3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This example clearly shows that completeness of $X$ is necessary to get $T$ surjective in Theorem 2. But I am having trouble understanding why $Y$ needs to be complete. I am pretty sure Theorem 2 is true as stated. Suppose we add the hypothesis that $X$ is complete and $Y$ is not complete. Then $T(X)$ will be complete because of the conclusion that $T$ is an isometric isomorphism. Hence $\overline{T(X)} = T(X)$. Combining this with the conclusion that $T(X)$ is dense in $Y$, we get that $Y = \overline{T(X)} = T(X)$. But this will mean that $Y$ is complete. Contradiction. Continued in next... $\endgroup$ – MichaelGaudreau Dec 18 '18 at 20:25
  • $\begingroup$ So is this saying that when $X$ is complete and $Y$ is not complete, then $T^{*}:Y^{*} \to X^{*}$ cannot be an isometric isomorphism? This is confusing because the duals $Y^{*}$ and $X^*$ cannot detect whether $X$ and $Y$ are complete. What is going on here? $\endgroup$ – MichaelGaudreau Dec 18 '18 at 20:35
  • 1
    $\begingroup$ I don't really know how to answer. This may help: even though $X^*$ and $Y^*$ can't detect whether or not $X$ and $Y$ are complete, the map $T^*:Y^*\to X^*$ is constructed from the map $T:X\to Y$. $\endgroup$ – Aweygan Dec 18 '18 at 23:52
  • $\begingroup$ Yes, I've come to the same understanding. Thanks for your excellent example. $\endgroup$ – MichaelGaudreau Dec 19 '18 at 1:33
  • $\begingroup$ You're welcome. Glad to help! $\endgroup$ – Aweygan Dec 19 '18 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.