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Let $f: [0,\infty)\to [0,\infty)$ be bounded and continuous. Find if exists : $$\displaystyle \lim_{n\to \infty}n\left(\sqrt[n]{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x} \;-\;\sqrt[n]{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}\right)$$ Note : $f^n(x)=(f(x))^n$ This problem was given in a facebook group and everyone claimed it was right. I have made zero progress so far. Can someone solve it? Thanks a lot.

hrkrshnn
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shadow10
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1 Answers1

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Notice that

$$ n\left(\sqrt[n]{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x} \;-\;\sqrt[n]{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}\right) = n \left[ \exp \left( \frac1{n}\ln\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x\right)- \exp\left( \frac1{n}\ln\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x\right) \right]$$

Using the mean value theorem

$$ n\left(\sqrt[n]{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x} \;-\;\sqrt[n]{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}\right) = e^{\xi_n}\ln \left(\frac{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x}{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}\right)$$

where $\xi_n$ is some number between the two arguments of the $\exp$ function in the first set of equations.

Because $f$ is bounded, the integrals converge

$$\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x \leq (\sup f(x))^n\int_{0}^{\infty}e^{-x}\,\mathrm{d}x=(\sup f(x))^n$$

Also

$$\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x \leq \sup f(x)\int_{0}^{\infty}f^n(x)e^{-x}\,\mathrm{d}x$$

and

$$\frac{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x}{\int_{0}^{\infty}f^n(x)e^{-x}\,\mathrm{d}x} \leq \sup f(x)$$

which provides an upper bound for the limit of the LHS.

To get a lower bound first change variables to $u=e^{-x}$ and apply Holder's inequality to obtain

$$\int_{0}^{1}f^{n}(-\ln u)\,\mathrm{d}u \leq \left(\int_{0}^{1}f^{n+1}(-\ln u)\,\mathrm{d}u\right)^{n/(n+1)}\left(\int_{0}^{1}\,\mathrm{d}u\right)^{1/(n+1)}=\left(\int_{0}^{1}f^{n+1}(-\ln u)\,\mathrm{d}u\right)^{n/(n+1)}$$

Rearranging and changing variables back to x we find

$$\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x \leq \left(\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x\right)\left(\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x\right)^{-1/(n+1)}$$

and

$$\left(\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x\right)^{1/(n+1)} \leq \frac{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x}{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}$$

It can be shown that the limit of the LHS is $\sup f(x)$ -- providing a lower bound for the limit of the RHS, that is identical to the upper bound.

Therefore,

$$ \lim_{n \rightarrow \infty}\frac{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x}{\int_{0}^{\infty}f^n(x)e^{-x}\,\mathrm{d}x} = \sup f(x)$$

We can also show $e^{\xi_n}$ converges to $\sup f(x)$ and the limit of the original expression is $\sup f(x) \ln(\sup f(x)).$

RRL
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    This proves an upper bound on the limsup, not the limit. – Did Jun 14 '14 at 12:49
  • Sir Did can you prove this? I have no idea about this problem, I was trying to write this as $\frac{g(x+h)-g(x)}{h}$ for some suitable $g,h$. I tried it like the above, but didn't get as far as he did. Sorry for not mentioning it. – shadow10 Jun 14 '14 at 14:21
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    @Joybangla: I added some more details to establish the limit – RRL Jun 14 '14 at 15:55
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    +1 Cool. I go through the details and this actually works. – achille hui Jun 14 '14 at 17:27
  • @achille hui: Thanks for checking. – RRL Jun 14 '14 at 23:00
  • Sir RRL can you tell me why $e^{{\xi}_n}$ converges to $\sup f(x)$ ? I am sorry if this is trivial, I don't really see it. Thanks a lot. – shadow10 Jun 15 '14 at 04:37
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    @Joybangla $e^{\xi_n}$ is sandwiched between two pieces both converge to $\sup f$. – achille hui Jun 15 '14 at 04:55
  • Thank you sir achille hui but why do those terms converge to $\sup f$ ? sorry for being impossible. Can you tell me? I see they both converge to same point, but couldn't see where. Thank you, for your kind information. – shadow10 Jun 15 '14 at 05:14
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    @Joybangla Let $\displaystyle;I_n = \int_0^\infty f^n(x) e^{-x} dx;$, using mean value theorem, the limit we want to evaluate can be rewritten as

    $$n\left[\left(\frac{I_{n+1}}{I_n}\right)^{1/n}-1\right] I_n^{1/n} = \underbrace{\left(\frac{I_{n+1}}{I_n}\right)^{\rho_n/n} I_n^{1/n}}{e^{\xi_n}} \log\left(\frac{I{n+1}}{I_n}\right) $$ for some $\rho_n \in (0,1)$. The convergence of $I_n^{1/n}$ to $\sup(f)$ is well known. RRL has use Holder's inequality to show $\frac{I_{n+1}}{I_n}$ also converges to $\sup(f)$. Can you see why $e^{\xi_n}$ converges to $\sup(f)$ now?

     – achille hui Jun 15 '14 at 05:26
  • Oh thanks a lot sir achilles hui, but the convergence you call well known, can you give me a link or something to prove it? I am sorry if I am being too foolish, I see sir RRL has also used this fact but I have tried and couldn't prove it, I mean the convergence of $I_n^{1/n}$ to $\sup f$. Thanks a lot for taking your time to help me. Sorry again for being impossible. – shadow10 Jun 15 '14 at 05:30
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    @Joybangla look at answer of this for a proof and the wiki page of $L^p$ space for background material. – achille hui Jun 15 '14 at 05:41