1
$\begingroup$

I'm dealing with the following problem and I would appreciate any suggestions.

Let $F, G:[1,\infty)\to\mathbb{R}$ piecewise constant on every interval of the form $[N,N+1)$, with $N$ positive integer and $F(s)=F(N)$ and $G(s)=G(N)$, for all $s\in[N,N+1)$.

Suppose that $G(s)\leq 0$ for all $s\geq 1$, $\lim_{s\to\infty}G(s)=-1$ and that $$\int_{1}^{x}\left(x^{1/2}F(s)s^{-3/2}-G(s)s^{-1/2}\right)\,ds=0,$$ for all $x>1$.

I want to show that $F(s)\leq 0$, for all sufficiently large $s$.

My idea: for each positive integer $N$ we have that $$\int_{1}^{N+1}\left((N+1)^{1/2}F(s)s^{-3/2}-G(s)s^{-1/2}\right)\,ds=0,$$ and $$\int_{1}^{N}\left(N^{1/2}F(s)s^{-3/2}-G(s)s^{-1/2}\right)\,ds=0.$$ Therefore, from this two previous identities we obtain $$\int_{1}^{N}((N+1)^{1/2}-N^{1/2})F(s)s^{-3/2}-G(s)s^{-1/2})\,ds+\int_{N}^{N+1}((N+1)^{1/2}F(s)s^{-3/2}-G(s)s^{-1/2})\,ds=0.$$ Since $$\lim_{N\to\infty}(N+1)^{1/2}-N^{1/2}=0$$ and since $F$ and $G$ are piecewise constant on each $[N,N+1)$ we have that

$$\lim_{N\to\infty}\,\left( -\int_{1}^{N}G(s)s^{-1/2}\,ds+(N+1)^{1/2}F(N)\int_{N}^{N+1}s^{-3/2}-G(N)\int_{N}^{N+1}s^{-1/2}\,ds\right)=0.$$ That is, $$\lim_{N\to\infty}\,\left( -\int_{1}^{N}G(s)s^{-1/2}\,ds+2(N+1)^{1/2}F(N)(N^{-1/2}-(N+1)^{-1/2}) -G(N)((N+1)^{1/2}-N^{1/2})\right)=0.$$ So, since $$(N+1)^{1/2}F(N)(N^{-1/2}-(N+1)^{-1/2})\sim \frac{1}{2}N^{-1} F(N),\,\,N\to\infty,$$ and $$\lim_{N\to\infty}G(N)((N+1)^{1/2}-N^{1/2})=0,$$ because $\lim_{N\to\infty}G(N)=-1$, we may conclude that $$\lim_{N\to\infty}\left(-\int_{1}^{N}G(s)s^{-1/2}\,ds+\frac{1}{2N}F(N) \right)=0.\,\,\,\,\,\,(1) $$ Again, since $G(s)\leq 0$ and $\lim_{N\to\infty}G(N)=-1$, we have that $$\int_{1}^{N}G(s)s^{-1/2}\,ds\leq 0$$ for all $N>1$ and therefore we see from (1) that $F(N)\leq 0$ for all $N$ sufficiently large.

$\endgroup$
  • $\begingroup$ Do you mean "on every interval of the form $[N,N+1)$?" Otherwise they're just constant. $\endgroup$ – Dzoooks Jun 4 at 23:34
  • 1
    $\begingroup$ Yes, I do. My original question was edited (by someone else) but I re-edited and now it is in its original form. $\endgroup$ – GoldSoundz Jun 5 at 0:43

Your Answer

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

Browse other questions tagged or ask your own question.