Trivial Kernel and density of range

Question

On the wiki page for unbounded operators, it states that since: $$ \text{ker}~T = (\text{range} T^*)^{\perp} $$ Then we may conclude that if $T^*$ has trivial kernel then $T$ has dense range. Does this hold for all operators $T$? Further, why does it actually hold? I don't see the connection between denseness and the above equivalence. I'm interested in the general Banach space case, not just operators defined on Hilbert spaces.

Answer 1

If $T:X\to Y$ is abounded operator such that $T^{*}$ is injective then, for any $y^{*} \in Y^{*}$ such that $y^{*}=0$ on the range of $T$ we have $T^{*}y^{*}(x)=y^{*}(Tx)=0$ for all $x$. But $T^{*}$ is injective so we get $y^{*}=0$. This proves that the range of $T$ is dense.