Show that sequence must be in $ℓ^\infty$ Given a sequence $(a_n) \in \mathbb{R}$ so that for every sequence $(x_n) \in c_0 : (a_nx_n) \in c_0$. Show that this implies that $(a_n)$ has to be in $ℓ^\infty$.
My thoughts: $(a_nx_n) \in c_0$ means that $\lim a_n \lim x_n=0$ and I know that $\lim x_n=0$ for $n \rightarrow \infty$.
$ℓ^\infty$ is the space of all bounded sequences with the supremum norm. The hint is that I need to use Banach-Steinhaus. How could I show this?
 A: Suppose $a$ is not in $\ell^\infty$. Then there's some subsequence of $a$ approaching $\infty$ (in absolute value). Let $x_n=1/|a_n|$ for $n$ indexing that subsequence, and $x_n=0$ for all other $n$. It follows that $x$ converges to $0$. But $a\cdot x$ does not converge to $0$, because $a_nx_n=\pm1$ for $n$ indexing that subsequence. This is a contradiction.
A: The Banach-Steinhaus theorem is not needed in this case, but if you insist it can be applied. For every $N$ consider an operator $$T_N(x_n)=(a_1x_1,a_2x_2,\dots, a_Nx_N, 0,0,0,\ldots)$$ The norm of this operator from $c_0$ to $c_0$ is equal $$\|T_N\| =\max\{|a_n|\,:\, 1\le n\le N\}$$ By assumptions for every $(x_n)\in c_0$ we have $T_N(x_n)\to T(x_n)$ in $c_0$ In particular the sequence of elements $\{T_N(x_n)\}_N$ is bounded in $c_0.$ By the Banach-Steinhaus theorem the norms $\|T_N\|$ are uniformly bounded. By the formula for $\|T_N\|$, the sequence $(a_n)$ is bounded.
A: 
$(a_nx_n) \in c_0$ means that $\lim a_n \lim x_n=0$ and I know that $\lim x_n=0$ for $n \rightarrow \infty$.

You have to be careful while saying this. The limits can be "distributed" only when we know that the two sequences are convergent. For instance, here we don't know whether the sequence $(a_n)$ is convergent or not.
We shall prove the statement by contradiction. Suppose $(a_n) \notin l^{\infty}$.
Then we claim "there exists a sequence of natural numbers $(n_k)$ such that,

*

*$n_k < n_{k+1}$

*$a_{n_k} >n^2$
Proof of the claim: Since $(a_n)$ is assumed to be a unbounded sequence, there exists $n_1 \in \mathbb{N}$  such that $a_{n_1} > 1$. Now suppose $k \in \mathbb{N}$ and there exists natural numbers $n_1 < n_2 <.....< n_{k-1}$ such that
\begin{equation*}
a_{n_i}>i^2 \hspace{1cm} \forall i \in \{1,2,...,k-1\}
\end{equation*}
If for all $m > n_{k-1}$,  $a_m \leq k^2$ then:
\begin{equation*}
||(a_n)||_{\infty} \leq \sup \{|a_1|,|a_2|,....|a_{n_{k-1}}|, k^2\} < \infty
\end{equation*}
which will contradict our assumption that $(a_n)$ is a bounded sequence. So there exists at least one $n_k > n_{k-1}$ such that $a_{n_k} >k^2$. From principle of Mathematical induction we can conclude that our claim is proved.
Set $n_0 =0$. Now let us define a real sequence as
\begin{align*}
&x_n = \frac{1}{k} \hspace{1cm} \text{if } n_{k-1} < n \leq n_k
\end{align*}
Note that $x_n$ is defined for all natural number $n$. Also it is clear to see that $\lim x_n = 0$. Hence $(x_n) \in c_0$. Now let $y_n = a_n y_n$ for all $n \in \mathbb{N}$. By the assumption in the question $(y_n) \in c_0$ and in particular $(y_n)$ is a bounded sequence.
But, note that for each $k \in \mathbb{N}$,
\begin{equation*}
y_{n_k} = a_{n_k} x_{n_k} > k^2. \frac{1}{k} = k
\end{equation*}
Which means that $(y_n)$ is not a bounded sequence. This is a contradiction. So our assumption is wrong. Hence $(a_n) \in l^{\infty}$.
