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As usual denote $L^p$ the quotient space where two integrable functions are identified if they are equal almost everywhere. So I'm using the definition written here:

http://en.wikipedia.org/wiki/Lp_space

Then we have the following result: for each $p \geq 1$ we have $L^{\infty}(X) \subseteq L^{p}(X)$ where X is a finite measure space and $L^{\infty}$ denotes the set of all essentially bounded functions endowed with the $||f||_{\infty}$ pseudonorm.

So I took $f \in L^{\infty}(X)$ then by definition there is some bounded function $g$ such that $g=f$ a.e. But then $f=g$. So:

$\int |f|^{p} = \int |g|^{p} \leq \int (||g||_{\sup})^{p} < \infty$.

Questions: Is the above correct? Why do we need $p \geq 1$. Why wouldn't $p>0$ work? Is it because we need $p \geq 1$ in the case q is not $\infty$ or where exactly?

Thank you.

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    @user:Technically, $g=f$ a.e. means that *the equivalence classes* of $f$ and $g$ are equal, rather than $f$ and $g$ necessarily being equal. We usually abuse notation in $L^p$ spaces and use the function to denote its class, but here it might be somewhat confusing to say "$f=g$ a.e., so $f=g$".2010-11-10

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If p<1, then the integral $(\int |f|^p d \mu)^{\frac{1}{p}}$ no longer defines a norm. That is why we need $p \geq 1$ typically. However, for the conclusion you want to draw, I guess it will still work if $p \geq 0$.

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    Ah cheers, it is clear now. Is my attempt of proof correct? thanks again.2010-11-10
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    Also I'm trying to show the case when q is not equal to infinity. Would I need Holder in this case?2010-11-10
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    @user: What are you trying to show about $q$?2010-11-10
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    @Jonas: the inclusions $L^{q}(X) \subseteq L^{p}(X)$ for each $p \geq 1$ and $0 < p \leq q \leq \infty$. Not so sure if I did the right thing when $q=\infty$. I still need to work the case q not equal to $\infty$.2010-11-10
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    I am assuming that you want to prove $L^q \subseteq L^p$ whenever $p \leq q$. Your attempt to prove when $q= \infty$ and $p \geq 0$ is right. For the case, when $q < \infty$, you do not need Holder. You would need Jensen's inequality (convex function thing).2010-11-10
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    For $0\lt p\leq q\lt\infty$ you could just break up integral into the sets where $|f|\leq1$ and $|f|\gt1$. You still don't need $p\geq1$.2010-11-10
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    So basically they tend to state $p \geq 1$ to then be able to say $L^{p}$ is a normed space?2010-11-10
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    @user: Yes. For $0\lt p\lt1$, $L^p$ is an F-space with complete translation invariant metric $d(f,g)=\int|f-g|^p$. But $\int|f|^p$ is not a norm, because $\int|\lambda f|^p=|\lambda|^p\int|f|^p$. Scalars would come out nice if you took the $1/p$ power, but then the triangle inequality would fail.2010-11-10
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    Yes. we need $p \geq 1$ just to ensure that it defines a norm. Some results can be proved (like this one for instance) can be proved for $p \geq 0$ as well.2010-11-10
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    Thank you very much to both! you guys really know how to explain. I will try your suggestions.2010-11-10
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    @Arturo: so I assume that if we are rigorous then we shall say assume f = g a.e then [f] = [g]. But then: $\int |[f]|^{p} = \int |[g]|^{p} \leq \int (||[g]||_{\sup})^{p} < \infty$. Is this what you meant?2010-11-10
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    @user: Since Arturo didn't comment on this answer, he may not get your comment to him here. He would get it if you posted the comment with "@Arturo" on your question where he commented.2010-11-10
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    @Jonas: what I wrote (i.e the equivalence classes things) makes sense?2010-11-10
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    @user: $\|[g]\|_\sup$ doesn't really make sense, strictly speaking. Also, using $[f]$ and $[g]$ in the integrals isn't necessary and possibly a little confusing. When it comes down it, we're integrating functions, we just also keep in mind that it doesn't matter which representative of the equivalence class we integrate. The way you wrote the integrals in your question is actually good. Arturo's point is only about saying $f=g$; it is true that they represent the same element of $L^\infty$, but since you're dealing explicitly with representatives it's best not to abuse notation too much.2010-11-10