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32) in the following table and then give some examples. 1 R(λ; L) exists and is bounded R(λ; L) exists and is unbounded R(λ; L) does not exist R(λI − L) = X λ ∈ ρ(L) ———— λ ∈ σp (L) R(λI − L) = X λ ∈ ρ(L) λ ∈ σc (L) λ ∈ σp (L) R(λI − L) = X λ ∈ σr (L) λ ∈ σr (L) λ ∈ σp (L) Observe that the case in the first row and second column cannot occur in a Banach space X, by the closed graph theorem. 6) (on a possibly “thin” domain of definition) as “algebraic” inverse of λI − L. This will be done in several examples in the sequel.

Let X = Y = R and 1 − 2 x 1 F (x) = 4 x x − 43 if x ≤ 0, if 0 ≤ x ≤ 1, if x ≥ 1. Then [F ]lip = 0, [F ]b = 41 , [F ]q = 21 , [F ]Q = [F ]B = [F ]Lip = 1. 27. Let X = Y = R and 1 − x− 21 −8x F (x) = x 1 3 4x + 4 3 8 if x ≤ −1, if −1 ≤ x ≤ 0, if 0 ≤ x ≤ 1, if x ≥ 1. Then [F ]lip = 0, [F ]b = 18 , [F ]q = 41 , [F ]Q = 21 , [F ]B = [F ]Lip = 1. 28. Let X = Y = R and 3 − 4 x 1x F (x) = 23 x−1 2 x if x ≤ 0, if 0 ≤ x ≤ 1, if 1 ≤ x ≤ 2, if x ≥ 2. Then [F ]lip = 0, [F ]b = 21 , [F ]q = 43 , [F ]Q = [F ]B = 1, [F ]Lip = 23 .

Consequently, the set N1 = {x1 , x2 , x3 . . } is precompact, and hence xnk → x∗ (k → ∞) for some subsequence (xnk )k . But the limit x∗ belongs then to the intersection M∞ , and so M∞ = ∅. The compactness of M∞ follows from the assumption that α(Mn ) → 0 as n → ∞, and the fact that M∞ is closed. Now let L : X → Y be a bounded linear operator. 25) for any bounded subset M ⊂ X, where at least k = L . In fact, if {z1 , . . , zm } is a finite ε-net for M, then obviously {Lz1 , . . , Lzm } is a finite L ε-net for L(M).

### Analysis of Ctl Rod Guide Tube, Subpile Room of ANS Reactor

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