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Let $(y_n)\subseteq\mathbb{C}$ a sequence such that $\sum_{n=1}^{\infty}x_n\overline{y_n}$ converges for every $(x_n)\in c_0$. ($c_0$ is the space of complex sequences converging to zero with the $\infty$-norm.) Show that $\sum_{n=1}^{\infty}|y_i|<\infty$. Hint: uniform boundedness theorem.

Since $c_0$ is a Banach space, in order to use the uniform boundedness theorem we need linear bounded operators $T_n:c_0\to \mathbb{C}$. My first guess was:

$$T_n(x)=\sum_{i=1}^nx_i\overline{y_i}$$

Clearly every $T_n$ is linear but to prove $T_n$ is continuous seems to be the same as to prove what we actually desire.

Would anyone give me a hint about how to define the family of bounded operators, in order to use the theorem?

Thank you.

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  • $\begingroup$ $x \to T(x) = \sum_i x_i y_i$ is bounded on $c_0 \subset l^\infty$ whenever $y \in l^1$, and the uniform boundedness merely says that $\|T_n - T\| \to 0$ iff $y \in l^1$ $\endgroup$ – reuns Sep 22 '16 at 2:21
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You've already made the perfect choice of bounded operators. By hypothesis, for all $x\in c_0$, the sequence $\{T_n(x)\}_{n=1}^\infty$ is convergent. Since convergent sequences in $\Bbb C$ are bounded, $\{T_n\}_{n = 1}^\infty$ is a point-wise bounded sequence. Hence, by the uniform boundedness principle, there exists $C > 0$ such that $\|T_n\| \le C$ for all $n$.

For each $n$, let $x^{(n)} := (\lambda_1, \ldots, \lambda_n,0,0,\dots)$ where the $\lambda_i$ are complex numbers of unit modulus such that $\lambda_i \overline{y_i} = \lvert y_i\rvert$. Use the estimate $\sup_n \|T_n\| \le C$ to obtain $$\sum\limits_{i = 1}^n \lvert y_i\rvert \le C\quad (n = 1,2,3,\ldots)$$

It follows from this that the series $\sum\limits_{i = 1}^\infty \lvert y_i\rvert$ converges .

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