0
$\begingroup$

As a part of a work of mine I wanna use this claim (which I hope is true), and don't know why I can:

Assume I have for every $i\in \mathbb N$ a series $\{a_i^n\}_{n\in\mathbb N}\subset\mathbb R$ which converges to $a_i\in \mathbb R$ (as $n$ tends to $\infty$). We can assume $\forall i>n: a_i^n=0$.

Let $\{b_i\}_{i\in\mathbb N}\subset \mathbb R$ be a converging sequence with limit $b$. Define ${b_i^*=\sup_{j\geq i}}b_i $.

I want to prove, as a part of a proof of mine, that $$\lim_{n\rightarrow\infty}\left(\sum_{i=1}^{n-1}a_i^nb_i+a_n^nb_n^*\right)=\sum_{i=1}^{\infty}a_ib_i+b\lim_{n\rightarrow\infty} a_n^n$$

Which seems reasonable to me since $\lim b_i^*=\limsup b_i=\lim b_i = b$, and the rest is just changing order of sum (which I have no clue how to justify, as I'm asking it here)

Thanks!

$\endgroup$
  • $\begingroup$ I don't see why $a_n^n$ converges as $n\to\infty$ Isn't that just the start of each sequence? $\endgroup$ – Empy2 Dec 27 '14 at 17:44
0
$\begingroup$

This claim actually can be false! Take for example $a_i^n=\delta_{i,n-1}$ (so $a_i^n=1$ if $i=n-1$, otherwise it is $0$) and $b_i=b_i^*=b\neq 0$. Then $a_i=0$ for any $i$ and $\lim a_n^n=0$, so the claim becomes $\lim_{n\to\infty}b=0$, which is false!

But there is a simple way of fixing this: choose instead $a_0:=1-\sum_{i=1}^\infty a_i$.
Observe that (in the original problem) we had $\sum_{i=1}^{n-1}a_i^n\le 1$, so we also have (for fixed $m$ and $n\ge m$) that $\sum_{i=1}^{m-1}a_i^n\le 1$ and letting $n\to\infty$ we deduce $\sum_{i=1}^{m-1}a_i\le 1$. As this holds for all $m$ we get $\sum_{i=1}^\infty a_i\le 1$, so $a_0\ge 0$.
(If you include also $a_n^n$ in these estimates you get $a_n^n+\sum_{i=1}^{m-1}a_i^n\le 1$ for $n\ge m$, so $(\lim_n a_n^n)+\sum_{i=1}^\infty a_i\le 1$ holds too, but as the counterexample showed the inequality could be strict!)
Referring to the original problem, fix again a positive integer $m$ and let $n\ge m$: the inequality $$f(x)\le\sum_{i=1}^{n-1}a_i^n\|x\|_i+a_n^n\|x\|_{(n)}\le\sum_{i=1}^{m-1}a_i^n\|x\|_i+(1-\sum_{i=1}^{m-1}a_i^n)\|x\|_{(m)}$$ (which holds as $\|x\|_i\le \|x\|_{(m)}$ for $m\le i\le n-1$ and $a_m^n+\dots+a_n^n=1-\sum_{i=1}^{m-1}a_i^n$) implies (as $n\to\infty$) $$f(x)\le\sum_{i=1}^{m-1}a_i\|x\|_i+(1-\sum_{i=1}^{m-1}a_i)\|x\|_{(m)}$$ and passing to the limit as $m\to\infty$ you obtain $$f(x)\le\sum_{i=1}^\infty a_i\|x\|_i+a_0\lim_i \|x\|_i$$

$\endgroup$

Your Answer

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