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: U u • H. su c To e skatral e T Proof. We show that Meo(T). By the spectral mapping theorem 7.4-2 the spectrum of T+ kI (k a real constant) is obtained from that of T by a translation, and Meo(T) M+ke a(T+kI). Hence we may assume 0smSM without loss of generality. Then by the previous theorem we have M = sup (Tx, x)= ||T||. l-1 By the definition of a supremum there is a sequence (x,) such that ||x,|= 1, (Tx, Xn)= M- 8n, 8, 20, →0. Then ||Tx,||T|||x.|=|T|| = M, and since T is self-adjoint, ||Tx, - Mx, = (Tx, – Mx, Tx, - Mx,) = ||Tx, – 2M(Tx, Xn)+ M² ||x,|P SM² -2M(M – 8,) + M² = 2M8, > 0. Hence there is no positive c such that ||Taxn||= ||Tx,- Mx,c = c x,|| (x,|= 1). Theorem 9.1-2 now shows that A M cannot belong to the resolvent set of T. Hence Meo(T). For A m the proof is similar.I %3D %3D Prof
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: U u • H. su c To e skatral e T Proof. We show that Meo(T). By the spectral mapping theorem 7.4-2 the spectrum of T+ kI (k a real constant) is obtained from that of T by a translation, and Meo(T) M+ke a(T+kI). Hence we may assume 0smSM without loss of generality. Then by the previous theorem we have M = sup (Tx, x)= ||T||. l-1 By the definition of a supremum there is a sequence (x,) such that ||x,|= 1, (Tx, Xn)= M- 8n, 8, 20, →0. Then ||Tx,||T|||x.|=|T|| = M, and since T is self-adjoint, ||Tx, - Mx, = (Tx, – Mx, Tx, - Mx,) = ||Tx, – 2M(Tx, Xn)+ M² ||x,|P SM² -2M(M – 8,) + M² = 2M8, > 0. Hence there is no positive c such that ||Taxn||= ||Tx,- Mx,c = c x,|| (x,|= 1). Theorem 9.1-2 now shows that A M cannot belong to the resolvent set of T. Hence Meo(T). For A m the proof is similar.I %3D %3D Prof
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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please proof with smallest detailsi iwant you to proof Underneath a red line.
Transcribed Image 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 <T s e skatral e t T
Proof. We show that Meo(T). By the spectral mapping theorem
7.4-2 the spectrum of T+ kI (k a real constant) is obtained from that
of T by a translation, and
Meo(T)
M+kea(T+kI).
Hence we may assume 0smSM without loss of generality. Then by
the previous theorem we have
M = sup (Tx, x)= ||T||.
l-1
By the definition of a supremum there is a sequence (x,) such that
||x|= 1,
(Tx, Xn)= M- 8n,
&, 20,
→0.
Then ||Tx,|T|||x.|= |T|| = M, and since T is self-adjoint,
||Tx, - Mx, = (Tx, - Mx, Tx, - Mx,)
= ||Tx, – 2M(Tx, Xn)+ M² ||x,
SM-2M(M- 8,) + M² = 2M8,
0.
Hence there is no positive c such that
||Taxn||= ||Tx,- Mx,c = c |x,||
(x|= 1).
Theorem 9.1-2 now shows that A M cannot belong to the resolvent
set of T. Hence Meo(T). For A= m the proof is similar. I
%3D
%3D
Prof
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