Suppose $\mu$ is a positive masure on $X$, $f:X\to [0,\infty]$ is measurable, $\int \limits_{X}fd\mu=c$, where $0<c<\infty,$ and $\alpha$ is a constant. Prove that $$\lim \limits_{n\to \infty}\int \limits_{X}n\log [1+(f/n)^{\alpha}]d\mu= \begin{cases} \infty, & \text{if} \quad0<\alpha<1, \\ c, & \text{if} \quad\alpha=1, \\ 0, & \text{if} \quad 1<\alpha<\infty. \end{cases}$$

Remark: $[\cdot]$ is not integer part!

Proof: $\color{blue}{Case \quad \alpha=1}$. Consider functions $f_n(x):X\to [0,\infty]$ defined by $f_n(x)=n\log \left[1+\dfrac{f(x)}{n}\right]$. It's easy to check that $0\leqslant f_1\leqslant f_2\leqslant \dots \leqslant f$ on $X-S$ where $S=\{x\in X:f(x)=\infty\}$ and note that $\mu(S)=0$ (otherwise $\int \limits_{X}fd\mu=\infty$ which is contradiction). Also $f_n$ is measurable for each $n$ since it's a compostion of continuous and measurable functions. Using Monotone Convergence Theorem we get: $$\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu=\int \limits_{X}\lim \limits_{n\to \infty}f_nd\mu=\int \limits_{X}fd\mu=c.$$

$\color{blue}{Case \quad 0<\alpha<1}$. Using Fatou's lemma to functions $f_n=n\log[1+(f/n)^{\alpha}]$ which is measurable in $X-S$ for each $n$ we get the following inequality:

$$\liminf \limits_{n\to \infty} \int \limits_{X}f_nd\mu\geqslant \int \limits_{X}\liminf \limits_{n\to \infty} f_nd\mu$$

Since $\int \limits_{X}fd \mu=\int \limits_{X\setminus S}fd \mu=c$ then the set $E=\{x\in X-S: f(x)>0\}$ has positive measure. And $$\int \limits_{X}\liminf \limits_{n\to \infty} f_nd\mu=\int \limits_{X\setminus S}\liminf \limits_{n\to \infty} f_nd\mu=\int \limits_{E}+\int \limits_{(X\setminus S)\setminus E}=$$ Since on $(X\setminus S)\setminus E$ we have that $f_n(x)=0$ $$=\int \limits_{E}\liminf \limits_{n\to \infty} f_nd\mu=+\infty$$ since $\liminf \limits_{n\to \infty} f_n=+\infty$ on $E$ and $\mu(E)>0.$

$\color{blue}{Case \quad \alpha>1}$. Our functions $f_n$ are measurable and non-negative on $X-S$ and $f_n(x)\to 0$ as $n\to \infty$ on $X-S$. Using derivative test we can show that $f_n\leqslant \alpha f$ for $n\in \mathbb{N}$ on $X-S$. Note that $\alpha f\in L^1(\mu)$. By Dominated Convergence theorem $$\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu=\int \limits_{X}\lim \limits_{n\to \infty}f_nd\mu=0.$$ Here we use that $\mu(S)=0$ because the measure of zero set in negligible in integration.

Sorry if this topic is repeated but I would be thankful if anyone checks out my solution.

  • $\begingroup$ Your proof looks good. I don't know what the derivative test is in the $\alpha>1$ case, but it is easy to establish by splitting into ${x \over n} \le 1$ and ${x \over n} > 1$ cases. $\endgroup$ – copper.hat Jun 12 '16 at 18:47
  • $\begingroup$ @copper.hat, I do not quite understand you. Can you demonstrate your approach? It would be great for me to know the new way of solution of this $\endgroup$ – ZFR Jun 13 '16 at 3:48
  • $\begingroup$ What is the 'derivative test'? If ${x \over n} \ge 1$, then $f_n(x) \le n \log ((1+{x \over n})^\alpha) = \alpha n \log (1+{x \over n}) \le \alpha x$, and if ${x \over n} < 1$ then $f_n(x) \le n \log (1+{x \over n}) = x$ and so $f_n(x) \le \alpha f(x)$. $\endgroup$ – copper.hat Jun 13 '16 at 5:29
  • $\begingroup$ @copper.hat, But you didn't show that $f_n(x)\leqslant \alpha f$ in the first case?! Where did you use that $x/n\geqslant 1$? $\endgroup$ – ZFR Jun 13 '16 at 10:33
  • $\begingroup$ @copper.hat, are you sure that you had not any typo? $\endgroup$ – ZFR Jun 13 '16 at 10:37

This is not an answer, just an elaboration of a comment:

Suppose $\alpha>1$.

(i) If ${f(x) \over n} \ge 1$ we have $f_n(x) \le n \log ((1+{f(x) \over n})^\alpha) = \alpha n \log (1+{f(x) \over n}) \le \alpha f(x) $.

(ii) If ${f(x) \over n} < 1$ we have $f_n(x) \le n \log (1+{f(x) \over n}) = f(x)$.

Hence $f_n(x) \le \alpha f(x)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.