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I have the following problem. I need to show that for a bounded sequence (of sequences) $\{a_n\}\subset \ell^q$, such that $\lim_{n\to\infty} a^n_i = 0$, i.e. the sequence converges coordinate-wise to zero, the following holds for all $x\in\ell^p$:

$$\lim_{n\to\infty} \sum_{i=1}^\infty a^n_i x_i = 0 $$

Now my first attempt was to use Hölder's inequality in the following way:

$$ \left| \sum_{i=1}^\infty a_i^n x_i \right| \leq \|a^n\|_q \|x\|_p $$

and then showing that $\lim_{n\to\infty} \|a^n\|_q =0$. However, this need not be true, take for example the sequence $\{e_n\}\subset \ell^q$ (where as always $e_n = (0,0,\ldots,1,0,\ldots)$ with the 1 in the $n$-th slot), this sequence satisfies our assumptions of boundedness and coordinate-wise convergence to zero, however the $\ell^q$-norm remains 1 for each $n$. So what other weapons do we have to attack this problem?

The Brainlet Exterminator
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1 Answers1

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You want to show coordinate convergence and boundedness implies weak convergence. So for $x \in \ell^p$ take $\tilde{x}^n = (x_1,x_2, x_3, \ldots x_n, 0,0,0,\ldots)$ to be the truncation after $n$ terms, such that $\|x - \tilde{x}^n\|_p <\epsilon$. Now

Let $f \in (\ell^q)^*$ be the functional corresponding to $x$, i.e. $f(y) = \sum y_ix_i$. Call $f_n$ the functional corresponding to $\tilde{x}^n$. Note that $f(0)=0$, and so on. So now we have $$|f(a^m)| \le |(f-f_n)(a^m)| + |f_n(a^m)|$$ $$\le \sum_{i=n+1}^\infty |a_i^mx_i| + \sum_{i=1}^n |a_i^m||x_i|$$ $$\le \|x-\tilde{x}^n\|_p \sup_{m} \|a^m\|_q+ \sum_{i=1}^n |a_i^m||x_i|.$$

Now take $n$ large so that the first bit is small, then take $m$ large for that $n$ so that the sum is small (which you can do since it's a finite sum and the $a_i^m \to 0$.

David Bowman
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