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I have reviewed Ayman Houreih's proof for the limit of the $L_p$ norm as $ p \rightarrow 0$ at "Scaled $L^p$ norm" and geometric mean.

While I have found the outline of the proof very helpful, I have a question regarding the applicability of LDCT:

The proof mentions $\frac{|f|^q - 1}{q}$ as the dominating function. A comment in the proof also mentions that $\frac{|f|^q - 1}{q}$ decreases as $x \rightarrow 0$. I have the following question:

I don't belive $\frac{|f|^q - 1}{q}$ decreases as $x \rightarrow 0$ when $|f| < 1$. This can easily be verified by computing the the value of $\frac{|f|^x - 1}{x} $ for different values of $x$ when $|f| < 1$. Am I missing something obvious? Can someone please share what the dominating function (say $g$) is and explain how $$ |\frac{f_n^{(1/n)} - 1}{(1/n)}| \leq g $$ where $g$ is a real Lebesgue integrable function?

Thanks.

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  • For a step of an answer on another page, why not ask directly for explanations by posting a comment to the answer itself? (Related: failing to signal this post to the author of the answer you are putting in doubt is bad form.) – Did Nov 18 '15 at 08:31
  • @Did would have been much easier to post a comment directly but I was not able to as I needed 50 privilege points to comment and I don't have any. Do you know of any way around that? Thanks. – e2pipi.crypto Nov 18 '15 at 14:51
  • In Ayman Houreih's proof, when he wrote "Since $\dfrac{|f|^{1/n} - 1}{1/n}$ is dominated by the integrable function $\dfrac{|f|^{q} - 1}{q}$ for large enough $n$", I think he meant " $\dfrac{|f|^{1/n} - 1}{1/n}$ is dominated by the integrable function $\left\vert\dfrac{|f|^{q} - 1}{q}\right\vert$ ... ". Otherwise, the statement would false for any $f$ such that $\vert f \vert <1$. – Ramiro Nov 18 '15 at 16:37
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    Yes, I assumed that was implicit in the proof. Even when one takes the absolute value, |(|f|^x - 1)|/x, it is not decreasing when |f| < 1. – e2pipi.crypto Nov 18 '15 at 16:47
  • To be clear, |(|f|^q - 1)|/q does not dominate |(|f|^1/n - 1)|/(1/n) when |f(x)I < 1. – e2pipi.crypto Nov 18 '15 at 17:39
  • I should have mentioned that |(|f|^q - 1)|/q does not seem to dominate |(|f|^1/n - 1)|/(1/n) when |f(x)I < 1. It will be great if someone can prove how it dominates |(|f|^1/n - 1)|/(1/n). In the proof, the reasoning that |(|f|^x - 1)|/x is decreasing does not seem to be accurate. Thanks. – e2pipi.crypto Nov 18 '15 at 17:59
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    Supose $0<r<q<1$, and $a<1$. Then, for all $x\in [a,1]$, $$ x^{q-1} \leqslant x^{r-1} $$ Integrating both sides fom $a$ to $1$, we get $$ \frac{1-a^q}{q} \leqslant\frac{1-a^r}{r}$$ Since $a^q<1$ and $a^r<1$, we have $$ \left\vert\frac{a^q -1}{q} \right \vert \leqslant \left\vert \frac{a^r -1}{r}\right\vert $$ So, if $\vert f\vert<1$ then $\dfrac{|f|^{1/n} - 1}{1/n}$ is NOT dominated by the integrable function $\left\vert\dfrac{|f|^{q} - 1}{q}\right\vert$ for large enough $n$ (and the value of $q$ stated in the assumptions). – Ramiro Nov 19 '15 at 00:58
  • Note that that the function $s \mapsto \frac{a^s -1}{s}$ is, in fact, increasing for $s\in(0,q]$ (decreasing as $s \to 0$). HOWEVER, if $a<1$, the function has negative values and then, in this case, $s \mapsto \left\vert \frac{a^s -1}{s}\right\vert$ is decreasing for $s\in(0,q]$ (incrasing as $s \to 0$).
    So, if $\vert f \vert <1$, the argument to apply the Dominated Convergence Theorem fails.     – Ramiro Nov 19 '15 at 01:34
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    @Ramiro Thanks for confirming. We can conclude that if |f| < 1 we cannot apply LDCT. I was however able to prove the theorem by making the following change. If |f| >= 1, we apply the LDCT since we have a dominating function in |(|f|^q-1)|/q. When |f| < 1, the sequence of functions {(1-|f|^1/n)/1/n} is positive and monotonically increasing and converge to -log|f|, we can apply the monotone convergence theorem to get the desired result. Combining the results we get the limit for 0 < |f|. The case of |f| = 0 is handled by examining whether the set of all x where |f(x)| = 0 has positive measure. – e2pipi.crypto Nov 19 '15 at 06:39
  • Yes. That is right. As I wrote if $a<1$, the function $s \mapsto \left\vert \frac{a^s -1}{s}\right\vert = \frac{1- a^s }{s}$ is decreasing for $s\in(0,q]$, so, incrasing as $s \to 0$. So we can, if $\vert f \vert<1$, correctly apply the Monotone Convergence Theorem to $\frac{1- \vert f \vert^{1/n} }{1/n}$. – Ramiro Nov 19 '15 at 12:47
  • It is worthy to mention that, given a single $f$, the cases $\vert f(x) \vert \geqslant 1$, $0<\vert f(x) \vert <1$ and $\vert f(x) \vert =0$ may all happen for $f$. – Ramiro Nov 19 '15 at 13:30
  • Absolutely. To be complete, when proving the result in the other direction, the special cases of |f(x)| = 0 and log|f| not being Lebesgue integrable have to be handled differently as Jensens inequality requires a real valued Lebesgue integrable function, but thats fairly trivial. – e2pipi.crypto Nov 19 '15 at 15:46
  • @dreammonger Would you be interested in editing my proof and adding this correction? I'd be delighted to accept it. Thanks! – Ayman Hourieh Nov 26 '15 at 16:45
  • @AymanHourieh I would be happy to do so, but I don't have enough points to add comments on others posts. – e2pipi.crypto Nov 26 '15 at 19:58

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