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I have a question regarding the Hilbert space $l_2(\mathbb{N,C})$ of all complex valued quadratically summable sequences.

Suppose we have a complex valued sequence $(b_n)_{n\in\mathbb{N}}$ and suppose that for all $(a_n)_{n\in\mathbb{N}}\in l_2(\mathbb{N,C})$ the series $\sum_{n\in\mathbb{N}}a_n\overline{b_n}$ converges (absolutely). I was wondering if it is then true that $(b_n)_{n\in\mathbb{N}}\in l_2(\mathbb{N,C})$.

My initial thought was that it is not true. I tried to choose $(b_n)_{n\in\mathbb{N}}$ as a sequence which is "almost" quadratically summable, for instance $b_n = \frac{1}{\sqrt n}$. Intuitively, all $l_2$ sequences are asymptotically upper bounded by this sequence. That's why I thought that previously mentioned series always converges. But I'm not sure at all if this works. If the statement is true, I'd love to know why.

Thanks in advance

Frederik vom Ende
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BabyWienerSpace
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2 Answers2

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Let $l^2 = l^2(\Bbb{N},\Bbb{C})$. For $N\ge 0$ and $(a_*)\in l^2$ let $T_N$ be the linear functional $$T_N((a_*)) := \sum_{k=0}^N a_k \overline{b_k}$$ By Cauchy-Schwarz $$|T_N((a_*))| \le \sqrt{\sum_{k=0}^N |a_k|^2}\sqrt{\sum_{k=0}^N |b_k|^2}\le \|(a_*)\|_2\sqrt{\sum_{k=0}^N |b_k|^2}$$ so each $T_N$ is a bounded linear functional on $l^2$. Furthermore, for a fixed $(a_*)\in l^2$ $$|T_N((a_*))| \le \sum_{k=0}^\infty |a_k \overline{b_k}| < +\infty$$ So $$\sup_N |T_N((a_*))| < +\infty$$ By the Banach-Steinhaus theorem, there exists some $C> 0$ such that $$\sup_{\|(a_*)\|_2 = 1}\sup_N |T_N((a_*))| \le C < +\infty$$ Now let $T = \lim_{N\rightarrow\infty} T_N$. Then $T$ is a linear functional on $l^2$ and its norm is bounded by $C$ because of the last inequality. Explicitly, $$T((a_*))=\sum_{k=0}^\infty a_k \overline{b_k}$$ Finally, by the Riesz representation theorem, since $T$ is a bounded linear functional on $l^2$, the sequence $(b_*)$ must be in $l^2$.

cs89
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Chad K
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No advanced theorems are needed.

Assume by contradiction that $\sum |b_n|^2=\infty.$ Then there is an increasing sequence $n_k$ of positive integers so that $\sum_{j=n_{k-1}+1}^{n_k}|b_j|^2\ge 1.$ Let $$a_n={\overline{b_n}\over \left(\sum_{j=n_{k-1}+1}^{n_k}|b_j|^2\right)^{1/2}}{1\over k}, \quad n_{k-1}<n\le n_k $$ Then $$\sum|a_n|^2=\sum {1\over k^2}<\infty$$ and $$\sum a_nb_n\ge \sum {1\over k}=\infty$$

Ryszard Szwarc
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