6
$\begingroup$

Let $(X,M, μ)$ be a measure space and $0 < p < q ≤ ∞$. Prove: $L^p(X)$ is not contained in $L^q(X)$ iff $X$ contains sets of arbitrarily small positive measure.

My work:

I proved the forward direction. For backward direction, considering $q<\infty$, if I take disjoint sets $E_n\subset X$ such that $0<\mu(E_n)<\frac{1}{2^n}$, then for $f=\sum_{n=1}^\infty \mu(E_n)^{(-1/q)}\chi_{E_n}$, I can show that $f\in L^p$ and $f\notin L^q$.

Now for the case $q=\infty$ how can I find a function $g$ so that $g\in L^p$ and $g\notin L^\infty$. Can anybody please give me a hint?

EDIT: Does $g=\sum_{n=1}^\infty \chi_{E_n}$ will work? I can prove that $g\in L^p$. But struggling to prove that $g\notin L^\infty$. Any ideas?

$\endgroup$
  • 2
    $\begingroup$ If your $E_n$ are disjoint, then $g$ will be bounded. You need $\sum c_n \chi_{E_n}$ with $c_n \to \infty$ such that $\sum c_n^p \mu(E_n) < \infty$. You know $\mu(E_n) < 2^{-n}$, so e.g. $c_n^p = 2^{n/2}$ will do. $\endgroup$ – Daniel Fischer May 2 '15 at 16:02
  • $\begingroup$ @DanielFischer: But I think we cannot prove that this function is in $L^p$ $\endgroup$ – Extremal May 2 '15 at 16:08
  • $\begingroup$ It was constructed to be in $L^p$. That's what the finiteness of the sum says (with measurability considered obvious). $\endgroup$ – Daniel Fischer May 2 '15 at 16:15
  • $\begingroup$ But $\sum 2^{n/2}\mu(E_n)<\sum{\sqrt{2}}$. We know that $\sum{\sqrt{2}}$ diverges. Does it follow that $\sum 2^{n/2}\mu(E_n)$ converges? $\endgroup$ – Extremal May 2 '15 at 16:18
  • 2
    $\begingroup$ You have $$2^{n/2}\mu(E_n) < 2^{n/2} 2^{-n} = 2^{-n/2} = \biggl(\frac{1}{\sqrt{2}}\biggr)^n.$$ $\endgroup$ – Daniel Fischer May 2 '15 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.