Real Analysis , Folland Problem 6.1.5

Question

Problem 6.1.5 - Suppose $0 < p < q < \infty$. Then $L^p \not\subset L^q$ if and only if $X$ contains sets of arbitrary small positive measure, and $L^q\not\subset L^p$ if and only if $X$ contains sets of arbitrarily large finite measure. (For the "if" implication: In the first case there is a disjoint sequence $\{E_n\}$ with $0 < \mu(E_n) < 2^{-n}$, and in the second case there is a disjoint sequence $\{E_n\}$ with $1\leq \mu(E_n) < \infty$. Consider $f = \sum a_n 1_{E_n}$ for a suitable constants $a_n$.) What about the case $q = \infty$?

Normally I do not just post a question without at least an attempted proof but I am completely lost on where to start

Answer 1

Take $f \in L^p\setminus L^q$ And define $E_n=\{x \in X: |f(x)|\geq n\}$ To show that $E_n$ has positive measure use Minkowski and the fact that $||f \chi_{E^c_n}||_q\leq ||f \chi_{E^c_n}||_p^{p/q} || f \chi_{E^c_n}||_{\infty}^{1-p/q}$ Showing that $\int|f|^{1/q}*\chi_{E_n} d\mu =\infty$, and consequently $E_n$ Has measure diferent then 0, ie. Strictly positive.

For the other implication take $f=\sum \mu_{E_n}^{-1/q}*\chi_{E_n}$ and show that f belongs to $L_p$ But not to $L_q$.

The second part is a bit diferent, but similar with $E_n=\{x \in X: |f(x)|\in (1/(2^{n+1}),1/2^n]\}.$

And $f$ the same as before with q instead of p where $E_n$ a sequence (disjoint) with $2^n<\mu(E_n)<\infty$.

As for the third part, asking about $q=\infty$ I would like to know myself, so if you find out, can you tell me?

I hope it helped.

Answer 2

To construct functions that negate these statements, have in mind $\{x_n\} \subset R$ such that one of the $\Sigma x_n^p$ and $\Sigma x_n^q$ converges but not the other.

On sequences of subsets with the properties assumed above, define $f$ piecewise. You will still have a measurable and quite well-behaved function, and you must be able to find bounds (lower/upper) for its $p$'th or $q$'th powers.

And the way you define $f$ is to make sure that the value of $f$ times the measure of the set where it is defined to have that value becomes the $x_n$ from 1st paragraph.