Linear operator on $\ell^{\infty}$ is not surjective – well-definition of inverse operator

Question

Let $T: \ell^{\infty} \rightarrow \ell^{\infty}$, such that $Tx = \left(\frac{1}{n} x_n\right)_{n\in \mathbb{N}}$. Claim: $T$ is not surjective.

I have been trying to come up with $y \in \ell^{\infty}$ such that there is no $x \in \ell^{\infty}$ with $Tx = y$.

Answer 1

Let $y$ be the constant sequence $1$. If there was $x\in\ell^\infty$ such that $T(x)=y$ then we would have $\frac{1}{n}x_n=1$ for all $n\in\mathbb{N}$ and hence $x_n=n$. But this means $x$ is not a bounded sequence, i.e not in $\ell^\infty$. A contradiction.