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I have come up with the following constrained minimization problem: \begin{eqnarray} \min\ \sum_{i=1}^\infty x_i^2\\ \sum_{i=1}^\infty a_ix_i=1 \end{eqnarray} If it were a finite-dimensional case it would be easily solved via Lagrange multipliers; in this case I ask your help since I don't know where to begin.

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2 Answers 2

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Assume wlog that there are no zero terms in the sequence by 'ommiting' them.

Assuming that the sequence $a_i $ is square summable, then one can easily use cauchy schwartz inequality to show that

$$x_i := \frac {a_i} {\sum_{i=1}^{\infty}a_i^2} $$

minimizes the quantity $\sum_{i=1}^{\infty}x_i^2$.


For the case where $a_i$ is not square summable i.e. $\sum_{i=1}^{\infty}a_i^2=\infty$, I will show that $$\inf\{\sum_{i=1}^{\infty}x_i^2: \sum_{i=1}^{\infty}a_ix_i=1\}=0$$

Consider the 'sequence' of sequences $\{x^{(n)}_i\}_{i\geq 1}$, where $x^{(n)}$ is defined as follows:

1) $x^{(n)}_i=\frac{a_i}{\sum_{j=1}^{n}a_j^2}\,\,\,$ if $i\leq n$

2) $x^{(n)}_i=0\,\,\,$ if $i> n$

It is easy to see the following that all the sequences $x^{(n)}$ satisfy the constraint i.e. $\sum_{i=1}^{\infty}a_ix^{(n)}_i=1$. Moreover :

$$ \lim_{n\rightarrow \infty} \sum_{i=1}^{\infty}[x^{(n)}_i]^2=\lim_{n\rightarrow \infty}\frac{1}{\sum_{j=1}^{n}a_j^2}=0$$

$\square$

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  • $\begingroup$ Thank you, unfortunately I must deal with the general case of a non-square-summable coefficient sequence. $\endgroup$ Commented Nov 27, 2015 at 15:26
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    $\begingroup$ @marco trevi my intuition tells me that in the non square summable case a minimum won't exist but the infimum will be zero $\endgroup$
    – Amr
    Commented Nov 27, 2015 at 15:36
  • $\begingroup$ I am thinking the same thing. Could you please point me to some reference please? $\endgroup$ Commented Nov 27, 2015 at 15:45
  • $\begingroup$ @Marco trevi i believe it should be easy to prove. I am not in my home now and I am typing from my mobile, once I get the chance to have a paper and a pen I ll edit my answer to include a proof $\endgroup$
    – Amr
    Commented Nov 27, 2015 at 16:45
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    $\begingroup$ It is difficult for me to think without a paper and a pen :) $\endgroup$
    – Amr
    Commented Nov 27, 2015 at 16:45
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Assuming the Cauchy's inequality still holds for infinite series, $$\left(\sum_{i=1}^{\infty}x_i^2\right)\left(\sum_{i=1}^{\infty}a_i^2\right)\geq \sum_{i=1}^{\infty}x_ia_i=1$$ $$\sum_{i=1}^{\infty}x_i^2\geq \frac{1}{\sum_{i=1}^{\infty}a_i^2}$$

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