1
$\begingroup$

Consider the $L^p$ spaces. is $\|\cdot\|_p\leq \|\cdot\|_{p'}$ for $p<p'$? is it true if the domain of $L^p$ is finite measure?

Thanks

$\endgroup$
  • $\begingroup$ no, one has a constant that is not equal to 1 (in general). but it is true if the domain has finite measure $\endgroup$ – Mister Benjamin Dover Jan 14 '15 at 21:56
  • $\begingroup$ So there is $C>0$ such that $\|\cdot\|_p\leq\|\cdot\|_{p'}$? Is it true generally or only in finite measure? also, why is it true generally? $\endgroup$ – JackTheRunner Jan 14 '15 at 21:57
  • $\begingroup$ in the answer I'll just post the correct result $\endgroup$ – Mister Benjamin Dover Jan 14 '15 at 21:57
  • $\begingroup$ Welcome to Math.SE! To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post. $\endgroup$ – apnorton Jan 15 '15 at 0:30
1
$\begingroup$

If $\mu(\Omega)<\infty$ then it follows from Hölder's inequality that $||f||_{L^p} \leq C||f||_{L^{p'}}$ whenever $f\in L^{p'}$, where $C$ depends only on $\mu(\Omega)$, $p$, and $p'$. The result is false if the measure is not finite: consider $\mathbf{R}$, it is easy to find an $L^2(\mathbf{R})$ function that does not lie in $L^1(\mathbf{R})$.

$\endgroup$
  • $\begingroup$ Thanks. Why does it follows from Holder's? isn't holder only relevant for $1/p+1/q=1$? and every then, it only gives estimation of $\|\cdot\|_1$ $\endgroup$ – JackTheRunner Jan 14 '15 at 22:02
  • $\begingroup$ @JackTheRunner: there is a more general version of Hölder which is used to prove this. It says if $1/p+1/q\leqslant 1$ and $1/r=1/p+1/q$ and $f\in L^p$, $g\in L^q$ then $fg\in L^r$. $\endgroup$ – Mister Benjamin Dover Jan 14 '15 at 22:03
  • $\begingroup$ Could you please link me to it? Can't seem to find it on the internet $\endgroup$ – JackTheRunner Jan 14 '15 at 22:05
  • $\begingroup$ @JackTheRunner: math.stackexchange.com/questions/159887/… $\endgroup$ – Mister Benjamin Dover Jan 14 '15 at 22:07
  • $\begingroup$ Thanks. one last question - is there a closed form of a possible $C$ in terms of $\mu(\Omega),p,p'$? Is there anything special that can't be said if $\mu$ is a probability measure? $\endgroup$ – JackTheRunner Jan 14 '15 at 22:20

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.