3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.