By S.M. Khaleelulla

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**Extra resources for Counterexamples in Topological Vector Spaces**

**Example text**

O. 2. Let F: lP' ....... C(H) be continuous on the closed chain lP'. Given c > 0, there exists a partition 7r such that for any partition {Qo, QI,' .. ,Qm} :J 7r 1 ~ j ~ m, provided (Qj-I,Qj) is not a jump of lP' which lies in PROOF. Put Po = O. '" P n }, we start by defining {p E lP' IIIF(P) - F(Po)1I 2: c/2}. take 7r = {O, I}, and we are done. Suppose ~o (Po, PI) is a jump of lP', take PI = if not, take If ~O = 0, = 7r 7r. A; =/: 0. PI = max{P E lP' I P ~ PI, IIF(P) - F(Po)1I ~ Let PI c/2}. = min ~o.

Let ii be the space of all complex-valued absolutely continuous functions defined on [0,1] with 1-f(O) = 0 and f' E L2([0, 1]). Define the inner product on H by (I,g) = fa f'(s)g'(s)ds, and let R(t,s) (2) o ~ t, = t /\ s = min{t,s}, s ~ 1. The inner product space H is complete since the map U defined by J t (UJ)(t) = (3) o~ t f(s)ds, ~ 1, a is a unitary operator mapping L2([0, 1]) onto J ii. )) = f'(s)ds = f(t), o~ t ~ 1, a for each f E ii. Thus ii is a RKHS. A reproducing kernel Hilbert space H has only one reproducing kernel.

Now, take v> O. , if v -=f:. IL, then SOv is not similar to Sal"' So on a separable Hilbert space there is a continuum of non-similar unicellular operators. In order to prove that the operators SOv (v > 0) are mutually non-similar, we first note that the j-th singular value of the operator So is equal to aj. 3, we have the following inequalities: 11F-11I-111F1I- 1 ~ s'(F-IS F) J sj(S3 ~ IIF-11111F1I, j = 1,2, .... Now j = 1,2, ... , and the latter sequence is bounded and bounded away from zero if and only if v = IL.