By de la Harpe P., Jones V.
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15, which constitutes an epsilon of spectral theory. 13. Observation. 14. Lemma. Let a be a compact self-adjoint operator on H: Then one at least of the numbers kak ;kak is an eigenvalue of a: Proof. 1 j j a(a )i 2 f1 ;1g: j Upon replacing ( j )j N by a subsequence, one may assume by compacity of a that the sequence (a ( j ))j N converges to some 2 H(1): Then 2 2 h j a i 2 f1 ;1g and the proof follows from the previous observation. 15. Proposition. Let a be a compact self-adjoint operator on H: Then there exists an orthonormal basis ( j )j J consisting of eigenvectors of a and the corresponding sequence ( j )j J of eigenvalues converges to 0: Proof.
Kf k for all f 2 A: But A is not a C -algebra indeed, for f de ned by f (t) = 0 for t 0 and f (t) = t for t 0 one has kf k = 1 and f ! f = 0: P On the convolution algebra A = `1( ) consider the norm de ned by kck1 = n2Z jc(n)j and the involution de ned by c (n) = c(;n): Then A is also a Banach algebra with an involution such that kc k1 = kck1 for all c 2 A: But A is not a C -algebra indeed, for c de ned by c(1) = c(0) = ;c(;1) = 1 and c(n) = 0 when jnj 2 one has kck1 = 3 and kc ck1 = 5: More generally, for any in nite locally compact abelian group G the convolution algebra L1(G) is not a C -algebra.
3. Suppose now that A has no unit. For each x = ( a) 2 A~ set kxk = sup kxbk : b2A kbk 1 Then x 7! 4. Let us check for example that kxk = 0 implies x = 0: If = 0 this is because a 7! kak is a norm on A:; If one had kxk = 0 for x = ( a) with 6= 0 one would have b + ab = 0 hence also ; ;1a b = b for all b 2 A so that ; ;1a would be a left-unit for A hence ; ;1a would be a right-unit for A and it would follow that A has a unit, in contradiction with the hypothesis. The norm just de ned makes A~ complete, because A~ contains a one-codimensional complete subspace A: Let us nally check that kx xk = kxk2 ~ We may assume without loss of generality that kxk = 1 and we for all x = ( a) 2 A: have to check that kx xk = 1: For each real number r < 1 there exists b 2 A such that kbk 1 and kxbk r: As xb 2 A one has kx xk kb k kx xk kbk k(xb) xbk = kxbk2 r2 : As this holds for all r < 1 one has kx xk 1 and consequently kx xk = 1: In conclusion, A~ is naturally a C -algebra whenever A is a C -algebra.