Can we show that if $T'$ is injective (surjective), then $T$ is surjective (injective)?

Question

If $T$ is a linear operator between vector spaces $X,Y$ and $T^\ast$ denotes the algebraic adjoint, i.e. $T^\ast:Y^\ast\to X^\ast,y^\ast\mapsto y^\ast\circ T$, it is easy to see that if $T$ is injective (surjective), then $T^\ast$ is surjective (injective). In particular, if $T$ is bijective, then $T^\ast$ is bijective.

Now I've read that if $X,Y$ are Banach spaces and $T':Y'\to X'$ denotes the topological adjoint, then $T$ is bijective if and only if $T'$ bijective. How can we show this and why is it important that we are dealing with Banach spaces and that we've replaced the algebraic with the topological duals?

And are we only able to show the equivalence or does it even hold that if $T'$ is injective (surjective), then $T$ is surjective (injective)?

Answer 1

The implication: ``$T$ surjective $\Rightarrow$ $T'$ or $T^*$ is injective'' is quite easy to prove:

Assume $T'f=T'g$. Then $(T'f)(x)=(T'g)(x)$ and $f(Tx)=g(Tx)$ for all $x\in X$. Since $T$ is surjective, $f=g$ follows. Same for $T^*$. Note that when dealing with continuous operators here, surjectivity of $T$ can be replaced by: image of $T$ is dense in $Y$.

The implication: ``$T$ injective $\Rightarrow$ $T'$ surjective'' is false in general: take $T$ to be an injective compact operator between infinite-dimensional Banach spaces. Then $T'$ is compact as well, and the range of $T'$ is not closed. I have no idea/intuition about the algebraic dual $T^*$ for this implication.

The implication: ``$T'$ injective $\Rightarrow$ $T$ surjective'' is false as well. Take $T$ compact with dense range, then $T'$ is injective, but $T$ has not closed range.

The implication ``$T$ bijective $\Rightarrow$ $T'$ or $T^*$ bijective'' on the other hand is true, and can be proven by showing that $(T')^{-1}=(T^{-1})'$ and $(T^*)^{-1}=(T^{-1})^*$.

The implication ``$T'$ bijective $\Rightarrow$ $T$ bijective'' is a consequence of the closed range theorem: By the closed range theorem, the range of $T$ is closed and the annihilator of the nullspace of $T'$, so $T$ is surjective. Injectivity of $T$ is clear. Here, $X,Y$ have to be Banach spaces.