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This question already has an answer here:

I am working on the following problem.

Suppose that $\{a_j\}_{j=1}^{\infty}$ is a sequence with the property that, whenever $\{b_j\}_{j=1}^{\infty} \in \ell^2$, one has $\sum_{j=1}^{\infty}|a_jb_j| < \infty$. Define $T : \ell^2 \to \ell^1$ by $T(b_1, b_2, \dots) = (a_1b_1, a_2b_2, \dots)$. Show that $\{a_j\}_{j=1}^{\infty} \in \ell^2$.

a) First under the assumption that $T$ is bounded.

b) Then without making the assumption that $T$ is bounded.

I have been able to do a) but not b).

If $T$ is continuous, then there is $C > 0$ such that $\|T(x)\|_1 \leq C\|x\|_2$. Let $x_k = (a_1, \dots, a_k, 0, \dots)$ and $x = (a_1, a_2, \dots)$. We have $$\|T(x_k)\|_1 = \|(a_1^2, \dots, a_k^2, 0, \dots)\|_1 = \sum_{n=1}^ka_n^2$$ and $$C\|x_k\|_2 = C\|(a_1, \dots, a_k, 0, \dots)\|_2 = C\left(\sum_{n=1}^ka_n^2\right)^{\frac{1}{2}}$$ so we see that $$\left(\sum_{n=1}^ka_n^2\right)^{\frac{1}{2}} \leq C.$$ Taking the limit as $k$ goes to infinity, we see that $\|x\|_2 \leq C < \infty$ so $x \in \ell^2$.


Any hints for part b), or suggestions for a better proof of part a) which may give some indication of how to approach part b), would be greatly appreciated.

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marked as duplicate by Nate Eldredge, drhab, AlexR, Shuchang, Ivo Terek Aug 14 '14 at 10:03

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  • $\begingroup$ Note that the condidition $\sum a_n b_n < \infty$ is equivalent to $\sum |a_n c_n| < \infty$ with $b'_n = \mathrm{sgn}(a_n) \cdot c_n$ as replacement sequence. $\endgroup$ – AlexR Aug 14 '14 at 8:08
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Hint: Use the Uniform boundedness Principle to show that $T$ is bounded, and the use a)

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For part a) you could also use the fact that $(\ell^2)'\cong\ell^2$. If $T$ is bounded, so is the functional $\overline{T}$ defined by $\overline{T}(b_1,b_2,\ldots)=\sum_{n=1}^{\infty}a_nb_n$. From the above isomorphism you get $(a_n)\in \ell^2$.

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