A NOTE ON TAUBERIAN OPERATORS

In this note we prove the existence of operators which are not Tauberian even though they satisfy properties about restrictions being Tauberian. The operators are defined on Banach spaces which contain a somewhat reflexive, non-reflexive subspace. This gives an answer to a question proposed by R. Neidinger [1].


Given T
L(E,F) the notation TIZ denotes the restriction of T to the subspace Z of E.
Recall that an operator T E L(E,F) is said to be semi-Fredholm if its null space N(T), is finite-dimenslonal and its range space R(T) is closed.Also, a Tauberian operator, as defined by D. Garllng and A. Wllansky in [2], is a bounded linear operator T E L(E,F) such that T" preserves the natural embedding of E into its double dual, i.e., T"x" F implies x" E. Some relationships between these tw classes of operators have been studied in [i], [3], [4] and [5].In particular, if R(T) is closed, then T is Tauberian if only if N(T) is reflexive.
It is well-known that the restriction of a semi-Fredholm operator to any closed subspace is again a semi-Fredholm operator.
In the opposite direction it is x)rthwhile to mention the following result that is basically due to T. Kato [6], THEOREM I (c.f.[6]).
Let E,F be inflnlte-dlmenslonal Banach spaces.Assume that T:E----+ F is an operator such that every inflnlte-dlmenslonal closed subspace Z of E contains an inflntle-dlmenslonal closed subspace W for which TIW is semi- Fredholm.Then T is semi-Fredholm.
It follows that in order to see that a given operator T is semi-Fredholm, it is enough to assure that its restriction to every closed subspace with a Schauder basis is semi-Fr edholm.
Another related result is the following theorem due to R. Neldlnger in which Banach spaces with no inflnlte-dlmenslonal reflexive subspace are called "purely non- reflexive" spaces.
THEOREM 2 ([I], p. 26).Let E be a weakly sequentially complete Banach space and let T L(E,F).Then T is Tauberlan if (and only if) TIZ is semi-Fredholm for all purely non-reflexlve closed subspaces Z of E.
In view of the preceding theorem, R.
Neldlnger raised the following question ([I], p. 139): If T E L(E,F), restricted to any purely non-reflexlve closed subspace is semi-Fredholm, is T Tauberlan?.
Indeed, the answer is positive if E is reflexive.
Then, we assne that E is not reflexive.
In this case there are some trivial situations for which the answer is negative (e.g., let E be a somewhat reflexive space, that Is, every inflnlte-dlmenslonal subspace of E contains an Inflnlte-dlmenslonal reflexive subspace, and let T be a finite rank operator).Our next example gives a negative answer to the question raised by R. Neldlnger in a non- trivial situation.
EXAFPLE.Let J be the James space and let T:J x ii----+ I I be the operator defined by T(x,y) y.
Since R(T) is closed and N(T) J is not reflexive then, T is not Tauberlan.Now, let Z be a purely non-reflexlve closed subspace of J x i I. Since J is somewhat N(TIZ) N(T) O Z is flnlte-dlmenslonal; otherrlse, N(TIZ would contain reflexive, an inflnlte-dlmenslonal reflexive subspace, which contradicts our assumption over Z.
Also, N(T) and Z are totally incomparable Banach spaces (i.e., there exists no inflnlte-dlmenslonal Banach space which is isomorphic to a subspace of N(T) and to a i subspace of Z).This implies that N(T) + Z is closed in J x 1 [7] and hence, R(TIZ is closed by the open mapping theorem.

T(Z)
TIZ is semi-Fredholm for all purely non-reflexive Thus, closed subspaces.

MAIN RESULTS.
Another related problem is as follows; we know that the restriction of a Tauberlan operator to any closed subspace is again Tauberlan.
So, is Theorem 1 true for Tauberlan operators instead of semi-Fredholm operators?.The answer is obvlously positive if, for instance E is reflexlve or E is purely non-refelxlve.However, we have, THEOREM 3.
Let E be an Inflnlte-dlmenslonal Banach space which contains an Inflnlte-dlmenslonal somewhat reflexlve closed subspace M which is not reflexive.