9.2-3 Theorem (m and M as spectral values). Let H and T be as in Theorem 9.2-1 and H# {0}. Then m and M defined in (1) are spectral values of T. T: UA u M. su


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9.2-3 Theorem (m and M as spectral values). Let H and T be as in<br>Theorem 9.2-1 and H# {0}. Then m and M defined in (1) are spectral<br>values of T. T: UA u<br>M. su <T s e skatral e t T<br>Proof. We show that Meo(T). By the spectral mapping theorem<br>7.4-2 the spectrum of T+ kI (k a real constant) is obtained from that<br>of T by a translation, and<br>Meo(T)<br>M+kea(T+kI).<br>Hence we may assume 0smSM without loss of generality. Then by<br>the previous theorem we have<br>M = sup (Tx, x)= ||T||.<br>l-1<br>By the definition of a supremum there is a sequence (x,) such that<br>||x|= 1,<br>(Tx, Xn)= M- 8n,<br>&, 20,<br>→0.<br>Then ||Tx,|T|||x.|= |T|| = M, and since T is self-adjoint,<br>||Tx, - Mx, = (Tx, - Mx, Tx, - Mx,)<br>= ||Tx, – 2M(Tx, Xn)+ M² ||x,<br>SM-2M(M- 8,) + M² = 2M8,<br>0.<br>Hence there is no positive c such that<br>||Taxn||= ||Tx,- Mx,c = c |x,||<br>(x|= 1).<br>Theorem 9.1-2 now shows that A M cannot belong to the resolvent<br>set of T. Hence Meo(T). For A= m the proof is similar. I<br>%3D<br>%3D<br>Prof<br>

Extracted text: 9.2-3 Theorem (m and M as spectral values). Let H and T be as in Theorem 9.2-1 and H# {0}. Then m and M defined in (1) are spectral values of T. T: UA u M. su

Jun 05, 2022
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