Followings are from Wikipedia.
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?
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?
- 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.