2
$\begingroup$

This is from Bass real analysis page 57 problem 9. For convenience: $$ \int_0^n \left(1-\frac{x}{n}\right)^n \log\left[2+\cos\left(\frac{x}{n}\right)\right]\mathrm dx $$ So I have bounded the easy part: $$ \left|\chi_{[0,n]}\left(1-\frac{x}{n}\right)^n \log\left[2+\cos\left(\frac{x}{n}\right)\right]\right |\\ \leq\log(3)\left|\left(1-\frac{x}{n}\right)^n\right | $$ but here is where I am stuck. I can certainly bound this by noting that it decreases in $x$, but not by something integrable.

I should note that I am just pretty sure that DCT is what is necessary here, but only because the monotone convergence theorem seems useless here.

$\endgroup$
2
  • $\begingroup$ So what is the question? To evaluate the integral? $\endgroup$ – Idonknow Aug 21 '17 at 12:56
  • $\begingroup$ @Idonknow How to use dominated convergence to bound the integral. $\endgroup$ – operatorerror Aug 21 '17 at 13:06
3
$\begingroup$

$\left(1-\frac{x}{n}\right)^n \le e^{-x}$, for $x \le n$

$\endgroup$
4
  • $\begingroup$ Yes by convexity of $x \mapsto e^{-x}$ we have $1-u \le e^{-u}$ take $u = \frac{x}{n}$ then $1-\frac{x}{n} \le e^{-\frac{x}{n}}$ $\endgroup$ – Youem Aug 5 '17 at 14:58
  • $\begingroup$ $1-x/n\le e^{-x/n}$ does not imply $(1-x/n)^n\le e^{-x}$ for all $x.$ For example, try $x=12, n = 4.$ You need to be careful with where $x$ is. The inequality will work for $x\in [0,n].$ $\endgroup$ – zhw. Aug 5 '17 at 15:58
  • $\begingroup$ I implicitly suppose that $x \le n$. To have a positive left member $\endgroup$ – Youem Aug 5 '17 at 16:03
  • $\begingroup$ I think you should say that then. It wasn't clear to me. $\endgroup$ – zhw. Aug 5 '17 at 16:32
3
$\begingroup$

The initial integral equals $$ \int_{0}^{1}n(1-x)^n \log(2+\cos x)\,dx\stackrel{\text{IBP}}{=}\frac{n}{n+1}\log(3)-\frac{n}{n+1}\int_{0}^{1}(1-x)^{n+1}\frac{\sin x}{2+\cos x}\,dx$$ hence we may prove the convergence to $\log 3$ just by squeezing, since the last integral is positive but bounded by $\int_{0}^{1}x(1-x)^{n+1}\,dx=\frac{1}{(n+2)(n+3)}$.

$\endgroup$
1
  • $\begingroup$ Clever!! Thanks a lot $\endgroup$ – operatorerror Aug 5 '17 at 15:01
2
$\begingroup$

Note that

$$\tag 1 \left | 1-\dfrac{x}{n}\right|^n \to \infty \text { as } x\to \infty.$$

Thus $(1)$ cannot be bounded by anything useful on all of $[0,\infty).$

However it is true that $( 1-x/n)^n\chi_{[0,n]}(x) \le e^{-x}$ on $[0,\infty)$ for every $n.$ We can see this by observing $\ln (1-u) = -(u+u^2/2 + u^3/3 + \cdots)\le -u$ for $u\in [0,1).$ Thus $\ln (1-x/n) \le -x/n $ for $x\in [0,n),$ which leads to $(1-x/n)^n \le e^{-x}$ for $x\in [0,n].$

Now your dominated convergence argument can go through.

$\endgroup$

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.