Prove that for every sequence $\{a_n\}$ such that $\sum\limits_{n=1}^{\infty}a_n^2=+\infty$ there exists sequence $\{b_n\}$ such that $\sum\limits_{n=1}^{\infty}b_n^2$ converges but $\sum\limits_{n=1}^{\infty}a_nb_n$ diverges. How should I approach this problem?

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    $\begingroup$ What is your background and what tools do you have? $\endgroup$ – Ted Shifrin Jun 5 '13 at 15:19
  • $\begingroup$ I can use any tools from standard university calculus course. $\endgroup$ – erza Jun 5 '13 at 15:36
  • $\begingroup$ If this helps, I can also use some results from functional analysis. $\endgroup$ – erza Jun 5 '13 at 15:59
  • $\begingroup$ Aha ... Try the Uniform Boundedness Principle! $\endgroup$ – Ted Shifrin Jun 5 '13 at 16:15
  • $\begingroup$ Post a solution! $\endgroup$ – Ted Shifrin Jun 5 '13 at 20:20

A UBP-based proof by contradiction works fine, but does not produce $b_n$. I'd rather give constructive proofs when possible and practical.

Divide the sequence $a_n$ into blocks $m_{k}\le n<m_{k+1}$. I think of each block as a vector $\vec v_k$ (they don't have to be of the same dimension). Since $\sum a_n^2=\infty$, there is enough material for arbitrarily long vectors: let's make them so that $|\vec v_k|\ge 2^k$.

Now we need another sequence of vectors $\vec w_k$ such that $\sum |\vec w_k|^2<\infty$ but $\sum \vec v_k\cdot \vec w_k=\infty$. So, they should be pretty short but with large inner product: such as $\vec v_k\cdot \vec w_k=|\vec v_k|\,|\vec w_k|\ge 1$. Of course, $w_k=2^{-k}\vec v_k/|\vec v_k|$ does the job.


Consider the duality for the norm in $\ell^2$ (it holds even for sequences not in $\ell^2$, but we apply it here only to finite sequences): $$\tag{1} \sum_n|a_n|^2=\sup\left\{\left|\sum_na_nb_n\right|^2:\ \sum_n|b_n|^2=1\right\}. $$ Now divide the sequence $\{a_n\}$ in chunks $E_k$ such that $$ \sum_{n\in E_k}a_n^2> 2^{3k/2}. $$ So, applying $(1)$ in each block $E_k$, we find numbers $c_n$ with $$ \sum_{n\in E_k}c_n^2=1,\ \ \ \sum_{n\in E_k}a_nc_n>2^{3k/2}. $$ If we now let $b_n=c_n/2^{k/2}$, then $$ \sum b_n^2=\sum_k\sum_{n\in E_k}\frac{c_n^2}{2^k}=\sum_k\frac1{2^k}=1, $$ and $$ \sum_na_nb_n=\sum_k\sum_{n\in E_k}\frac{a_nc_n}{2^{k/2}}>\sum_k\frac{2^{3k/2}}{2^{k/2}}=\sum_k2^k=+\infty. $$


For any arbitrary positive termed series,$a_{n}$,we have three possibilities:

1.$a_{n}$ converges to a limit,$l<1$.
2.$a_{n}$ converges to a limit,$l>1$.
3.$a_{n}>1,\forall n \in\mathbb{N}$,diverges.
4.$a_{n}=1,\forall n\in \mathbb{N}$

Case 1:

If $l<1$
$\Rightarrow 0<a_{n}<1$. We know that $\infty<\sum a^2_{n}<\sum a_{n}\Rightarrow \sum a_{n}$ diverges. Also,$\forall n\in\mathbb{N},\sum a^n_{n}$ converges since by the root test, $\lim (a^n_{n})^{1/n}=a_{n}<1$.

Now,to prove the existence of $b_{n}$,it suffices to show that atleast one such $b_{n}$ exists.Let us define $b_{n}:=a^n_{n}$.As shown above,by the root test $\sum b^2_{n}$ converges. Now,$\sum a_{n}b_{n}=\sum a_{n}a^n_{n}$
Also since we had $0<a_{n}<1$,therefore ,$\exists c\in [0,1]$,such that $c<a_{n}\Rightarrow r=c^{n}<a^{n}_{n}$.

Therefore,$r\sum a_{n}<\sum a_{n}a^n_{n}$ Now,since $\sum a_{n}$ diverges therefore $r\sum a_{n}$ diverges,and by the comparison test,$\sum a_{n}a^n_{n}$ diverges.

Case 4: If $a_{n}=1,\forall n \in \mathbb{N}$,let $b_{n}:=1/n$, then we know that $\sum a^2_{n}$ diverges,$b^2_{n}=\sum \frac{1}{n^2}$ converges, and $\sum a_{n}b_{n}=\sum \frac{1}{n}$,diverges.

Case 2 and 3:

I'll work these out later.


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