Some New Results on Strictly Singular Operators and Pre- Compact Operators

INTRODUCTION

A compact administrator is only explicit, despite the fact that all in all the opposite isn't substantial. Van Dulst first concentrated carefully remarkable operators on Ptak (or B-complete) spaces and summed up Hilbert spaces as far as lcs. For the class of Br-complete spaces, Wrobel recognized carefully particular operators on lcs's. In the event that the pair (X, Y) has a place with the class of Banach spaces, Ls(X, Y) is a proficient administrator. This isn't the situation, as observed in [5], since it has a place with the overall class of lcs. However, each class of purely special bounded operators LBs (E, F) and compact operators Lc(E, F) is an ideal operator in lcs. By [22], if F has the property (y), for Fréchet spaces E and F, LBs (E, F) = Ls(E, F); If, as a quotient, E includes l1, then LBs(E, F) Ls(E, F).

Proof. Suppose that T : E → F and S : E → F are strictly singular operators. Then, for any M ≤ E, by [28, Theorem 1-IV], find N ≤ M such that T |N is pre-compact. Then find P ≤ N such that S|P is pre-compact. The ideal property of pre-compact operators on lcs’s yields the result. We give the accompanying recommendation as a use of administrator ideal property of carefully particular operators on Br-complete lcs's. It is a speculation of [1, 2010 Mathematics Subject Classification. 46A03, 46A11, 46A32, 46A45.Problem 4.5.2], and in particular, it is also true when re-stated for bounded strictly singular operators acting on general lcs’s.

Proof. Assume that each Tji is strictly singular. Let πi : E → Ei be the canonical projection and define 0, for which yj is the j-th summand. Consider and write where Tji is the j-th summand. By Lemma 1 and rewriting

Where M places in the i-th summand, M^ is a vector subspace of X. Nrs(τMˆ) > ε, for all r ∈ I. Yet, substitute compactness with absolute uniqueness, the results reveal that a purely singular operator T: E ⁇ F between lcs may not be unbounded, as such an inference contributes to an inconsistency with Bessaga, Pelczynski, and Rolewicz's well-known outcome. In Section 2 we survey the additional assumptions in which a bounded operator is immediately purely singular between two lcs.

In this segment, the extra speculations for a limited administrator's exacting peculiarity are included. Our beginning stage would be the Banach Spaces class. Utilizations of [7, Lemma 2] can bring about any of these discoveries getting significant for the lcs class. In a given case, it is important to accomplish a characterization (Theorem 3). For L(X, Y) = Ls(X, Y) in Banach spaces, the most generally perceived non-insignificant case is when X = lp and Y = lq with the end goal that 1 = p < q < ⁇.This concept was important in the isomorphic classification of Cartesian power series spaces and in the issue of whether the sum of two supplemented subspaces is also supplemented[15] (if E subspaces are supplemented by X and Y, X+Y is supplemented by E if LB(X, Y) = Ls(X, Y)). If a weakly convergent sequence in X converges in norm, a Banach space X is said to have the Schur property (SP). On the off chance that a compelled arrangement is feebly Cauchy, X is viewed as practically reflexive. X is viewed as pitifully successively complete (wsc) if any feebly joined Cauchy arrangement in X is powerless. X is said to have the Dieudonné property (DP) if operators on X are feebly compact while changing frail Cauchy successions into pitifully joined groupings. A couple of Banach spaces (X,Y) is considered completely unique if there is no Banach space Z isomorphic to a subspace of X and Y. A property P is viewed as innate on a Banach space X if any M ⁇ X cherishes it. X is expected to have P no place if the property P has no subspace. By Lw and Lv respectively we denote the groups of weakly compact and wholly continuous (or complete) operators. The Dunford-Pettis property (DPP) is said to have a Banach space X if Lw(X, Y) ⁇ Lv(X, Y), for each Banach space Y. X, if Lv(X, Y) ⁇ Lw(X, Y), is said to have the proportional Dunford-Pettis property (rDPP). Lemma 2. Leave X and Y single Banach spaces. 1. Let X be practically reflexive.Then, for any wsc Banach space Y, L (X, Y) = Ls(X, Y). 2. Let X have DP, and let Y be wsc. Then, L(X, Y) = Lw(X, Y). 4. Let X be a Banach space with SP. Then, for every M ≤ X, 1 → M.

1. Because X is practically reflexive, at that point (Txn) has a feebly Cauchy grouping in Y if (xn) is a limited arrangement in X. But Y is wsc, that is, any weak sequence of Cauchy weakly converges in Y. T is weakly compact, along these lines. 2. Let (xn) x be Cauchy weakly, and let T be L(X, Y). Then (Txn) is Cauchy weakly in Y. Since Y is believed to be wsc, (Txn) weakly converges. X, though, has DP, so T x Lw(X, Y). 3. See [16, 1.7], Theorem. 4. Let X has the SP and concludes that M does not contain. Any bounded sequence (xn) in M then has a weak Cauchy subsequence because M is almost reflexive equivalently. M should, though, inherit SP. The poor Cauchy series of (xk) then converges into X. Therefore, M is dimensionally finite Disagreement.

1. X and Y are absolutely exceptional. 2. X is nowhere reflexive, Y is reflexive. 3. X is nowhere reflexive, Y is quasi-reflexive. 4. X is practically reflexive and no place reflexive, Y is wsc (see Example 1). 5. Y has hereditary P, X has nowhere P. 6. Y is almost reflexive, X is hereditarily- 1. 7. X has SP, Y is almost reflexive. 8. X is reflexive, Y has SP. 9. X has the hereditary DPP, Y is reflexive. 10. L(X, Y ) = Lw(X, Y ) and X has DPP. 11. L(X, Y ) = Lv(X, Y ) and X has rDPP. 12. X is a Grothendieck space with DPP, Y is separable.

1. Assume T : X → Y is a non-strictly singular operator. So find M ≤ X on which. Since X and Y are totally incomparable, this is impossible. 2. See Theorem This result is generalized in part 5. 3. Suppose there exists a non-carefully particular administrator T ∈ L(X, Y ). At that point, T is an isomorphism when limited to M ≤ X, so M is semi reflexive. In any case, by [12, Lemma 2] there exists a reflexive N ≤ M. This repudiates the presumption X is no place reflexive. 4. By section 1 of Lemma 2, L(X, Y ) = Lw(X, Y). Presently let T : X → Y which has a limited backwards on M ≤ X. In the event that (xn) is a limited arrangement in M, at that point there exists (Txkn) a feebly joined aftereffect of (Txn) in Y . Henceforth (xkn) is feebly united in M, since T has a limited backwards on M. In this manner, each limited grouping in M has a feebly merged aftereffect in M. So M is reflexive Contradiction. 5. For some M ≤ X assume there exists T: X → Y with the end goal that M ≃ T (M). Be that as it may, T (M) acquires P. Consequently M has P. This repudiates X has no place P. Presently let S: Y → X be with the end goal that for some N ≤ Y. Since X has no place P, S(N ) abhors P . Logical inconsistency. 6. A specific instance of section 5. 7. Any administrator T with extends Y maps limited groupings into feebly Cauchy arrangements, since Y is practically reflexive. Then again, any such T de-fined on X maps pitifully Cauchy groupings into standard focalized successions by the SP. That infers L(X, Y) = Lv(X, Y). Subsequently by section 3 of Lemma 2 the outcome follows. 8. Let T ∈ L(X, Y) have a limited reverse on some M ≤ X, that is, (M). Since Y have SP, it likewise has the inherited DPP [6]. Thus does as well M. In any case, M is reflexive. By [14, Theorem 2.1], reflexive spaces have no place DPP. Logical inconsistency. Henceforth M can't have DPP. Inconsistency. 10. Since X has DPP, L(X, Y) ∈ Lv(X, Y). At that point, by [16, Theorem 2.3], the outcome follows. 11. Since X has the rDPP and L(X, Y) = Lv(X, Y), T ∈ Lv ∩ Lw. By [16, Theorem 2.3], we are finished. 12. By [21, Theorem 4.9], any such administrator T: X → Y is pitifully compact. Since X has the DPP, T is totally ceaseless. By section 10, we arrive at the outcome. 13. By section 2 of Lemma 2, L(X, Y) = Lw(X, Y). X has the DPP, so L(X, Y ) = Lv(X, Y ). By [16, Theorem 2.3], the evidence is finished.

1. If X′, Y ′, Z′ have SP and W is almost reflexive, every operator defined on

1. By [16, Corollary 1.6], L(W, Z′) = L (W, Z′). So by [9, Theorem 3] we deduce W ⊗πZ is almost reflexive. On the other hand, by [23, Theorem 3.3(b)] we reach that L(X, Y ′) has SP. But in [24] it is proved that 2. By [24, Theorem 4.21], is reflexive. By [18], SP respects injective tensor products. So has SP. Then, part 8 of Theorem 1 finishes the proof. 3. By [6], Y ′ has SP. Then by [16, Corollary 1.6], L(X, Y ′) = L (X, Y ′). Hence, [9, 4. Theorem 3] yields that is almost reflexive. It is clear that every operator defined from an almost reflexive space into ℓ1 is strictly singular. 5. Because X is almost reflexive, if (xn) in X is a small series, then (Txn) in Y has a poor Cauchy series. But Y is wsc, that is, any weak series of Cauchy converges in Y weakly. T is therefore weakly compact. 6. Let (xn) x be Cauchy weakly, and let T be L(X, Y). So (Txn) Cauchy is small in Y. Because Y is believed to be wsc, (Txn) is weakly converging. X, though, has DP, so T ⁇ Lw(X, Y).AlsoTheorem. 7. Let X have the SP and reason that M/X doesn't contain. Any limited arrangement (xn) in M at that point has a pitifully cauchy aftereffect since M is practically reflexive comparably. M acquires SP, however. The poor Cauchy arrangement of (xk) at that point unites with X. M is subsequently limited dimensional. The irregularity. Let λ1(A) ∈ (d2), and λp(A) ∈ (d1) as in [8]. Then, by [30], L(λ1(A), λp(A)) = LB(λ1(A), λp(A)). For 1 ≤ p < ∞, we know that λp(A) = proj lim ℓp(an). Since ℓp(an), 1 < p < ∞ has no subspace isomorphic to ℓ1, L(ℓ1, ℓp) = Ls(ℓ1, ℓp). Then, by [7, Lemma 2], L(λ1(A), λp(A)) = Ls(λ1(A), λp(A)). Resting on the same argument, to obtain several sufficient conditions for L(E, F ) = Ls(E, F ) is possible for the class of general lcs’s.

3. E is infra-Schwartz and F ∈ s(V) 4. E ∈ s(P¬) and F ∈ s(P)

duplicate of 1 with the end goal that F = proj lim Fm. Because E ∈ s(X), there exists a family of Banach spaces {Ek} such that every Mk ≤ Ek contains a subspace isomorphic to ℓ1. By part 6 of Theorem 1, any linear operator Tmk : Ek → Fm is strictly singular. Making use of [7, Lemma 2], we reach the result. 2. By [19, Theorem 6], F is locally Rosenthal. Since E ∈ s(V), by part 4 of Lemma 2, E ∈ s(X). Then, by part 1, we are done. 3. Since E is infra-Schwartz, any of its local Banach spaces Ek is reflexive. The assumption on F completes the conditions in part 8 of Theorem 1. Combined with [7, Lemma 2], we are done. 4. Since E ∈ s(P¬), one may rewrite E = proj lim E , where each E has no subspace having property P . Similarly, F = proj limm Fm where each Fm is hereditarily P . Hence, by part 5 of Theorem 1, L(Ek, Fm) = Ls(Ek, Fm) for every k, m. Applying [7, Lemma 2], we obtain LB(E, F ) = Ls(E, F ).

Proof. The adequacy aspect is quite close to the proof of Part 4 of Theorem 2. Let E be a Fréchet space, for requirement, and let F be a (FM)-space admitting a continuous norm. Let each linear T operator be purely special. And it is bounded by [29, Proposition 1]. Let N Y, which is isomorphic to G, live now. And I: N is enclosed, compact, thus. All N is Dimensional Finite. Disagreement.

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Associate Professor & Head of Department, B.N. college, Patna University, Patna