \|h\|^p_ {\rm sup}\int 1$, as above.

Now take the infimum of the right hand side.

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    That's precisely my question. I know I'm blind but why $|f| \leq ||f||^{p}$. My definition of $||f||_{\infty}$ is the infimum of the set $\{||h||_{\sup} : \textrm{h is bounded, measurable and h=f a.e}\}$.2010-11-10
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    @user: How do you define $\|h\|_ {\rm sup}$?2010-11-10
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    As sup $\{|h(x)| : x \in X\}$.2010-11-10
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    @user: So that means that $|f|\leq||f||\_{\infty}$ a.e.!2010-11-10