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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

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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    Thank you very much for the answer. In fact you have answred my question completely. But I am now confused by other things. As $n\log(1+(t/n)^\alpha)\leqslant\alpha t$, we have $\int_X n \log[1+(f/n)^{\alpha}]d \mu \leq \int_X \alpha f d \mu = \alpha c$. But $\alpha c \geq 0$. Why is $\lim_{n \rightarrow \infty} \int_X n \log[1+(f/n)^{\alpha}]d \mu$ equal to $0$ in the case when $1 < \alpha < \infty$? Thanks a lot.2012-10-08
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    Some sequence $(I_n)_n$ is such that $0\lt I_n\leqslant C$ for every $n$, for some positive $C$, and $I_n\to0$ when $n\to\infty$. No contradiction here (example: $I_n=1/n$, $C=1$).2012-10-08
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    Thanks very much for the comment. I am sure there is no contradiction. But I don't know how to prove when $1 < \alpha < \infty$, $\lim_{n \rightarrow \infty} \int_X n \log[1+(f/n)^{\alpha}]d \mu = 0$. Because what I know is only that it is less than or equal to a positive number. Thanks again.2012-10-09
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    Let $f_n=n\log(1+(f/n)^\alpha)$. Then $f_n\leqslant\alpha f$ for every $n$ and, when $n\to\infty$, $f_n\to0$ pointwise. Hence...2012-10-09
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    Suppose $f_n \leq \alpha f$ is known. As far as I can see, $ \alpha f$ has nothing to do with $n$. When $n \to \infty$, why $f_n \to 0$? I am sorry if I have been annoying. Thank you very much for the help and patience.2012-10-16
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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