2
$\begingroup$

Let $f:[0,1]\rightarrow [0,\infty)$ be in $L^1([0,1])$ such that for every measurable set $E\subset [0,1]$ $$ \int_E f\leq \sqrt[2]{m(E)}$$ where $m$ is the Lebesgue measure on $\mathbb{R}$.

a) Show $f\in L^p([0,1])$ for all $p\in[1,2)$

b) Provide an example to show that a) fails when $p=2$

I've just started learning about $L^p$ spaces and am wondering if part a) requires some more advances knowledge. At this point, I have no idea how to even start part a). I did find that the function $f(x)=\frac{1}{2\sqrt{x}}$ with $f(0):=0$, gives a counterexample for when $p=2$. Help and/or hints for part a) would much appreciated !

$\endgroup$
  • $\begingroup$ Sorry, I forgot to add that $f$ should be nonnegative. I've fixed this. Thanks $\endgroup$ – MAM Aug 10 '16 at 2:51
3
$\begingroup$

Let $S_c$ be the set $S_c= \{x | f(x) > c \}$. Note that $\int_{S_c} f \geq cm(S_c)$. We then have, by the hypothesis, that $\sqrt{m(S_c)} \geq cm(S_c)$, which implies that $m(S_c) \leq \frac{1}{c^2}$. Similarly, note that $\{ x | (f(x))^p > c\} = S_{c^{1/p}}$, hence $m(\{ x | (f(x))^p > c\}) \leq c^{-2/p}$.

Finally, note that $\int_{S_1} f^p = \int_1^\infty m(f^p > c) dc \leq \int_1^\infty c^{-2/p} dc$, which makes sense when $p<2$, but for $p=2$ will give a contradiction, as the function $(4x)^{-0.5}$ shows.

$\endgroup$
  • $\begingroup$ Can you explain the equality $\int_{E_1}f^p=\int_{1}^{\infty} m(f^p>c)dc$? Also should $c^{p/2}$ be $c^{-2/p}$. Thanks $\endgroup$ – MAM Aug 10 '16 at 3:09
  • $\begingroup$ I have edited it. Please verify the proof, it should be straightforward now. $\endgroup$ – астон вілла олоф мэллбэрг Aug 10 '16 at 3:15

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.