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?


  • $\begingroup$ 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
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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 $L^1(\mathbf{R})$.

  • $\begingroup$ 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

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