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This is on page 32 of Rudin's Real and Complex Analysis, 3rd Edition:

Suppose $\mu$ is a positive measure on $X$, $f: X \rightarrow [0, \infty]$ is measurable, $\int_X f d\mu = c$, where $0, and $\alpha$ is a constant. Prove that$\lim_{n \rightarrow \infty} \int_X n \log[1+(f/n)^{\alpha}]d \mu = \begin{cases} \infty & \text{ if } 0 < \alpha <1, \\ c & \text{ if } \alpha=1, \\ 0 & \text{ if } 1 < \alpha < \infty. \end{cases}$

The hint says "if $\alpha \geq 1$, the integrands are dominated by $\alpha f$". But why?

Thanks a lot.

1 Answers 1

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To be proven: $n\log(1+(t/n)^\alpha)\leqslant\alpha t$, for every $t\geqslant0$ and $n\gt0$, with $\alpha\geqslant1$.

Step 1 Replace $t$ by $nt$, hence it suffices to prove that $\log(1+t^\alpha)\leqslant\alpha t$, for every $t\geqslant0$.

Step 2 Show that, if $\alpha\geqslant1$, then $t^{\alpha-1}\leqslant1+t^\alpha$ for every $t\geqslant0$. (Hint: consider separately the cases $t\leqslant1$ and $t\geqslant1$.)

Step 3 Compute the derivative of the function $u:t\mapsto\log(1+t^\alpha)-\alpha t$.

Step 4 Compute $u(0)$ and conclude.

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    $\log(1+t)\leqslant t$ for every $t$ hence $0\leqslant f_n\leqslant n(f/n)^\alpha$. Since $\alpha\gt1$, ...2012-10-16