Surjectivity of the transpose of an operator

Question

I'm trying to show that if $T:X \rightarrow Y$ is a continuous injective linear map, then the transpose $T^*$ is not necessarily surjective. I was trying to use the map $(a_n)_{n=1}^{\infty} \mapsto (\frac{a_n}{n})_{n=1}^{\infty}$ from $l^1$ to itself. It is injective (and it is not bounded below). How can one show that its transpose is not surjective?

Answer 1

If you use $\ell^2$ instead, your map is selfadjoint.

If you consider $\ell^1$, the transpose is $T^*:\ell^\infty \to\ell^\infty $ given by $$(T^*x)y=x (Ty)=\sum x_n (Ty)_n=\sum \frac {x_n}n\,y_n. $$ Thus $T^*x=(x_n/n) $, and so $T^*$ is not surjective (it is compact, in fact).