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Again I am stuck with this problem in Rudin:

Assume that $|f|_{r}<\infty$ for some $r<0$. Prove that $$\lim_{p\rightarrow 0}|f|_{p}=e^{\int_{X}\log |f|d\mu}$$

Let $r and we have $K=\frac{r}{p}>1$. Denote its conjugate by $K'$. Then we have $$\int f^{p}d\mu=\int (f^{p})*1d\mu\le (\int [f^{p}]^{\frac{r}{p}})^{\frac{p}{r}}$$ since by assumption $\mu(X)=1$. So in particular we have $$|f|_{p}\le |f|_{r}<\infty$$ Since $|f|_{p}$ is monotonely decreasing with $p\rightarrow 0$, it must have a limit.

We now apply Jensen's inequality, which gives us $$\log^{\int_{X}Fd\mu}\ge \int_{X}\log[F]d\mu$$ Here $F=f^{p}$. So we have $$ \int_{X}f^{p}d\mu\ge (e^{\int_{X}\log[f]d\mu})^{p} $$ taking the $p$-th root on both sides we conclude that $$|f|_{p}\ge e^{\int \log|f|d\mu}$$

But then I got totally stuck. It is worth pointing that Jensen's inequality is only an equality when $f^{p}=c$ is a constant. Therefore $f$ has to be a constant as well.

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    This has already been asked on the site. See, for example, http://math.stackexchange.com/questions/158049/lp-norm-and-integral-equality-prove/158118#1581182012-12-25
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    .......:( sorry! I am not good at searching...2012-12-25
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    But there is a difference; here the $p$ is approaching from negative instead of positive.2012-12-25
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    No worries. It actually took me several minutes to find the duplicate, and that was because I vaguely remembered having answered a similar question.2012-12-25
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    I see. In your arguments you seem to use Hölder's inequality for negative exponents: what is your source for that?2012-12-25
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    I did not use the one for negative exponents; $\frac{r}{p}$ is positive.2012-12-25
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    I see. In any case, problem 5 in chapter 3 from Rudin asks your question with $r>0$. What's the source to believe that this works for $r<0$?2012-12-25
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    No, my book states the problem with $r<0$.2012-12-25
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    I see a problem. $p$ is negative, so taking $\frac{1}{p}$'s power on both sides at that step actually does not give us the desired relationship. Let me think about it.2012-12-25
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    I'm curious about which edition you are looking at. I have copies of both the first and third edition, and in both it says $r>0$.2012-12-25
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    The edition is the 1966 edition, with no "x edition" sign on it.2012-12-25
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    I'm fairly sure that it is a typo. In any case, $$|f|_r=\frac1{|1/f|_{-r}},$$ so one can get the result for $r<0$ from the case $r>0$.2012-12-25
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    I have not read your proof carefully, but the other proof clearly broke down when $p<0$. His proof will not work even for $f=0$.2012-12-25
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    But thank you! I need to read your proof carefully.2012-12-25
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    I should ask: Is it possible to prove this by Jensen's inequality alone?2012-12-25
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    This is possibly a duplicate of http://math.stackexchange.com/questions/156878/integrate-and-measure-problem2012-12-25
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    Yes, please close it. Thank you.2012-12-25

1 Answers 1

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Note that for $u>0$, $$ k(u)=u\log(u)-u+1\ge0\tag{1} $$ This is because $k'(u)=\log(u)$ and $k''(u)=1/u$ show that $k(u)$ has a minimum at $u=1$.

Now, by L'Hospital, we have $$ \lim_{p\to0}\frac{x^p-1}p=\log(x)\tag{2} $$ Furthermore, applying $(1)$ yields $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}p}\frac{x^p-1}p &=\frac{p\log(x)x^p-x^p+1}{p^2}\\ &=\frac{k(x^p)}{p^2}\\ &\ge0\tag{3} \end{align} $$ Thus, by Dominated Convergence, we have $$ \begin{align} \log\left(\lim_{p\to0}\left(\int_X|f(x)|^p\,\mathrm{d}\mu\right)^{1/p}\right) &=\lim_{p\to0}\frac1p\log\left(\int_X|f(x)|^p\,\mathrm{d}\mu\right)\\ &=\lim_{p\to0}\frac1p\log\left(1+p\int_X\frac{|f(x)|^p-1}{p}\,\mathrm{d}\mu\right)\\ &=\lim_{p\to0}\frac1p\log\left(1+p\int_X\log|f(x)|\,\mathrm{d}\mu\right)\\ &=\int_X\log|f(x)|\,\mathrm{d}\mu\tag{4} \end{align} $$ Therefore, $$ \lim_{p\to0}\left(\int_X|f(x)|^p\,\mathrm{d}\mu\right)^{1/p} =\exp\left(\int_X\log|f(x)|\,\mathrm{d}\mu\right)\tag{5} $$

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    Thank you so much for the help!2012-12-25
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    You're welcome. Each time I prove this theorem, I see some new underlying idea that makes it simpler.2012-12-25
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    How can you jump your lines to (4)? by Dominated Convergence?2018-03-13
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    @mnmn1993: yes. For any $p\gt0$, $\frac{|u|^p-1}p\ge\log|u|$.2018-03-13
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    I think it shows that $\dfrac{\int|f|^p-1}{p}\geq \int \log|f|$, but what next? How can you change $\int\dfrac{\int|f|^p-1}{p}$ to $\int \log|f|$?2018-03-13
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    $\lim\limits_{p\to0^+}\frac{x^p-1}p=\log(x)$ and the left side dominates the right. I don't know where the second integral sign comes from.2018-03-13
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    I am sorry that I write it not clearly. Since $|f(x)|^p \rightarrow 1$ pointwise if $f\neq0$, then $\lim\limits_{p\to 0}\dfrac{|f(x)|^p-1}{p} =\log f(x)$ pointwise. But how to find the dominator for $\int_X \dfrac{|f(x)|^p-1}{p}$? The inequality you mentioned is in reversed sign.2018-03-13
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    @mnmn1993: any one of the integrals $\int_X\frac{|f(x)|^p-1}{p}\,\mathrm{d}\mu$ dominates those for smaller $p$ and for the $\log$. That is what Dominated Convergence requires, not that the limit function dominates the sequence. That would be Monotone Convergence.2018-03-13
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    Thanks! But one more question, how can you take the limit inside since there are some $p$ multiplied in front of it $ \lim_{p\to0}\frac1p\log\left(1+p\int_X\frac{|f(x)|^p-1}{p}\,\mathrm{d}\mu\right) $, I am not sure that how we can take the limit inside.2018-03-13
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    We can find a $p$ small enough so $\left|\int_X\frac{|f(x)|^p-1}p\,\mathrm{d}\mu-L\right|\le\epsilon/2$ where $L=\int_X\log|f(x)|\,\mathrm{d}\mu$. Then we can use that $p$ or a smaller one so that $\left|\frac1p\log(1+pL)-L\right|\le\epsilon/2$.2018-03-13