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I have the following function $n \ln\left(1 + \frac{f(x)^{\alpha}}{n^{\alpha}}\right)$. If $\alpha > 1$ Then I am trying to show that $ \lim_{n \to \infty} \int_{X} f d\mu = 0$. Where by $f$ I mean the given function.

Update:

$f:X \rightarrow [0, \infty]$ is a measurable function such that $ 0 < c:= \int_{X}f d\mu < \infty$

I wish to apply the Dominated Convergence Theorem. However to do this I need to show two things:

  1. Pointwise convergence of a sequence of functions almost everywhere
  2. The existence of some other integrable function $g$ such that $|f_{i}| \leq g$ almost everywhere for all $i$.

I'm confused how to show point 1, because i'm not told what the pointwise limit would be, I also need to show that the measure where it does not converge is equal to 0, to meet the almost everywhere requirement.

For point2: I think I have done this by using the fact that $n^{1 - \alpha} \ln \left(1 + \frac{f(x)^{\alpha}}{n^{\alpha}}\right)^{n^{\alpha}} \leq \left(1 + \frac{f(x)^{\alpha}}{n^{\alpha}}\right)^{n^{\alpha}} \leq f(x)^{\alpha}$.

Thanks.

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    $\begingroup$ The inequality $ \left(1 + \frac{f(x)^{\alpha}}{n^{\alpha}}\right)^{n^{\alpha}} \leq f(x)^{\alpha}$ is false. Take $f(x)=1$, $\alpha>1$. Also, you did not say any thing about the integrability of $f$. Is $f$ itself integrable ? Is it at least bounded ? $\endgroup$
    – Medo
    Mar 16, 2020 at 11:22
  • $\begingroup$ In any case, your pointwise limit should be zero (the identically zero function) wherever $f$ is well-defined. To see this, consider the following exercise: Show that $\lim_{n\rightarrow +\infty}(1+c/n^\alpha)^{1/n}=1$. $\endgroup$
    – Medo
    Mar 16, 2020 at 11:26
  • $\begingroup$ @Medo updated to give details on the function. Thanks I will try your exercise. Can I then say that pointwise convergence implies pointwise almost everywhere convergence? Or do I need to show this separately? $\endgroup$
    – VBACODER
    Mar 16, 2020 at 11:31
  • $\begingroup$ @Medo any hints on how to prove your exercise? I have tried taking logarithms of both sides. Also the fact it looks somewhat like the exponential limit form. $\endgroup$
    – VBACODER
    Mar 16, 2020 at 11:52

1 Answer 1

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Point-wise convergence follows from the fact that $0 \leq n \ln (1+\frac {f(x)^{\alpha}} {n^{\alpha }}) \leq \frac {f(x)^{\alpha}} {n^{\alpha-1 }} \to 0$.

Now consider the function $g(t)=\frac {\ln (1+t^{\alpha})} t$ defined for $0<t<\infty$. $g$ is continuous, $g(t) \to 0$ as $ t \to 0+$ (since $0 \leq g(t) \leq t^{\alpha -1}$)) and $g(t) \to 0$ as $ t \to \infty$ (as seen by an application of L'Hopital's Rule). These fact imply that $g$ is bounded. If $g(t) \leq M$ for all $t$ then we get $\ln (1+t^{\alpha}) \leq Mt$ from which we get $n \ln (1+\frac {f(x)^{\alpha}} {n^{\alpha }})\leq Mf(x)$. Now we are ready to apply DCT.

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  • $\begingroup$ Thank you that makes sense! I will try to prove the results about $g(t)$ myself for practice. The pointwise argument is clever! Does pointwise convergence imply almost-everywhere convergence then? $\endgroup$
    – VBACODER
    Mar 16, 2020 at 12:12

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