2
$\begingroup$

I'm working on this exercise (not homework) and I would gladly welcome some hints for how to solve it!

Exercise: Let $1 \leq p,q \leq \infty$ be conjugate exponents. Let $a=(a_1,a_2,...)$ be a sequence such that $\sum_1^\infty a_n x_n$ converges for all $x=(x_1,x_2,...) \in l^p$. Prove that $a \in l^q$.

My idea is like this: It's sufficient to show that $a \in l^1$ since $l^1 \subset l^2 \subset... $ I define a family of operators $\{T_n \}_{n=1}^\infty$ by $T_n(x) = \sum_{k=1}^n a_k x_k$. It is clear that each $T_n$ is linear, bounded and that $\sup_{n}|T_n(x)| < \infty$ so by the Uniform Boundedness principle we get $\sup_n \| T_n \| < \infty$.

Is this a good approach? If so, I'd be grateful for some guidance on how to proceed to get to the conclusion:

$$\sum_1^\infty |a_k| < \infty$$

Otherwise, steer me in a better direction :)

Thanks in advance

  • 1
    This is a duplicate. I saw the exact same question the other day. Unfortunately, I can't remember its title.2012-12-17
  • 0
    I will search and see if I can find something, thank for the info!2012-12-17
  • 2
    In general $a \in \ell^1$ does not hold, you really have to show $a \in \ell^q$.2012-12-17
  • 1
    Come to think of it: I remember typing an answer when suddenly a wild answer by richard appeared. Let me check this.2012-12-17
  • 0
    I misremembered. But [this](http://math.stackexchange.com/questions/253072/show-that-x-n-is-in-ell2/253078#253078) might be of interest to you nonetheless.2012-12-17
  • 1
    elements of $l^q$ are linear functionals on $l^p$ , so i would say with the above given condition , elements seen as the elements of double dual ie in this case $l^p$, it seems quite clear that application of uniform boundedness principle gives you direct answer as you have done.2012-12-17
  • 0
    show that $||T_n||=(\sum_{k=1}^n|a_k|^q)^{\frac{1}{q}}$.2012-12-17
  • 2
    http://math.stackexchange.com/questions/183684/functional-analysis-banach-steinhaus-theorem?rq=12012-12-17
  • 0
    Thank you all for your answers. I will be writing an answer soon, so please check back and correct me if I did anything wrong :)2012-12-17

1 Answers 1

1

Thanks to the comments in my original post I believe I have a solution. Recall that we know that $\sup_n \|T_n\| < \infty$. First let $1 \leq p \leq \infty$ and fix $n \geq 1$.

Consider the sequence $x=\{x_k\}$ defined by $x_k = |a_k|^{q}/a_k$ when $1 \leq k \leq n$ and $x_k=0$ when $k > n$. Then $x\in l^p$ and

$$\| x \|_p = \left( \sum_{k=1}^n |a_k|^{p(q-1)} \right)^{1/p} = \left( \sum_{k=1}^n |a_k|^{p} \right)^{1/p}$$

Furthermore, using that $T_n$ is bounded, we get

$$T_n(x) = \sum_{k=1}^n |a_k|^{q} \leq \|T_n\|\left( \sum_{k=1}^n |a_k|^{p} \right)^{1/p}. $$ From this we get:

$$ \left( \sum_{k=1}^n |a_k|^{p} \right)^{1/q} \leq \|T_n\| \leq \|T \|$$

And if we let $n \rightarrow \infty$ we get

$$ \| a \|_q \leq \|T_n\| < \infty$$

and hence $a \in l^q$.

If $p=1$ we again fix $n\geq 1$ but instead consider the sequence $x=(0,...,0,1,0...)$ (the 1 on the n:th place). Then $x\in l^p$ and $\|x\|_1 = 1$. Furthermore we get

$$ |T(x)| = |a_n| \leq \|T_n\| \leq \|T\| < \infty.$$

Taking supremum over $n$ we get that $a \in l^\infty$.