Suppose $X$ and $Y$ are normed spaces. Let $T$ be a bounded linear map from $X$ to $Y$. Let $T^*$ be the adjoint map from $Y^{*}$ to $X^{*}$ defined by $T^{*}(y^*) = y^* T$.
A straightforward calculation shows:
Theorem 1. If $T$ is an isometric isomorphism from $X$ onto $Y$, then $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$.
I'm trying to prove the converse.
But the best I can get is the following. (It comes by applying the above theorem with $T^*$ in place of $T$ and using that $T^{**}$ extends $T$ [if $X$ is identified with a subspace of $X^{**}$ in the natural way]).
Theorem 2. If $T^*$ is an isometric isomorphism from $Y^*$ onto $X^*$, then $T$ is an isometric isomorphism from $X$ into $Y$ and $T(X)$ is dense in $Y$.
I cannot seem to strengthen the conclusion to $T$ is surjective.
If $X$ is complete, or, more generally, if $T(X)$ is closed in $Y$, then $T$ is surjective.
But what happens if $X$ is not complete or $T(X)$ is not closed?
In the discussion of the following question, the OP claims to be able to prove that $T^{∗}$ being an isomorphism implies $T$ is surjective, but I don't see how:
$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)
There are also some Hilbert space examples in the following links, but they don't address what I am asking about:
$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?
Example: operator injective, then the adjoint is NOT surjective
Best Answer
The statement does not hold unless you assume both spaces are Banach spaces. This is because the dual space is the dual space of a normed space is (naturally isomorphic to) the dual of its completion.
Indeed, let $X$ be an infinite-dimensional, not complete normed space, let $\tilde X$ denote its completion, and let $T:X\to \tilde X$ be the canonical inclusion. Then $T^*$ is an isometric isomorphism, but $T$ is not a surjective isometry.