I can't make up my mind whether this question is trivial, or simply wrong, so i decided to ask, just in case someone sees a fallacy in my reasoning:

Question: Suppose $V,W$ are two banach spaces, and $T:V\to W$ is an isomorphism. Is $T^*:W^*\to V^*$ an isomorphism?

On the one hand this seems trivial- it requires a little work, but one can show that $T^*$ is injective, given that $T$ is surjective without working too hard, so if $T^*$ is also surjective, the open mapping theorem should finish the work for us:

To show this, let $f\in V^*$ be arbitrary. Then $f\circ T^{-1}:W\to \mathbb{C}$ is bounded and linear (since $T^{-1}$ and $f$ both are), and $$T^*(f\circ T^{-1})(w)=f\circ T^{-1}(Tw)=f(w)$$

again, this apears (to me, at least) to be correct, but my little experience with Banach spaces has taught me to fear such immediate results, when discussing duals :-P...

anyhow, I would be very happy if someone could tell me if I'm correct, or otherwise, give a counter-example, or point to a mistake.

Additionally, assuming this isn't as immediate as I thought- does the assertion hold when $T$ is an isometric isomorphism?

Thank you very much :-)

(p.s i added the homework tag, as this question arose as part of a h.w assignment, but this isn't a h.w question per-se)

  • 1
    $\begingroup$ This should be fine. More generally, we have $(ST)^*=T^*S^*$ and so taking $S=T^{-1}$ will do $\endgroup$ – user16299 Apr 13 '12 at 15:38

For the first part, your argument is okay.

I'd argue:

  1. For every Banach space $V$ we have $(1_{V})^\ast = 1_{V^\ast}$
  2. For bounded linear maps we have $(ST)^\ast = T^\ast S^\ast$.

This implies that for $S: V \to W$ and $T: W \to V$ such that $ST = 1_{W}$ and $TS= 1_{V}$ that $T^\ast S^\ast = 1_{W^\ast}$ and $S^\ast T^\ast = 1_{V^\ast}$, in other words $(S^{-1})^\ast = (S^\ast)^{-1}$.

As for isometric isomorphisms, check that the inverse of an isometric isomorphism is isometric and that the adjoint of an isometric isomorphism is isometric, too.

No need to invoke the open mapping theorem anywhere.

  • $\begingroup$ great :) thanks $\endgroup$ – kneidell Apr 13 '12 at 15:45
  • $\begingroup$ The results here can of course be sharpened by considering more general maps than isomorphisms, this is nicely done in Rudin's Functional Analysis, chapter 4 on duality theory, for example. Some of the arguments there involve quite a bit more sophistication than what's needed here, in particular the closed range theorem. $\endgroup$ – t.b. Apr 13 '12 at 15:50

Some hints

1) Every functor preserves isomorphisms.

2) The map $$ \begin{align} {}^*&:\operatorname{Ban}\to\operatorname{Ban}&:&W\mapsto W^*\\ &:\mathcal{B}(U,V)\to\mathcal{B}(V^*,U^*)&:&T\mapsto T^* \end{align} $$ is a contravariant functor from category of Banach spaces into category of Banach spaces.

3) For the case of isometric isomorphism consider restriction of $^*$ functor on the category of Banach spaces with contractive maps.


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