Inequality for series with constraint

Question

I have difficulty with the following estimation.
Let $x_n$, $n=1,2,3,\dots$ satisfy $$\sum\limits_{n=1}^{\infty}x^2_n + \left(\sum\limits_{n=1}^{\infty}x_n\right)^2 = M < +\infty.$$ Let $a_n$, $n=1,2,3,\dots$ satisfy $1 > a_1 > a_2 > \dots > a_n> \dots >0$, $a_n \to 0$ as $n \to +\infty$. Can we find the maximum and minimum of $\sum\limits_{n=1}^{\infty}a_nx_n$?
I intend to use Cauchy-Schwarz inequality, but the sum includes infinitely many terms.
Thank you very much for your help.

Answer 1

Some thoughts:

For convenience, let $M = 1$.

Fact 1: Given $b_1, b_2, \cdots, b_n\ge 0$, the maximum of $\sum_{i=1}^n a_i x_i$ subject to $\sum_{i=1}^n x_i^2 + (\sum_{i=1}^n x_i)^2 \le 1$ is given by $$\sqrt{\sum_{i=1}^n b_i^2 - \frac{1}{n + 1}\left(\sum_{i=1}^n b_i\right)^2}$$ when $$x_j = \frac{b_j - \frac{1}{n+1}\sum_{i=1}^n b_i}{\sqrt{\sum_{i=1}^n b_i^2 - \frac{1}{n + 1}\left(\sum_{i=1}^n b_i\right)^2}}, \quad j = 1, 2, \cdots, n.$$ (Proof: Use Lagrange Multipliers.)

Fact 2: Given a sequence $\{b_i\}_{i=1}^\infty$ with $b_i \ge 0, \forall i\ge 1$, we have, for all $n \ge 1$, $$\sqrt{\sum_{i=1}^n b_i^2 - \frac{1}{n + 1}\left(\sum_{i=1}^n b_i\right)^2} \le \sqrt{\sum_{i=1}^{n+1} b_i^2 - \frac{1}{n + 2}\left(\sum_{i=1}^{n+1} b_i\right)^2}.$$ (Proof: We have $\mathrm{RHS}^2 - \mathrm{LHS}^2 = \frac{1}{(n+1)(n+2)}(\sum_{i=1}^n b_i - b_{n+1} - n b_{n+1})^2 \ge 0$.)

Discussion:

Let $$S = \left\{(x_1, x_2, \cdots) : ~ \sum_{n=1}^\infty x_n^2 + \left(\sum_{n=1}^\infty x_n\right)^2 = 1\right\}.$$

(1) If $\lim_{n\to \infty} \frac{1}{n + 1}\left(\sum_{i=1}^n a_i\right)^2 = 0$, we have $$\sup_{S}\sum_{n=1}^\infty a_n x_n = \sqrt{\sum_{n=1}^\infty a_n^2}. \tag{1}$$ (The proof is given at the end.)

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proof of (1):

By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\sum_{n=1}^\infty a_n x_n \le \sqrt{\sum_{n=1}^\infty a_n^2 \sum_{n=1}^\infty x_n^2} \le \sqrt{\sum_{n=1}^\infty a_n^2}.$$

On the other hand, for any $\epsilon > 0$, there exists $N$ such that $$\sum_{n=1}^\infty a_n x_n = \sqrt{\sum_{i=1}^N a_i^2 - \frac{1}{N + 1}\left(\sum_{i=1}^N a_i\right)^2} > (1 - \epsilon)\sqrt{\sum_{n=1}^\infty a_n^2}$$ when $x_{N + 1} = x_{N+2} = \cdots = 0$ etc. (See Fact 1).

We are done.