Linear operator on $\ell^1$ is not surjective.

Question

I was reading this question linked here and was wondering if the same holds true when we replace our domain and codomain.

Suppose we have $T: \ell^1 \rightarrow \ell^1$ such that $T(x)=\left(\frac{x_n}{n}\right)_{n \in \mathbb{N}}$. Is this function surjective? I have been trying to come up with a counter-example, but have not come up with anything that would work. For starters, we cannot use the sequence given in the link $(1)_{n \in \mathbb{N}}$ since this is not even in $\ell^1$. Any linear combination of the standard basis that is in $\ell^1$ does not work either I think. Is there some trivial sequence I am overlooking?

Answer 1

$T$ is not surjective.

Let $y = (y_n)_{n=1}^\infty$ be the sequence given by $y_n = \frac{1}{n^2}$, so that $y \in \ell^1$. If $T$ were surjective, then you would be able to find an $x \in \ell^1$ such that $T(x) = y$. But then $\frac{1}{n^2} = y_n = \frac{x_n}{n}$ for each $n$, or $x_n = \frac{1}{n}$, a contradiction, as $x \not\in \ell^1$.