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.
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.