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By Dr. rer. nat. Albrecht Böttcher, Prof. Dr. sc. nat. Bernd Silbermann (auth.)

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Since both (T(a) x;, Xk) and (AXI' Xt) equal ai-I' it fo11ows that A = T(a). 2 gives the first ":::;;:" in (1). The second ":::;" in (1) is trivial. (b) Since Aen E IP for a11 nE ZH it is obvious that the sequence {an}nez belongs to 1"(Z). After defining B as B: IO(Z) ~ 1"(Z) , {9'j}jell H- {E aHt9'a:}jeZ tEil the proof is completely analogous to (a). 8. Coro11ary. ) E l'(IQ) i8 eq:ual to T(ä). r(lI) 10t' all rE [p, q] < lIal/w, Ip - 21). Proof. 7. 9. Shift operators. Reca11 that Xn is given by Xn(t) = ,n (I E T).

The proof for the right invertibility is analogous. It is elear that aN is left invertible if a is so. , assume aN is left invertible in AN for a11 NE M(B) but let a have no left inverse in A. 16(a)). Put N:= L n B. 33, NE M(B). We claim that J N C L. Indeed, if x = 1: x/tak> where X/t E N = L n Band at E A, then x = 1: akxk (because Be Cen A), and hence xE L (because L is a left ideal). Thus, J N C L. By our assumption, aN is left invertible in AN, that is, there exists a b E A with ba - e E J N, and since J N c L, we have ba - e E L.

Then a E LP and the n-th Fourier coefficient of a is (a, Xn) = (AXo, Xn) = an' If I E Loo, then both AI and al are in LP. We claim that (1) AI =al Let {/n}neJZ denote the Fourier coefficient sequence of I. Then the j-th Fourier coefficient of al is E aj-klk' On the other hand, since the series E IkXk converges to I in the LPkeJZ norm, we deduce that the series E Ik(AXk> Xi) keJZ keJZ converges to (AI, Xi) for each jE Z. This shows that the j-th Fourier coefficient of AI equals E ai-kib too. keJZ Thus, AI = a,f.

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