Let $a_n$ be in $l^\infty$ . Prove that $f$ defined by \begin{equation} f(b_n) = \sum_{n\geq 0} a_nb_n \end{equation} is a continuous real valued function on $l^1$
My thoughts on the problem are that to show continuity, I must show that for any $\epsilon >0$, there exists a $\delta >0$ such that $\vert b_n-x_n\vert< \delta$ implies $\vert a_nb_n-a_nx_n \vert< \epsilon$.
I'm thinking about factoring out the $a_n$, and letting my $\epsilon = \delta/a_n$.
On
Let $(b_n)$ be a sequence in $l^1$, i.e. $\sum |b_n| < \infty$. Since $(a_n)$ is in $l^\infty$, there exists $a\in \mathbb{R}$ such that $|a_n|<a$ for all $n$. Therefore, $|\sum a_nb_n|\leq\sum |a_nb_n|\leq a\sum |b_n|\leq \infty$ so your function is well-defined.
To show that it is continuous, take $\epsilon>0$ and $(b_n) \in l^1$. If $(c_n)\in l^1$ is such that $||(b_n)-(c_n)||=\sum |b_n-c_n|<\frac{\epsilon}{a}$, show that $f(c_n) \in (x-\epsilon,x+\epsilon)$.
On
You will need to use the "norm" for space $\ell_1$. If $b = (b_1, b_2, \cdots)$, then define $$ \|b\|_1 = \sum_{n=1}^\infty |b_n| $$ Then prove: for any $\epsilon > 0$ there exists $\delta>0$ such that if $\|b\|_1 < \delta$, then $f(b)<\epsilon$. And of course you will have to use the definition of $f$ to find this $\delta$.
Strictly speaking, this shows continuity of $f$ at zero. Then use that to prove continuity elsewhere.
note
It is confusing to write
$$
f(b_n) = \sum_{n\geq 0} a_nb_n
$$
It is better to write
$$
f(b) = \sum_{n\geq 0} a_nb_n
$$
where $b = (a_1, b_2, \cdots)$
$$\vert f(\{b_n\}_{n=1}^\infty)\vert = \vert \sum_{n=1}^\infty a_nb_n\vert\leq \sum_{n=1}^\infty \vert a_n\vert \vert b_n\vert\leq \Vert a_n\Vert_\infty\sum_{n=1}^\infty \vert b_n\vert=\Vert a_n\Vert_\infty\Vert b_n\Vert_1$$ Note that $f$ is linear so it is enough find $c>0$ such that $\vert f(b_n)\vert<c\Vert b_n\Vert$, put $c=\Vert a_n\Vert_\infty$