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?
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!
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.