The version of the Uniform Boundedness Principle that I know goes like this: Let $E$ be a Banach space and $F$ a normed space, with $I$ an indexing set (e.g. $\mathbb{N}$) and $\{T_\alpha : \alpha \in I\}$ a subset of $\mathscr{B}(E, F)$ such that for any given $x \in E$, there is some $C_x \geq 0$ such that $||T_\alpha (x)||\leq C_x$ for all $\alpha \in I$. Then, according to the principle, the set $\{||T_\alpha ||: \alpha \in I\}$ is bounded.

That seems fine, but I'm not sure how to apply it to this problem about $\ell ^p$ and $\ell ^q$ spaces. Suppose we have $p\in (0, \infty)$ and $q$ such that $\frac{1}{p} + \frac{1}{q} = 1$. And let $a=(a_k)_{k=1}^\infty$ be a real sequence such that for every $x=(x_k)_{k=1}^\infty \in \ell ^p$, $\sum_{k=1}^\infty a_k x_k < \infty$.

How can I use the principle to show that $a \in \ell^q$? I've made the following start: Fix $x \in \ell^p$ and let $$\phi_n (x) = \displaystyle\sum_{k=1}^n a_k x_k.$$ Then since $\sum_{k=1}^\infty a_k x_k < \infty$, call the sum $C$, $C$ is a uniform bound on the $|\phi_n(x)|$, and so by the principle, there is some $M\geq 0$ with $||\phi_n|| \leq M$ for all $n\in \mathbb{N}$. To show that $a \in \ell^q$, though, we need $\sum_{k=1}^\infty |a_k|^q$ to exist, and I'm stuck trying to find some $x\in\ell^p$ such that $||x||_p = 1$ and $\phi_n(x)=\sum_{k=1}^n |a_k|^q$, which would give us what we want. Tips would be helpful.


You 've almost proved it !

Now consider the sequence $x_N=(y_1,...,y_N,0,0,...)$ where $$y_i = \begin{cases}\dfrac{|\alpha_i|^qsgn(\alpha_i)}{|\alpha_i|} &, \alpha_i \neq 0 \\ 0 &, \text{otherwise} \end{cases}$$

Then you have shown that $\Vert{\phi_N}\Vert\leq M < \infty$ so by hitting every $x_N$ in $\phi_N$ you get

$$ \tag{*} \sum_{n=1}^{N}|\alpha_i|^q=|\phi_N(x_N)|\leq M \Vert{x_N}\Vert_p$$

but $$\Vert{x_N}\Vert_p = \biggl(\sum_{n=1}^{N}|\alpha_i|^{p(q-1)}\biggr)^{1/p}= \biggl(\sum_{n=1}^{N}|\alpha_i|^{q}\biggr)^{1/p}$$

so from $(*)$ you get

$$\sum_{n=1}^{N}|\alpha_i|^q \leq M \biggl(\sum_{n=1}^{N}|\alpha_i|^{q}\biggr)^{1/p} \implies \biggl(\sum_{n=1}^{N}|\alpha_i|^{q}\biggr)^{1/q} \leq M$$

So for $N \to \infty$ $$\Vert{\alpha}\Vert_q \leq M < \infty$$

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    $\begingroup$ For that last implication, you're dividing both sides by the sum to the power of 1/p? That seems pretty neat, thanks. $\endgroup$ – SPS May 15 '18 at 15:30
  • $\begingroup$ Yes thats exactly what am doing. Since you cant find a fixed $x \in l^p$ to give you what you want, you are trying to estimate the partial sums of $|\alpha_i|^q$. $\endgroup$ – dem0nakos May 15 '18 at 15:41

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