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The space $\ell^p$ has a Schauder basis and so we can uniquely express any element of $\ell^p$ as: $$x= \sum^\infty_{i=1}{\alpha_i}{e_i}$$ Where $(e_i)$ is the Schauder basis for $\ell^p$.

My question is, if we apply a functional in the dual space of $\ell^p$ to $x$:

$$f(x)= f \left(\sum^\infty_{i=1}{\alpha_i}{e_i}\right)$$

My text then motivates the next step by saying "since $f$ is linear and bounded", we get:

$$f(x)= \sum^\infty_{i=1}{\alpha_i}{f(e_i)}$$

I am unsure whether boundedness really plays a role here with this step.

A non-rigorous way of thinking about the above step would be:

$$f(x)= f \left({\alpha_1}{e_1}+{\alpha_2}{e_2}+{\alpha_3}{e_3}+...\right)\implies$$

$$f(x)= f ({\alpha_1}{e_1})+f({\alpha_2}{e_2})+f({\alpha_3}{e_3})+...\implies$$

$$f(x)= {\alpha_1}f ({e_1})+{\alpha_2}f({\alpha_2}{e_2})+{\alpha_3}f({e_3})+... = \sum^\infty_{i=1}{\alpha_i}{f(e_i)}$$

All following from the linearity of $f\in \ell^{p'} $. The use of '$...$' is bothering me.

Is there a better way of thinking about distributing a function over an infinite sum ? Is boundedness playing a role that I am unaware of ?

Thanks in advance.

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    $\begingroup$ the definition of a Schauder basis is the convergence in norm, i.e. with $x_n = \sum_{i=1}^n \alpha_i e_i$, then $\|x-x_n\| \to 0$, so that $f(x) = f(x_n) + f(x-x_n) = \sum_{i=1}^n\alpha_i f(e_i) + f(x-x_n)$ . the last term $\to 0$ since $|f(x-x_n)| < \|x-x_n\| \ \|f_n\|$, hence the series $ \sum_{i=1}^\infty \alpha_i f( e_i) $ converges (conditionally) and is $= f(x)$ $\endgroup$ – reuns Jun 7 '16 at 6:56
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    $\begingroup$ As @user1952009 pointed out, boundedness of $f$ ensures convergence of the series $\sum_{i\ge 1}\alpha_if(e_i)$ $\endgroup$ – Samrat Mukhopadhyay Jun 7 '16 at 7:11
  • $\begingroup$ @user1952009 Thank you very much ! That is a more subtle step than one might expect. You are welcome to repost your comment as an answer so that I can accept it. $\endgroup$ – Walt van Amstel Jun 7 '16 at 7:11
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    $\begingroup$ nothing really complicated. think this way : $f$ continuous means that $x_n \to x \implies f(x_n) \to f(x)$. $\endgroup$ – reuns Jun 7 '16 at 9:06
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    $\begingroup$ so the problem reduces to : understanding the definition of continuity on a Banach space, and that bounded = countinuous, and recognizing where to apply all this in the your exercice $\endgroup$ – reuns Jun 7 '16 at 9:08
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The key point is to recognize that when you write $$x= \sum^\infty_{i=1}{\alpha_i}{e_i},$$ you mean $$ x=\lim_{n\to\infty} \sum^n_{i=1}{\alpha_i}{e_i} $$ (in other words, a series is not a sum, it is a limits of sums). This also requires that you recognize which topology you are using for the limit, which in this case is the norm topology.

So your equality is in reality $$ f(x)= \lim_{n\to\infty}\sum^n_{i=1}{\alpha_i}{f(e_i)}. $$ The equality is indeed true because, as $f$ is bounded, it is continuous.

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