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Given a measure space $(\Omega,\mathcal{F},\upsilon)$ and a $p>0$. What does the following mean? $$ \|f\|_p=(\upsilon|f|^p)^{1/p}$$

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I think they're trying to say:

$$ \|f\|_p=\left(\int_\Omega \vert f\vert^p dv\right)^{1/p} $$ One way to justify their shorthand is that integration is like "measuring" the function, so $v\vert f\vert^p$ could be thought of as shorthand for $\int_\Omega\vert f\vert^pdv$.

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  • $\begingroup$ Is $|f|^p$ the $p^{th}$ power of the absolute value of the function $f$ ? $\endgroup$ – N Zhang Apr 8 '13 at 18:04
  • $\begingroup$ Yes, that's correct. $\endgroup$ – icurays1 Apr 8 '13 at 18:07
  • $\begingroup$ so $|f|^p$ is another function, say, $g$, and $\upsilon|f|^p$ is the integral of $g$ with respect to $\upsilon$ ? $\endgroup$ – N Zhang Apr 8 '13 at 18:10
  • $\begingroup$ Correct, with the integration being over $\Omega$. $\endgroup$ – icurays1 Apr 8 '13 at 18:12
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Indeed, for every measure $v$ and every integrable function $g$, one can use indifferently $$ v(g),\qquad\int g\mathrm dv,\qquad\int_\Omega g(\omega)\mathrm dv(\omega), $$ and a few other combinations of the above to denote the integral of the function $g$ with respect to the measure $v$. In particular, $$ \|f\|_p=\left(\int_\Omega |f(\omega)|^p\mathrm dv(\omega)\right)^{1/p}=(v(|f|^p))^{1/p}. $$

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