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Let $X, Y$ be normed spaces and $T : X \to Y$ bounded and linear, such that its adjoint $T^* : X^* \to Y^*$ is boundedly invertible.

If $X$ and $Y$ are Banach spaces, then $T$ is also boundedly invertible, see, e.g., the answer here: Show $T$ is invertible if $T'$ is invertible where $T\in B(X)$, $T'\in B(X')$

However, I suspect that this is not true if $X$ or $Y$ are not complete.

Is there a simple example for such a non-invertible $T$ with invertible $T^*$?

Edit: I found some simple examples in case $X$ is not complete: Let $X$ be a dense, proper subspace of an reflexive space $Z$. Let $T = i_{X}: X \to X^{**} = Z^{**}$ and $S = i_{X^*} : X^*=Z^* \to X^{***}=Z^{***} $ be the canonical embeddings into the biduals. Then, one can show $T^* = S^{-1}$. Hence, $T^*$ is invertible, but $T$ is not.

Therefore, only the case $X$ complete, $Y$ not complete is left. If I do not miss something, we get that $T$ is (not necessarily boundedly) invertible in this case.

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If $X$ is complete and $Y$ is not, then $T'$ can not be invertible.

Assume $T'$ is invertible, then it is bijective operator between Banach spaces. Hence $T'$ is bounded below and open mapping.

1) Since $T'$ is bounded below, then by result of this answer $T$ is open mapping.

2) Since $T'$ is open mapping by result of this answer operator $T''$ is bounded below. By $i_E$ we denote standard isometric embedding into second dual, then $i_Y T=T'' i_X$. From here we derive that $T$ is also bounded below.

From 1) and 2) we see that $T$ is bounded below and open mapping, hence not only a bijection but an isomorphism. Since $T$ is an isomorphism and $X$ is complete, then so does $Y$. Contradiction.

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  • $\begingroup$ So putting it together, we get the following theorem: If $X$ is complete and $T'$ is boundedly invertible, then $T$ is boundedly invertible. Thank you for your effort! $\endgroup$ – gerw Nov 12 '13 at 19:15
  • $\begingroup$ You are wellcome! $\endgroup$ – Norbert Nov 12 '13 at 19:17
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Yes: as you say, $T$ is invertible but the inverse will be unbounded if $Y$ is not complete.

Because one can apply the same proof, to conclude that $T$ is injective and has dense image. The inverse cannot be bounded because if it were we would be able to extend it from $Y$ to its closure, with inverse $T$, and then the image of $T$ would be the closuse of $Y$.

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  • $\begingroup$ Our answers contradict to each other. I can't find a mistake neither in yours nor in mine answer. Could you double check them? $\endgroup$ – Norbert Nov 8 '13 at 17:50
  • $\begingroup$ I will. Right now I'm going to "enjoy" a seminar, but I'll check when I come back. $\endgroup$ – Martin Argerami Nov 8 '13 at 17:51
  • $\begingroup$ Have you found mistakes in your or in my answer? $\endgroup$ – Norbert Nov 9 '13 at 16:04
  • $\begingroup$ Sorry for the late reply. I cannot immediately see how you argue that $T$ is surjective. $\endgroup$ – Martin Argerami Nov 9 '13 at 22:42
  • $\begingroup$ Yes this is the hard part. I've proved this earlier in one of my answers, follow this link. For credibile resourse see p 333 in the book Operator spaces by E. G. Effros and Z.J. Ruan $\endgroup$ – Norbert Nov 10 '13 at 6:14

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