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Followings are from Wikipedia.

  1. Why is it that:

    Colloquially, if $1 ≤ p < q ≤ ∞$,

    • $L^p(S, μ)$ contains functions that are more locally singular,
    • while elements of $L^q(S, μ)$ can be more spread out?

    I was also wondering what "locally singular" and "spread out" mean mathematically?

  2. Why is it that:

    Consider the Lebesgue measure on the half line $(0, ∞)$.

    • A continuous function in $L^1$ might blow up near 0 but must decay sufficiently fast toward infinity.
    • On the other hand, continuous functions in $L^∞$ need not decay at all but no blow-up is allowed?

    I was also wondering what "blow up (near 0)" and "decay sufficiently fast (toward infinity)" mean mathematically?

  3. Although it is stated in the following, I don't understand how this "precise technical result" related to the above two quotes?

    The precise technical result is the following:

    • Let $0 ≤ p < q ≤ ∞$. $L^q(S, μ)$ is contained in $L^p(S, μ)$ iff $S$ does not contain sets of arbitrarily large measure, and
    • Let $0 ≤ p < q ≤ ∞$. $L^p(S, μ)$ is contained in $L^q(S, μ)$ iff $S$ does not contain sets of arbitrarily small non-zero measure.
Thanks and regards!
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    I believe this approach is far too abstract. As I suggested in a previous question of yours: do you have some toy examples of functions which belong to $L^1[0,1]$, but not $L^2[0,1]$ or $L^2[0,1]$ but not $L^\infty[0,1]$? Do you have some examples of sequences in $\ell^2$, but not in $\ell^1$ or $\ell^\infty$, but not $\ell^2$? If not, find them! If so, try to understand their features and how they meet the descriptions in your post. I would strongly recommend to answer these explicit questions on the simplest of measure spaces, i.e., $[0,1]$ and $\mathbb{N}$, first.2012-12-28
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    @Martin: Thanks! http://math.stackexchange.com/a/18399/12812012-12-28

1 Answers 1

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I guess it suffices to answer the first question. One observation is

If $|x|>1$, then $|x|^p<|x|^q$. If $|x|<1$, then $|x|^p>|x|^q$.

If $f$ is integrable, then $|f|$ have to decay at infinity (you can take this literally as on $(0,\infty)$, but I think in many other cases this can also be understood). So $|f(x)|<1$ for large $x$. But $|f|^p>|f|^q$ there, so "$f$ is $q$-integrable" puts a weaker restriction on $|f|$ for large $x$ than "$f$ is $p$-integrable".

That is why $\mathcal{L}^q$ functions can be more spread-out than $\mathcal{L}^p$ functions.

Turn this argument around, you see $\mathcal{L}^p$ functions can have more local blow ups than $\mathcal{L}^q$ functions.

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    Thanks, +1! (1) Does "singular" mean the same as "blow-up"? (2) "Turn this argument around, you see $L^p$ functions can have more local blow ups than $L^q$ functions." Is it because that $L^q$ functions may be less bounded than $L^p$ functions at infinity, to even out their integrals over the domain, $L^q$ functions may have less spikes than $L^p$ functions locally?2012-12-28
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    (3) I think in your reply, $p, q \in (0, \infty]$ not just $\in [1, \infty]$?2012-12-28
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    (4) in Part 3, how is the blow-up and spread out of $f$ at infinity and locally related to "S does not contain sets of arbitrarily large measure" and "S does not contain sets of arbitrarily small non-zero measure"?2012-12-28