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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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    Yes, please close it. Thank you.2012-12-25

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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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    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