prove that there is no sequence of nonzero reals $\{u_{n}\}$ such that prove that there is no sequence of nonzero reals $\{u_{n}\}$ such that  for every real sequence  $\{t_{n}\}$, we have 
$$\sum_{i=1}^{\infty}|t_{n} |< \infty \Longleftrightarrow  \sup_{n}|u_{n}t_{n}| < \infty$$
It seems that if this is right, then it give a bijection from $l^{1}$ to $l^{\infty}$, but I can not conclude a contradiction.
 A: Assume that $(u_n)_n$ is such a sequence. Then notice that the map $F : \ell^1 \to \ell^\infty$ given by $F(t_n)_n = (u_nt_n)_n$ is well-defined and linear.
Furthermore, $F$ is bounded. Since $\ell^1$ and $\ell^\infty$ are Banach spaces, we can show it using the closed graph theorem: assume that $\left(t_n^{(m)}\right)_n \xrightarrow{\ell^1 } \left(t_n^{(0)}\right)_n$ and that $\left(u_nt_n^{(m)}\right)_n \xrightarrow{\ell^\infty } \left(y_n\right)_n$.
We know that $\ell^1$ and $\ell^\infty$ convergence both imply coordinate-wise convergence so $t_n^{(m)} \xrightarrow{m\to\infty} t^{(0)}_n$ for all $n\in\mathbb{N}$. Multiplying this with $u_n$ gives $u_nt_n^{(m)} \xrightarrow{m\to\infty} u_nt^{(0)}_n$ for all $n\in\mathbb{N}$. But we already know that $u_nt_n^{(m)} \xrightarrow{m\to\infty} y_n$ so we conclude $y_n = u_nt^{(0)}_n, \forall n \in \mathbb{N}$.
Notice that $F$ is invertible, namely $F^{-1} : \ell^\infty \to \ell^1$ is given by $F^{-1}(u_nt_n)_n = (t_n)$. By the bounded inverse theorem, $F^{-1}$ is bounded as well.
Hence, it follows that $F$ is a linear homeomorphism of $\ell^1$ and $\ell^\infty$. But this is impossible as $\ell^1$ is separable and $\ell^\infty$ is not.
Therefore, such a sequence $(u_n)_n$ does not exist.
