# Adjoint of bounded linear map is isometric isomorphism implies original map is isometric isomorphism?

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

## 1 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.

• 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... – MichaelGaudreau Dec 18 '18 at 20:25
• 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? – MichaelGaudreau Dec 18 '18 at 20:35
• 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 $T:X\to Y$. – Aweygan Dec 18 '18 at 23:52
• Yes, I've come to the same understanding. Thanks for your excellent example. – MichaelGaudreau Dec 19 '18 at 1:33
• You're welcome. Glad to help! – Aweygan Dec 19 '18 at 1:35