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$\ell^2$ here is real, not complex. Consider a sequence $x:\mathbb{N}\to\mathbb{R}$. For any element $y\in\ell^2$, we have $(x,y)_{\ell^2}<\infty$. How do we prove that $x\in\ell^2$? We can assume all the entries of $x$ and $y$ above are nonnegative. If we can show $(x,\cdot)$ is bounded on $\ell^2$, the statement follows from Riesz rep. theorem, though.

J. Doe
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1 Answers1

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Let $B_k = \{y \in \ell^2: |(x,y)| \le k\}$. These are closed, and their union is $\ell^2$. By the Baire Category Theorem some $B_k$ has nonempty interior. This implies that $(x, \cdot)$ is bounded.

Robert Israel
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    The use of the Baire Category theorem always surprises me... (+1 though) – nicomezi Dec 11 '18 at 15:02
  • Thank you very much for an elegant answer! I extensively searched for the duplicate question on SE before posting this question, but I just found it here: https://math.stackexchange.com/questions/37647/if-sum-a-n-b-n-infty-for-all-b-n-in-ell2-then-a-n-in-ell2 – J. Doe Dec 11 '18 at 15:09