# Problem of the Week #231 - Nov 01, 2016

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Consider a sequence of real numbers $(x_n)_{n = 1}^\infty$ such that $\sum\limits_{n = 1}^\infty \lvert x_n y_n\rvert$ converges for every real sequence $(y_n)_{n = 1}^\infty$ such that $\sum\limits_{n = 1}^\infty y_n^2$ converges. Prove that $\sum\limits_{n = 1}^\infty x_n^2$ converges.

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#### Euge

##### MHB Global Moderator
Staff member
This week's problem was solved correctly by Kokuhaku . Here is the solution.

Let $x = \{x_n\}_{n=1}^\infty$ be fixed and define $A_n(y)=\sum_{k=1}^n x_k y_k$ for $y = \{y_n\}_{n=1}^\infty \in \ell^2(\mathbb{N})$. We have that $A_n$ is pointwise bounded, since $\sup_n |A_n(y)| \leqslant \sum_{n=1}^\infty |x_n y_n| < +\infty$, for every $y \in \ell^2(\mathbb{N})$. Thus, by uniform boundedness principle we have $\sup_n \|A_n\| < +\infty$.

Also, Cauchy-Schwarz inequality (or Hölder inequality with $p,q=2$) gives us $|A_n(y)| \leqslant \sqrt{\sum_{k=1}^n x_k^2} \cdot \|y\|_2$ for $y \in \ell^2(\mathbb{N})$, that is $\|A_n\| \leqslant \sqrt{\sum_{k=1}^n x_k^2}$. Choosing $y_k=x_k$ for $1 \leqslant k \leqslant n$ and $y_k=0$ for $k>n$ we obtain $A_n(y)=\sum_{k=1}^n x_k^2$, and so $\sqrt{\sum_{k=1}^n x_k^2} = \frac{A_n(y)}{\sqrt{\sum_{k=1}^n x_k^2}} \le \|A_n\|$.

So we find that $\|A_n\| = \sqrt{\sum_{k=1}^n x_k^2}$ and then from $\sup_n \|A_n\| < +\infty$ we have that $x \in \ell^2(\mathbb{N})$, since $\|x\|_2 = \sup_n \sqrt{\sum_{k=1}^n x_k^2}$.

Remark: If we have complex sequences, then we would choose $y_k=\operatorname{sign} x_k \cdot |x_k|$ for $k \leqslant n$ and $y_k=0$ for $k>n$. Also, if we have $y \in \ell^p$ and same condition $\sum_{n=1}^\infty |x_n y_n| < +\infty$, then we would had $x \in \ell^q$ (use general Hölder and $y_k=\operatorname{sign} x_k \cdot |x_k|^{q-1}$).

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