Operator whose spectrum is different from the spectrum of its dual

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It's mentioned in our functional analysis course that for any operator $T$ on a Banach space, T has the same spectrum as its dual. A friend asked for an example where this fails and I can't think of one for the life of me.

As mentioned, it's clear that the space can't be complete. My thoughts are that if $T$ is a bdd linear operator on $X$, $T$ extends uniquely to a bdd operator $\tilde{T}$ on the completion $\tilde{X}$. Further, $X$ and $\tilde{X}$ have the same dual space, and also $T$ and $\tilde{T}$ have the same dual operator on this space.

Because of this, the spectrum of $T^{*}$ coincides with the spectrum of $\tilde{T}$, so really what we want is an operator $T$ on a non-complete space whose unique extension to the completion has a different spectrum.

My best attempt was to look at some $\ell^{p}$ space ($p<\infty$) as that has the canonical Schauder basis $(e_n)_n$ and take $X$ to be the linear span of this basis. Then try to define an operator that isn't surjective on $X$, but does extend to an invertible operator on $\ell^p$. Clearly I was unsuccessful :)

It has crossed my mind that the result about Banach spaces uses the axiom of Choice (through the use of Hahn-Banach) so maybe to find an example where it fails I also need Choice, but I didn't manage to come up with and example even when allowing for "non-canonical" ones (e.g starting with an algebraic basis of a Banach space and trying to work from there).

If anyone has a construction for this, it would be greatly appreciated.

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Let $\{e_n\}_{n\in\mathbb{Z}}$ be an orthonormal basis in a Hilbert space $\mathcal{H}$. Let $V$ consists of finite linear combinations of the elements of the basis. The operator $S$ on $V $defined by $Se_n=e_{n+1}$ extends to the unitary operator $\tilde{S}$ on $\mathcal{H}.$ The spectrum of $S$ is equal $\mathbb{C}\setminus\{0\},$ while the spectrum of $\tilde{S}$ coincides with the unit circle.

Concerning the particular question in OP, the operator $2I-S$ is not surjective on $V$ but $2I- \tilde{S}$ is invertible, therefore surjective.