$\sum_n |x_n||y_n| < \infty$ for all $(x_n) \in l^3$ implies $(y_n) \in l^{3/2}$

Question

Let $(y_n) \subset \mathbb{R}$ be a fixed sequence. Suppose that $\sum_n |x_n||y_n| < \infty$ holds for all $(x_n) \in l^3$. Show that $(y_n) \in l^{3/2}$ using the uniform boundedness principle.

First, for $x = (x_n) \in l^3$, we can set $T_n(x) = \sum_{k=1}^n x_ny_n$ and note that $T_n(x)$ is pointwisely bounded as $T_n(x)$ is increasing in $n$ and is bounded above by $\sum_n |x_n||y_n|$, which is finite. As $l^3$ and $\mathbb{R}$ are Banach, we can claim the uniform boundedness principle, but I don't see the point of this for now; maybe we should use a different operator? I guess we want to plug in $y = (y_n)$ as a parameter somehow, but in order to do that we need to show $y \in l^3$ as well.

Answer 1

Define a sequence of functionals $T_n :\ell^3 \to \mathbb{R} \mbox{ ( or } \mathbb{C} ),$ by $$T_n (x) =\sum_{k=1}^{n} x_n y_n .$$ Observe that $$|T_n (x) | =\left|\sum_{k=1}^{n} x_n y_n\right| \leq \left(\sum_{k=1}^{n} |y_n |^{\frac{3}{2}}\right)^{\frac{2}{3}}\left(\sum_{k=1}^{n} |x_n |^{3}\right)^{\frac{1}{3}}$$ from which follows that $$||T_n ||\leq \left(\sum_{k=1}^{n} |y_n |^{\frac{3}{2}}\right)^{\frac{2}{3}}$$ and taking $ v=(v_i )_{i\in\mathbb{N}}$ where $v_i = \frac{\overline{y_i }}{|y_i |^{\frac{1}{2}} }$ for $i\leq n $ and $v_i =0 $ for $i>n $ we get $$T_n (v) =\sum_{i=1}^{n} |y_i |^{\frac{3}{2}}=\left(\sum_{k=1}^{n} |y_n |^{\frac{3}{2}}\right)^{\frac{2}{3}}\left(\sum_{k=1}^{n} |v_n |^{3}\right)^{\frac{1}{3}}$$ thus $$||T_n ||= \left(\sum_{k=1}^{n} |y_n |^{\frac{3}{2}}\right)^{\frac{2}{3}}.$$ Now by Uniform Boundness Principle we get $$\left(\sum_{k=1}^{\infty} |y_n |^{\frac{3}{2}}\right)^{\frac{2}{3}} =\sup_n ||T_n ||<\infty$$ therefore $$y=(y_i )_{i\in\mathbb{N}} \in \ell^{\frac{3}{2}} .$$

Answer 2

Your $T_n$ works fine. $(T_n)$ is pointwise bounded, as $$ \sup_n |T_n(x)| \le \sum_n |x_ny_n| < \infty $$ for every $x \in \ell^3$. Uniform boundedness gives $\sup_n \|T_n\| < \infty$.

Now note that, due to Hölder we have $$ |T_n(x)| \le \|(x_n)\|_3\|(y_1, \ldots, y_n)\|_{3/2} $$ and choosing $x_i = \operatorname{sgn} y_i |y_i|^{1/2}$ gives $$ |T_n(x)| = \sum_i |y_i|^{3/2} = \|(y_1, \ldots, y_n)\|_{3/2}^{3/2} = \|(y_1, \ldots, y_n)\|_{3/2} \|x\|_3 $$ hence $\|T_n\| = \|(y_1, \ldots, y_n)\|_{3/2}$. Therefore $$ \|y\|_{3/2} = \sup_n \|(y_1, \ldots, y_n)\|_{3/2} = \sup_n \|T_n\| < \infty $$