Let $c_{0} $ be a space of all real or complex sequences converging to $0$ and $l_{\infty} $ be space of real or complex bounded sequences with norm: $|| \{ x_{n} \} _{n=1}^{\infty} || = sup_{n \in \mathbb{N} }|x_{n} | $. Then $c_{0} $ is a vector subspace of $l_{\infty} $.
a) Let $ \{ a_{n} \}_{n=1}^{\infty} $ be a sequence of real or complex numbers such that for every $\{ x_{n} \} _{n=1}^{\infty} \in c_{0} $ sequence $\{a_{n} x_{n} \} _{n=1}^{\infty} \in c_{0} $. Prove that $\{ a_{n} \} _{n=1}^{\infty} \in l_{\infty} $.
b) Let $ \{ a_{n} \}_{n=1}^{\infty} $ be a sequence of real or complex numbers such that for every $\{ x_{n} \} _{n=1}^{\infty} \in l_{\infty} $ sequence $\{a_{n} x_{n} \} _{n=1}^{\infty} \in c_{0} $. Prove that $\{ a_{n} \} _{n=1}^{\infty} \in c_{0} $.
Can someone help to prove this, especially the part b)?
I tried to prove a) (not sure if it's correct):
$ \{ a_{n}x_{n} \}_{n=1}^{\infty} \in c_{0} $ so there is $M$ such that $ \sup_{n \in N} |a_{n}x_{n}| \leq M $. Similarly, there is $K $ such that $ |x_{n}| \leq K $, for every $n$.
Could we use Uniform boundedness theorem in the following way: define $\{ A_{n} \}_{n=1}^{\infty} $ , which is sequence of functionals on $c_{0}$ such that $A_{n}(\{ x_{n} \}_{n=1}^{\infty}) = a_{n} x_{n}$. All operators $A_{n}$ are bounded, $|| A_{n} || \leq |a_{n} | $ for every $n$.
$\sup_{n \in N} |A_{n} x| = \sup_{n \in N} |a_{n} x_{n}| \leq M $. Then from theorem, it would follow that $\sup_{n \in N } ||A_{n} || < \infty $. For sequence $ x =\{x_{n} \} = \{0,0,...,1,0,...\}$, where $1$ is on the nth position, we have: $ \frac{ |A_{n} x |} { ||x|| } = a_{n} \leq ||A_{n} || \leq \sup_{n \in N } ||A_{n} || < \infty $, so it follows that $ \{ a_{n} \}_{n=1}^{\infty} \in l_{\infty} $
I don't know how to finish part b). Thanks in advance.
Your proof may be a little unclear in terms of indices but, in general it seems fine. The operators $$A_n: c_0 \to \mathbb R \quad , \quad A_n(x) = a_nx_n$$ verify $\|A_n\|=|a_n|$, so they are continuous. Moreover, for a fixed $x\in c_0$,
$$\sup_n |A_n(x)|\leq \sup_n |a_nx_n| < \infty $$
so $\sup_n \|A_n\| = \sup_n |a_n| < \infty$, which gives that $(a_n) \in \ell_\infty$.
However, functional analysis is not needed: if $(a_n)$ were not bounded, for every natural $k$ there is $a_{n_k}$ such that $|a_{n_k}|>k$. The sequence $$x_n = \begin{cases} \frac1k \cdot \frac{|a_{n_k}|}{a_{n_k}} & \text{ if } n=n_k \\ 0 & \text{ otherwhise } \end {cases} $$ converges to 0, but $(a_nx_n)$ does not.