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Let $f:[0,1]\to\mathbb{R}$ be a bounded (Lebesgue) measurable function.

Consider the function $$w(p)=\int_0^1|f|^p\,d\mu$$.

Is $w(p)$ differentiable at any $0<p<\infty$? I.e. does $w'(p)$ exist for all (not just almost all) $0<p<\infty$?

I hope this is not too easy a question. I have asked on Math.SE (few days ago with bounty), classmates, none of them know how to prove or disprove it. It is kind of a research question since it is not known whether this result is true or false.

Thanks.


My attempt: I can prove it with an additional assumption that $|f|\geq\epsilon$ for some $\epsilon>0$. But this technique does not work for the general case.

First let $E=\{x\in [0,1]: f(x)>0\}$. Then $w(p)=\int_E |f|^p\,d\mu$.

Basically I hope to apply "differentiation under the integral" Theorem 2 in http://planetmath.org/differentiationundertheintegralsign.

We check the conditions:

  1. $|f|^p$ is measurable since $f$ is, it is clearly integrable since it is bounded, and $|E|<\infty$.

  2. $\frac{\partial}{\partial p}|f|^p=|f|^p\ln|f|$ exists since on $E$, $|f|>0$.

  3. $|f|^p\ln|f|$ is also bounded (since $f$ is bounded, and $|f|\geq\epsilon$) so it is dominated by its upper bound, which is integrable over $|E|<\infty$.

So we apply "Differentiation under the integral", $w'(p)=\int_E |f|^p\ln|f|\,d\mu$, which exists again since $|f|^p\ln|f|$ is measurable (composition of $f$ with continuous $\phi=x^p\ln x$), and also bounded.

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    $\begingroup$ Yes, this is true, and the argument you outline does work in general. Since $t^p\log t\to 0$ as $t\to 0+$, your extra assumption is not needed anywhere. $\endgroup$ – Christian Remling Sep 13 '16 at 5:27
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    $\begingroup$ Answered at math.stackexchange.com/questions/1921032/… $\endgroup$ – Robert Israel Sep 13 '16 at 5:30
  • $\begingroup$ @ChristianRemling Very nice, thanks. $\endgroup$ – yoyostein Sep 13 '16 at 5:39
  • $\begingroup$ I just realized a problem: We need $|f(x)|^p\ln|f(x)|\leq g(x)$ for all $p$ in order for the "dominating" to work. I think my previous argument on boundedness of $|f(x)|^p\ln|f(x)|$ does not apply, we need boundedness as $p$ varies not as $x$ varies. $\endgroup$ – yoyostein Sep 16 '16 at 1:48
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Yes, we need the boundedness of $|f|^p \ln|f|$ when both $x$ and $p$ vary, but important note is that $p$ may vary on a given segment $[p_1,p_2]$, $1<p_1<p_2$ (this would imply that the derivative $d/dp(\int)$ exists at all points $p\in (p_1,p_2)$, and since the segment is arbitrary, it exists for all $p>1$.) Now this boundedness is pretty clear.

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In fact, if $f$ is any measurable function on a measure space $(X,\mathcal{S},\mu)$, and wlog $f(x)>0$ on $X$, if for real numbers $a<b$ one has $\int_X f^a d\mu<+\infty$ and $\int_X f^b d\mu<+\infty$ then $$w(p):= \int_X f^p d\mu$$ is finite and continuous on $[a,b]$ (by Dominated Convergence, since $f^p\le f^a+f^b$), and analytic on $]a,b[$: indeed for any $a<p<b$ and $|s|<\min\{p-a,b-p\}$ we have

$$w(p+s)=\sum_{k=0}^\infty \Big(\int_X f^p\log^k f\, d\mu\Big)\frac{s^k}{k!}$$ as it follows immediately by standard integration by series, since one has

$$ \sum_{k=0}^\infty \int_X\Big|\, f^p\log^k f\, \frac{s^k}{k!} \bigg|\, d\mu\le \int_X f^a d\mu+\int_X f^b d\mu<+\infty\, .$$

(the latter follows plainly splitting the integration domain as $X=\{f\le1\}\cup\{f>1\}$).

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  • $\begingroup$ Note that $f$ bounded, $\mu(X)$ finite, $a$ positive, are not needed. $\endgroup$ – Pietro Majer Sep 16 '16 at 11:21

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