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Square summable

Let \{a_n\}_{n \in \mathbb{N}} be a real sequence such that

If \{b_n\}_{n \in \mathbb{N}} is a real sequence that is square summable; i.e \sum \limits_{n=1}^{\infty} b_n^2 < +\infty the sequence \sum \limits_{n=1}^{\infty} a_n b_n converges.

Prove that \{a_n\}_{n \in \mathbb{N}} is also square summable.

Solution

Let f_N:\ell_2 \rightarrow \mathbb{R} be defined as

    \[f_N(b)=\sum_{n=1}^{N} a_nb_n\]

where b = (b_n) \in \ell_2. We note that

    \begin{align*} |f_N(b)|^2 &\leq \sum_{n=1}^N|a_n|^2 \sum_{n=1}^N|b_n|^2 \\ & \leq ||b||^2\sum_{n=1}^N|a_n|^2 \end{align*}

Equality holds when b=(a_1, \dots, a_N, 0, 0, \dots ). Hence, \displaystyle ||f_N||= \left( \sum_{n=1}^N|a_n|^2\right) ^{1/2}. From the hypothesis, it follows that \{f_n\}_{n \in \mathbb{N}} is pointwise bounded. It follows from the Uniform boundedness principle ( Banach – Steinhaus ) that \displaystyle ||f_N||= \left( \sum_{n=1}^N|a_n|^2\right) ^{1/2} are bounded. Hence, \{a_n\}_{n \in \mathbb{N}} is square summable.

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