By Meglicki Z.

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**Example text**

The natural operations (An)+ (B~) := (A,~ + B,~), (An)(Bn) := (AnB,~), A(An) := (AAn) make 5r to an algebra. This algebra is unital with identity (I[ImL,), and can be normedon defining II(A~)ll:-- supn_>0IIA~L~ll. 2. 13 9~ is a Banach algebra: Proof. Let ((A~m))n>0),~>o be a Cauchy sequence in ~-, that is, e > 0 there is an N such that supllA(~m)Ln-A(~k)L~ll~_¢ forall k,m>_N. 8) n In particular, (A(~m)),~_>0is a Cauchysequence for every fixed n and, thus, convergent. Set An~ := limm-~oA(~m). 8) one obtains sup~ IIA~Ln - A~)Lnll <_ ~ for all k _> N.

7). 10 Let H be an infinite-dimensional separable Hilbert space. (a) If A is an invertible operator on H, then there exists an orthonormal basis of H such that the finite section method with respect to this basis applies to A. (b) IrA = B+iS where B is positive definite and S is self-adjoint, then the finite section method with respec* to every orthonormal basis of H applies to A. Proof. 1. ,en}. For all x E H one has (PnAPnz, Pnx) = (APnx, Pnx) = (BPnx, Pnx) + i(SPnx, Since (BPnx, Pax) and (SPnx, P,~x) are real, this identity implies ](P~APnx, P,,x)] >_ I(ePnx, Pnx)[.

Choose elements Xn ~- X with Axn -~ Yn. The inequalities imply that (Xn) is a Cauchy sequence in X, hence convergent. Set x lira xn. Since A is continuous, one has y,~ = Axn ~. e. the limit of (Yn) belongs to Im A. Thus, A is normally solvable, and the injectivity of A is obvious. 2. 20. Let (An) E re be a st able se quence an d se t C := supn>,~o IIA~ILnlI. Then, for every x E X and all sufficiently large n, IILnxll = IVA’(~IAnLnxll <_ C IIAnLaxlI. e. 21. Repeating these arguments with the adjoint sequence (A~) in place of (An), we get that the kernel of W(A~) = W(An)* is trivial, too.