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I've got a hand-in question in a pure analysis course that I was hoping I might get a hint on - having difficulty coming up with a decent approach.

The question:

Let $(X,\Sigma,\mu)$ be a measure space and let $f:X\rightarrow [0,\infty]$ be a measurable function such that $$\int_X f(x)d\mu(x)=A,$$ for some $0<A<\infty$. If $\alpha>0,$ show $$\lim_{n\rightarrow\infty} \int_X n\log\left(1 + \left(\frac{f(x)}{n}\right)^{\alpha} \right)d\mu(x)=\begin{cases} \infty&\mbox{if }0<\alpha<1\\\ A&\mbox{if } \alpha=1\\\ 0&\mbox{if }\alpha>1. \end{cases}$$

My attempt at a solution only comes as far as the first part:

\begin{align*} g(x,n)&=n\log(1 + [f(x)/n]^{\alpha})\\ &=n\cdot \sum_{m=1}^{\infty}(-1)^{m+1}[f(x)/n]^{\alpha m}/m\\ &=\sum_{m=1}^{\infty}(-1)^{m+1}\cdot \frac{f(x)^{\alpha m}}{m\cdot n^{\alpha*m-1}} \\ &= \frac{f(x)^\alpha}{n^{\alpha-1}}+\sum_{m=2}^{\infty}(-1)^{m+1}\frac{f(x)^{\alpha m}}{m\cdot n^{\alpha m-1}}, \end{align*} which is increasing in $n$ for $\alpha<1$ (this is a bit handwavy, but I can't seem to figure out how to show it in a strict manner). Thus, we can apply the Monotone Convergence Theorem to move the limit inside the integrand, transform $n=1/t$, use a bit of L'hopitals rule, and get that this limit is diverging for any $f(x)$ and $\alpha<1$ (and $f(x)$ for $\alpha=1$, zero for $\alpha>1$). But how do I go about proving that I can switch limit and integrand in these other cases, or is there any other simple way to prove it? Any hints would be much appreciated!

Many thanks in advance

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1 Answer

Let $I_n(\alpha):=\int_X n\log\left(1+\left(\frac{f(x)}n\right)^{\alpha}\right)d\mu (x)$. If $0<\alpha<1$, since $\int_X fd\mu>0$ we can find $\beta>0$ such that $\mu(\{f\geq \beta\})>0$. Therefore we have \begin{align*} I_n(\alpha)&\geq\mu(f\geq \beta)n\log\left(1+\left(\frac{\beta}n\right)^{\alpha}\right) \\ &\geq \mu(f\geq \beta)n\left(\left(\frac{\beta}n\right)^{\alpha}-\frac 12\left(\frac{\beta^2}{n^2}\right)^{\alpha}\right)\\ &=\mu(f\geq \beta)n^{1-\alpha}\left(\beta^{\alpha}-\frac{\beta^{2\alpha}}{2n^{\alpha}}\right) \end{align*} and it converges to $+\infty$.

If $\alpha=1$, the use of the inequality $x-\frac{x^2}2\leq \ln(1+x)\leq x$ allow us to apply the dominated convergence theorem.

If $\alpha>1$, use the measure $\nu=f\mu$ which is a probability measure. We have $$I_n(\alpha)=\int_X\phi\left(\frac n{f(x)}\right)d\nu,$$ with $\phi(x)=x\ln (1+x^{-\alpha})$. Since $\phi'(x)=\ln(1+x^{-\alpha})-\alpha\frac{x^{-\alpha}}{1+x^{-\alpha}}$ and $\phi''(x)=\frac{-\alpha x^{-1-\alpha}}{1+x^{-\alpha}}-\frac 1{(1+x^{-\alpha})^2}(-\alpha x^{-\alpha-1})\alpha=-\alpha\frac{x^{-1-\alpha}}{1+x^{-\alpha}}x^{-\alpha}<0$ for $x>0$, $\phi$ is concave. Jensen's inequality yields $$I_n(\alpha)=\int_X\phi\left(\frac n{f(x)}\right)d\nu\leq \phi\left(\int_X\frac n{f(x)}d\nu\right)\leq \left(\int_X\frac n{f(x)}d\nu\right)^{1-\alpha},$$ so it works if $(X,\Sigma,\mu)$ is finite. I guess we can solve the case $\sigma$-finite using the dominated convergence theorem for the counting measure.

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The function $\phi = x\ln(1+x^{-\alpha})$ is not concave. You may plot the case where $\alpha = -2$, so there is probably something wrong in your differentiation. –  Hawii Mar 5 '12 at 4:48
I will check my differentiation. But why $\alpha=-2$? We assumed $\alpha>1$? –  Davide Giraudo Mar 5 '12 at 20:46
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