1 多項式のスペクトル写像と連続汎関数計算 命題 1.1. H ≠ { 0 } H\ne\{0\} H = { 0 } T ∈ B ( H ) T\in B(H) T ∈ B ( H ) p ( z ) = ∑ k = 0 n c k z k p(z)=\sum_{k=0}^nc_kz^k p ( z ) = ∑ k = 0 n c k z k T 0 = I T^0=I T 0 = I p ( T ) = ∑ k = 0 n c k T k p(T)=\sum_{k=0}^nc_kT^k p ( T ) = ∑ k = 0 n c k T k 0 ∈ ρ ( T ) 0\in\rho(T) 0 ∈ ρ ( T ) T − 1 = − R ( 0 , T ) T^{-1}=-R(0,T) T − 1 = − R ( 0 , T ) n ∈ N ≥ 0 n\in\N n ∈ N ≥ 0 c − n , … , c n ∈ C c_{-n},\dots,c_n\in\C c − n , … , c n ∈ C q ( z ) = ∑ k = − n n c k z k q(z)=\sum_{k=-n}^nc_kz^k q ( z ) = ∑ k = − n n c k z k q : C ∖ { 0 } → C q\colon\C\setminus\{0\}\to\C q : C ∖ { 0 } → C k < 0 k<0 k < 0 T k = ( T − 1 ) − k T^k=(T^{-1})^{-k} T k = ( T − 1 ) − k q ( T ) = ∑ k = − n n c k T k q(T)=\sum_{k=-n}^nc_kT^k q ( T ) = ∑ k = − n n c k T k
すべての多項式p p p σ ( p ( T ) ) = { p ( λ ) : λ ∈ σ ( T ) } \sigma(p(T))=\{p(\lambda):\lambda\in\sigma(T)\} σ ( p ( T )) = { p ( λ ) : λ ∈ σ ( T )}
0 ∈ ρ ( T ) 0\in\rho(T) 0 ∈ ρ ( T ) q q q σ ( q ( T ) ) = { q ( λ ) : λ ∈ σ ( T ) } \sigma(q(T))=\{q(\lambda):\lambda\in\sigma(T)\} σ ( q ( T )) = { q ( λ ) : λ ∈ σ ( T )} 証明. 主張 (1) を示す。μ ∈ C \mu\in\C μ ∈ C p p p c c c μ I − p ( T ) = ( μ − c ) I \mu I-p(T)=(\mu-c)I μ I − p ( T ) = ( μ − c ) I μ ≠ c \mu\ne c μ = c ( μ − c ) − 1 I (\mu-c)^{-1}I ( μ − c ) − 1 I B ( H ) B(H) B ( H ) μ = c \mu=c μ = c H ≠ { 0 } H\ne\{0\} H = { 0 } V ∈ B ( H ) V\in B(H) V ∈ B ( H ) 0 ⋅ V = 0 ≠ I 0\cdot V=0\ne I 0 ⋅ V = 0 = I μ I − p ( T ) = 0 \mu I-p(T)=0 μ I − p ( T ) = 0 B ( H ) B(H) B ( H ) σ ( p ( T ) ) = { c } \sigma(p(T))=\{c\} σ ( p ( T )) = { c } §E12.1 定理 4.1 によりσ ( T ) ≠ ∅ \sigma(T)\ne\emptyset σ ( T ) = ∅ { p ( λ ) : λ ∈ σ ( T ) } = { c } \{p(\lambda):\lambda\in\sigma(T)\}=\{c\} { p ( λ ) : λ ∈ σ ( T )} = { c }
p p p n n n 1 1 1 c n ≠ 0 c_n\ne0 c n = 0 §E5.9 系 4.2 により多項式p − μ p-\mu p − μ α \alpha α z − α z-\alpha z − α n n n α 1 , … , α n ∈ C \alpha_1,\dots,\alpha_n\in\C α 1 , … , α n ∈ C
p ( z ) − μ = c n ( z − α 1 ) ⋯ ( z − α n ) p(z)-\mu=c_n(z-\alpha_1)\cdots(z-\alpha_n) p ( z ) − μ = c n ( z − α 1 ) ⋯ ( z − α n ) を満たすものを得る。p ( λ ) = μ p(\lambda)=\mu p ( λ ) = μ λ ∈ C \lambda\in\C λ ∈ C { α 1 , … , α n } \{\alpha_1,\dots,\alpha_n\} { α 1 , … , α n } S j = α j I − T S_j=\alpha_jI-T S j = α j I − T T T T S 1 , … , S n S_1,\dots,S_n S 1 , … , S n
μ I − p ( T ) = ( − 1 ) n + 1 c n S 1 ⋯ S n \mu I-p(T)=(-1)^{n+1}c_nS_1\cdots S_n μ I − p ( T ) = ( − 1 ) n + 1 c n S 1 ⋯ S n である。
互いに可換なS 1 , … , S n ∈ B ( H ) S_1,\dots,S_n\in B(H) S 1 , … , S n ∈ B ( H ) R = S 1 ⋯ S n R=S_1\cdots S_n R = S 1 ⋯ S n B ( H ) B(H) B ( H ) S j S_j S j B ( H ) B(H) B ( H ) S j S_j S j V j V_j V j V n ⋯ V 1 V_n\cdots V_1 V n ⋯ V 1 R R R R R R V V V j j j S j S_j S j n − 1 n-1 n − 1 R j R_j R j n = 1 n=1 n = 1 R j = I R_j=I R j = I S j R j = R j S j = R S_jR_j=R_jS_j=R S j R j = R j S j = R S j ( R j V ) = R V = I S_j(R_jV)=RV=I S j ( R j V ) = R V = I ( V R j ) S j = V R = I (VR_j)S_j=VR=I ( V R j ) S j = V R = I
V R j = V R j ( S j R j V ) = ( V R j S j ) R j V = R j V VR_j=VR_j(S_jR_jV)=(VR_jS_j)R_jV=R_jV V R j = V R j ( S j R j V ) = ( V R j S j ) R j V = R j V であるから、R j V R_jV R j V S j S_j S j B ( H ) B(H) B ( H )
c n ≠ 0 c_n\ne0 c n = 0 μ ∈ ρ ( p ( T ) ) \mu\in\rho(p(T)) μ ∈ ρ ( p ( T )) j j j α j ∈ ρ ( T ) \alpha_j\in\rho(T) α j ∈ ρ ( T ) p ( λ ) = μ p(\lambda)=\mu p ( λ ) = μ λ \lambda λ σ ( T ) \sigma(T) σ ( T ) σ ( p ( T ) ) = { p ( λ ) : λ ∈ σ ( T ) } \sigma(p(T))=\{p(\lambda):\lambda\in\sigma(T)\} σ ( p ( T )) = { p ( λ ) : λ ∈ σ ( T )}
主張 (2) を示す。0 ∈ ρ ( T ) 0\in\rho(T) 0 ∈ ρ ( T ) q ( z ) = ∑ k = − n n c k z k q(z)=\sum_{k=-n}^nc_kz^k q ( z ) = ∑ k = − n n c k z k μ ∈ C \mu\in\C μ ∈ C s ( z ) = ∑ k = − n n c k z k + n − μ z n s(z)=\sum_{k=-n}^nc_kz^{k+n}-\mu z^n s ( z ) = ∑ k = − n n c k z k + n − μ z n z ≠ 0 z\ne0 z = 0 s ( z ) = z n ( q ( z ) − μ ) s(z)=z^n(q(z)-\mu) s ( z ) = z n ( q ( z ) − μ ) T T − 1 = T − 1 T = I TT^{-1}=T^{-1}T=I T T − 1 = T − 1 T = I s ( T ) = T n ( q ( T ) − μ I ) s(T)=T^n(q(T)-\mu I) s ( T ) = T n ( q ( T ) − μ I ) T n T^n T n ( T − 1 ) n (T^{-1})^n ( T − 1 ) n μ I − q ( T ) = − ( T − 1 ) n s ( T ) \mu I-q(T)=-(T^{-1})^ns(T) μ I − q ( T ) = − ( T − 1 ) n s ( T ) B ( H ) B(H) B ( H ) s ( T ) s(T) s ( T ) B ( H ) B(H) B ( H ) 0 ∉ σ ( s ( T ) ) 0\notin\sigma(s(T)) 0 ∈ / σ ( s ( T )) 主張 (1) により、これはs ( λ ) = 0 s(\lambda)=0 s ( λ ) = 0 λ \lambda λ σ ( T ) \sigma(T) σ ( T ) 0 ∉ σ ( T ) 0\notin\sigma(T) 0 ∈ / σ ( T ) λ ∈ σ ( T ) \lambda\in\sigma(T) λ ∈ σ ( T ) s ( λ ) = 0 s(\lambda)=0 s ( λ ) = 0 q ( λ ) = μ q(\lambda)=\mu q ( λ ) = μ σ ( q ( T ) ) = { q ( λ ) : λ ∈ σ ( T ) } \sigma(q(T))=\{q(\lambda):\lambda\in\sigma(T)\} σ ( q ( T )) = { q ( λ ) : λ ∈ σ ( T )} ▨
補題 1.2. H ≠ { 0 } H\ne\{0\} H = { 0 } T ∈ B ( H ) T\in B(H) T ∈ B ( H ) p p p q q q p ( T ) p(T) p ( T ) q ( T ) q(T) q ( T ) 命題 1.1 のとおりとする。このとき、すべての多項式p p p p ( T ) p(T) p ( T )
∥ p ( T ) ∥ = max { ∣ p ( λ ) ∣ : λ ∈ σ ( T ) } \|p(T)\|=\max\{|p(\lambda)|:\lambda\in\sigma(T)\} ∥ p ( T ) ∥ = max { ∣ p ( λ ) ∣ : λ ∈ σ ( T )} が成り立つ。0 ∈ ρ ( T ) 0\in\rho(T) 0 ∈ ρ ( T ) q q q q ( T ) q(T) q ( T ) ∥ q ( T ) ∥ = max { ∣ q ( λ ) ∣ : λ ∈ σ ( T ) } \|q(T)\|=\max\{|q(\lambda)|:\lambda\in\sigma(T)\} ∥ q ( T ) ∥ = max { ∣ q ( λ ) ∣ : λ ∈ σ ( T )} A ∈ B ( H ) A\in B(H) A ∈ B ( H ) p p p ∥ p ( A ) ∥ = max { ∣ p ( λ ) ∣ : λ ∈ σ ( A ) } \|p(A)\|=\max\{|p(\lambda)|:\lambda\in\sigma(A)\} ∥ p ( A ) ∥ = max { ∣ p ( λ ) ∣ : λ ∈ σ ( A )} U ∈ B ( H ) U\in B(H) U ∈ B ( H ) 0 ∈ ρ ( U ) 0\in\rho(U) 0 ∈ ρ ( U ) q q q ∥ q ( U ) ∥ = max { ∣ q ( λ ) ∣ : λ ∈ σ ( U ) } \|q(U)\|=\max\{|q(\lambda)|:\lambda\in\sigma(U)\} ∥ q ( U ) ∥ = max { ∣ q ( λ ) ∣ : λ ∈ σ ( U )}
証明. q q q 0 ∈ ρ ( T ) 0\in\rho(T) 0 ∈ ρ ( T ) q ( z ) = ∑ k = − n n c k z k q(z)=\sum_{k=-n}^nc_kz^k q ( z ) = ∑ k = − n n c k z k q q q k < 0 k<0 k < 0 c k = 0 c_k=0 c k = 0 T T T T ∗ T^* T ∗ 0 ∈ ρ ( T ) 0\in\rho(T) 0 ∈ ρ ( T ) V = T − 1 V=T^{-1} V = T − 1 §E12.1 命題 1.3 (7) によりV ∗ T ∗ = T ∗ V ∗ = I V^*T^*=T^*V^*=I V ∗ T ∗ = T ∗ V ∗ = I
V T ∗ = V T ∗ ( T V ) = V ( T ∗ T ) V = V ( T T ∗ ) V = ( V T ) T ∗ V = T ∗ V VT^*=VT^*(TV)=V(T^*T)V=V(TT^*)V=(VT)T^*V=T^*V V T ∗ = V T ∗ ( T V ) = V ( T ∗ T ) V = V ( T T ∗ ) V = ( V T ) T ∗ V = T ∗ V である。この等式の両辺の共役作用素をとると、§E12.1 命題 1.3 (3) と§E12.1 命題 1.3 (4) によりT V ∗ = V ∗ T TV^*=V^*T T V ∗ = V ∗ T V V ∗ = V ∗ ( T ∗ V ) V ∗ = V ∗ ( V T ∗ ) V ∗ = V ∗ V VV^*=V^*(T^*V)V^*=V^*(VT^*)V^*=V^*V V V ∗ = V ∗ ( T ∗ V ) V ∗ = V ∗ ( V T ∗ ) V ∗ = V ∗ V 0 ∈ ρ ( T ) 0\in\rho(T) 0 ∈ ρ ( T ) T , T ∗ , V , V ∗ T,T^*,V,V^* T , T ∗ , V , V ∗ §E12.1 命題 1.3 (1) 、§E12.1 命題 1.3 (2) 、§E12.1 命題 1.3 (3) により、k ≥ 0 k\ge0 k ≥ 0 ( T k ) ∗ = ( T ∗ ) k (T^k)^*=(T^*)^k ( T k ) ∗ = ( T ∗ ) k k < 0 k<0 k < 0 ( T k ) ∗ = ( V ∗ ) − k (T^k)^*=(V^*)^{-k} ( T k ) ∗ = ( V ∗ ) − k
q ( T ) ∗ = ∑ k = 0 n c k ‾ ( T ∗ ) k + ∑ k = − n − 1 c k ‾ ( V ∗ ) − k q(T)^*=\sum_{k=0}^n\overline{c_k}\,(T^*)^k+\sum_{k=-n}^{-1}\overline{c_k}\,(V^*)^{-k} q ( T ) ∗ = k = 0 ∑ n c k ( T ∗ ) k + k = − n ∑ − 1 c k ( V ∗ ) − k である。q q q q ( T ) q(T) q ( T ) q ( T ) ∗ q(T)^* q ( T ) ∗ q ( T ) q(T) q ( T ) §E12.1 定理 4.7 と命題 1.1 により
∥ q ( T ) ∥ = max { ∣ μ ∣ : μ ∈ σ ( q ( T ) ) } = max { ∣ q ( λ ) ∣ : λ ∈ σ ( T ) } \|q(T)\|=\max\{|\mu|:\mu\in\sigma(q(T))\}=\max\{|q(\lambda)|:\lambda\in\sigma(T)\} ∥ q ( T ) ∥ = max { ∣ μ ∣ : μ ∈ σ ( q ( T ))} = max { ∣ q ( λ ) ∣ : λ ∈ σ ( T )} である。
自己共役作用素は§E12.1 定義 2.2 により正規作用素である。ユニタリ作用素U U U U ∗ U = U U ∗ = I U^*U=UU^*=I U ∗ U = U U ∗ = I − U ∗ -U^* − U ∗ 0 I − U 0I-U 0 I − U B ( H ) B(H) B ( H ) 0 ∈ ρ ( U ) 0\in\rho(U) 0 ∈ ρ ( U ) ▨
補題 1.3. K K K H H H A ⊆ C ( K ) \mathcal A\subseteq C(K) A ⊆ C ( K ) 1 1 1 C ( K ) C(K) C ( K ) φ : A → B ( H ) \varphi\colon\mathcal A\to B(H) φ : A → B ( H ) f ∈ A f\in\mathcal A f ∈ A ∥ φ ( f ) ∥ = ∥ f ∥ ∞ \|\varphi(f)\|=\|f\|_\infty ∥ φ ( f ) ∥ = ∥ f ∥ ∞
単位的 *-準同型Φ : C ( K ) → B ( H ) \Phi\colon C(K)\to B(H) Φ : C ( K ) → B ( H ) A \mathcal A A φ \varphi φ Φ \Phi Φ f ∈ C ( K ) f\in C(K) f ∈ C ( K ) ∥ Φ ( f ) ∥ = ∥ f ∥ ∞ \|\Phi(f)\|=\|f\|_\infty ∥Φ ( f ) ∥ = ∥ f ∥ ∞
S ∈ B ( H ) S\in B(H) S ∈ B ( H ) f ∈ A f\in\mathcal A f ∈ A S φ ( f ) = φ ( f ) S S\varphi(f)=\varphi(f)S S φ ( f ) = φ ( f ) S f ∈ C ( K ) f\in C(K) f ∈ C ( K ) S Φ ( f ) = Φ ( f ) S S\Phi(f)=\Phi(f)S S Φ ( f ) = Φ ( f ) S 証明. 主張 (1) を示す。§E10.1 定理 3.5 によりC ( K ) C(K) C ( K ) §E10.2 定理 2.2 によりB ( H ) B(H) B ( H ) A \mathcal A A C ( K ) C(K) C ( K ) φ \varphi φ §E10.4 系 2.2 により、有界線形写像Φ : C ( K ) → B ( H ) \Phi\colon C(K)\to B(H) Φ : C ( K ) → B ( H ) A \mathcal A A φ \varphi φ A \mathcal A A f n → f f_n\to f f n → f Φ ( f ) = lim n φ ( f n ) \Phi(f)=\lim_n\varphi(f_n) Φ ( f ) = lim n φ ( f n )
f , g ∈ C ( K ) f,g\in C(K) f , g ∈ C ( K ) A \mathcal A A ( f n ) (f_n) ( f n ) ( g n ) (g_n) ( g n ) f n → f f_n\to f f n → f g n → g g_n\to g g n → g ∥ Φ ( f ) ∥ = lim n ∥ φ ( f n ) ∥ = lim n ∥ f n ∥ ∞ = ∥ f ∥ ∞ \|\Phi(f)\|=\lim_n\|\varphi(f_n)\|=\lim_n\|f_n\|_\infty=\|f\|_\infty ∥Φ ( f ) ∥ = lim n ∥ φ ( f n ) ∥ = lim n ∥ f n ∥ ∞ = ∥ f ∥ ∞ ∥ f n g n − f g ∥ ∞ ≤ ∥ f n ∥ ∞ ∥ g n − g ∥ ∞ + ∥ f n − f ∥ ∞ ∥ g ∥ ∞ → 0 \|f_ng_n-fg\|_\infty\le\|f_n\|_\infty\|g_n-g\|_\infty+\|f_n-f\|_\infty\|g\|_\infty\to0 ∥ f n g n − f g ∥ ∞ ≤ ∥ f n ∥ ∞ ∥ g n − g ∥ ∞ + ∥ f n − f ∥ ∞ ∥ g ∥ ∞ → 0 §E10.2 命題 2.4 によりB ( H ) B(H) B ( H )
Φ ( f g ) = lim n → ∞ φ ( f n g n ) = lim n → ∞ φ ( f n ) φ ( g n ) = Φ ( f ) Φ ( g ) \Phi(fg)=\lim_{n\to\infty}\varphi(f_ng_n)=\lim_{n\to\infty}\varphi(f_n)\varphi(g_n)=\Phi(f)\Phi(g) Φ ( f g ) = n → ∞ lim φ ( f n g n ) = n → ∞ lim φ ( f n ) φ ( g n ) = Φ ( f ) Φ ( g ) である。∥ f n ‾ − f ‾ ∥ ∞ = ∥ f n − f ∥ ∞ → 0 \|\overline{f_n}-\overline f\|_\infty=\|f_n-f\|_\infty\to0 ∥ f n − f ∥ ∞ = ∥ f n − f ∥ ∞ → 0 §E12.1 命題 1.3 (1) 、§E12.1 命題 1.3 (2) 、§E12.1 命題 1.2 により∥ φ ( f n ) ∗ − Φ ( f ) ∗ ∥ = ∥ φ ( f n ) − Φ ( f ) ∥ → 0 \|\varphi(f_n)^*-\Phi(f)^*\|=\|\varphi(f_n)-\Phi(f)\|\to0 ∥ φ ( f n ) ∗ − Φ ( f ) ∗ ∥ = ∥ φ ( f n ) − Φ ( f ) ∥ → 0 Φ ( f ‾ ) = lim n φ ( f n ‾ ) = lim n φ ( f n ) ∗ = Φ ( f ) ∗ \Phi(\overline f)=\lim_n\varphi(\overline{f_n})=\lim_n\varphi(f_n)^*=\Phi(f)^* Φ ( f ) = lim n φ ( f n ) = lim n φ ( f n ) ∗ = Φ ( f ) ∗ Φ ( 1 ) = φ ( 1 ) = I \Phi(1)=\varphi(1)=I Φ ( 1 ) = φ ( 1 ) = I Φ \Phi Φ
Ψ : C ( K ) → B ( H ) \Psi\colon C(K)\to B(H) Ψ : C ( K ) → B ( H ) A \mathcal A A φ \varphi φ §E12.7 補題 6.1 (2) により、すべてのf ∈ C ( K ) f\in C(K) f ∈ C ( K ) ∥ Ψ ( f ) ∥ ≤ ∥ f ∥ ∞ \|\Psi(f)\|\le\|f\|_\infty ∥Ψ ( f ) ∥ ≤ ∥ f ∥ ∞ Ψ \Psi Ψ φ \varphi φ §E10.4 系 2.2 の一意性によりΨ = Φ \Psi=\Phi Ψ = Φ
主張 (2) を示す。f ∈ C ( K ) f\in C(K) f ∈ C ( K ) A \mathcal A A f n → f f_n\to f f n → f §E10.2 命題 2.4 により
S Φ ( f ) = lim n → ∞ S φ ( f n ) = lim n → ∞ φ ( f n ) S = Φ ( f ) S S\Phi(f)=\lim_{n\to\infty}S\varphi(f_n)=\lim_{n\to\infty}\varphi(f_n)S=\Phi(f)S S Φ ( f ) = n → ∞ lim S φ ( f n ) = n → ∞ lim φ ( f n ) S = Φ ( f ) S である。▨
定理 1.4. H ≠ { 0 } H\ne\{0\} H = { 0 } A ∈ B ( H ) A\in B(H) A ∈ B ( H ) ι : σ ( A ) → C \iota\colon\sigma(A)\to\C ι : σ ( A ) → C
単位的 *-準同型Φ : C ( σ ( A ) ) → B ( H ) \Phi\colon C(\sigma(A))\to B(H) Φ : C ( σ ( A )) → B ( H ) Φ ( ι ) = A \Phi(\iota)=A Φ ( ι ) = A Φ \Phi Φ Φ A \Phi_A Φ A f ∈ C ( σ ( A ) ) f\in C(\sigma(A)) f ∈ C ( σ ( A )) f ( A ) = Φ A ( f ) f(A)=\Phi_A(f) f ( A ) = Φ A ( f ) f ↦ f ( A ) f\mapsto f(A) f ↦ f ( A ) A A A f ∈ C ( σ ( A ) ) f\in C(\sigma(A)) f ∈ C ( σ ( A )) ∥ f ( A ) ∥ = ∥ f ∥ ∞ \|f(A)\|=\|f\|_\infty ∥ f ( A ) ∥ = ∥ f ∥ ∞ p p p σ ( A ) \sigma(A) σ ( A ) f f f f ( A ) = p ( A ) f(A)=p(A) f ( A ) = p ( A )
S ∈ B ( H ) S\in B(H) S ∈ B ( H ) S A = A S SA=AS S A = A S f ∈ C ( σ ( A ) ) f\in C(\sigma(A)) f ∈ C ( σ ( A )) S f ( A ) = f ( A ) S Sf(A)=f(A)S S f ( A ) = f ( A ) S 証明. §E12.1 定理 4.1 と§E12.1 定理 3.5 によりσ ( A ) \sigma(A) σ ( A ) §E12.1 命題 4.4 (1) によりσ ( A ) ⊆ R \sigma(A)\subseteq\R σ ( A ) ⊆ R σ ( A ) \sigma(A) σ ( A ) A \mathcal A A
多項式p , p ′ p,p' p , p ′ σ ( A ) \sigma(A) σ ( A ) 補題 1.2 により∥ p ( A ) − p ′ ( A ) ∥ = max { ∣ p ( λ ) − p ′ ( λ ) ∣ : λ ∈ σ ( A ) } = 0 \|p(A)-p'(A)\|=\max\{|p(\lambda)-p'(\lambda)|:\lambda\in\sigma(A)\}=0 ∥ p ( A ) − p ′ ( A ) ∥ = max { ∣ p ( λ ) − p ′ ( λ ) ∣ : λ ∈ σ ( A )} = 0 p ( A ) = p ′ ( A ) p(A)=p'(A) p ( A ) = p ′ ( A ) p p p f f f p ( A ) p(A) p ( A ) φ : A → B ( H ) \varphi\colon\mathcal A\to B(H) φ : A → B ( H ) 補題 1.2 によりすべてのf ∈ A f\in\mathcal A f ∈ A ∥ φ ( f ) ∥ = ∥ f ∥ ∞ \|\varphi(f)\|=\|f\|_\infty ∥ φ ( f ) ∥ = ∥ f ∥ ∞ A \mathcal A A 1 1 1 φ \varphi φ φ ( 1 ) = I \varphi(1)=I φ ( 1 ) = I p ( z ) = ∑ k c k z k p(z)=\sum_kc_kz^k p ( z ) = ∑ k c k z k p ♯ ( z ) = ∑ k c k ‾ z k p^\sharp(z)=\sum_k\overline{c_k}z^k p ♯ ( z ) = ∑ k c k z k σ ( A ) ⊆ R \sigma(A)\subseteq\R σ ( A ) ⊆ R λ ∈ σ ( A ) \lambda\in\sigma(A) λ ∈ σ ( A ) p ( λ ) ‾ = p ♯ ( λ ) \overline{p(\lambda)}=p^\sharp(\lambda) p ( λ ) = p ♯ ( λ ) A \mathcal A A A ∗ = A A^*=A A ∗ = A §E12.1 命題 1.3 (1) 、§E12.1 命題 1.3 (2) 、§E12.1 命題 1.3 (3) によりp ( A ) ∗ = p ♯ ( A ) p(A)^*=p^\sharp(A) p ( A ) ∗ = p ♯ ( A ) φ \varphi φ A \mathcal A A ι \iota ι σ ( A ) \sigma(A) σ ( A ) §E2.32 系 3.5 によりC ( σ ( A ) ) C(\sigma(A)) C ( σ ( A ))
主張 (1) を示す。補題 1.3 (1) により、φ \varphi φ Φ A : C ( σ ( A ) ) → B ( H ) \Phi_A\colon C(\sigma(A))\to B(H) Φ A : C ( σ ( A )) → B ( H ) f f f ∥ Φ A ( f ) ∥ = ∥ f ∥ ∞ \|\Phi_A(f)\|=\|f\|_\infty ∥ Φ A ( f ) ∥ = ∥ f ∥ ∞ Φ A ( ι ) = φ ( ι ) = A \Phi_A(\iota)=\varphi(\iota)=A Φ A ( ι ) = φ ( ι ) = A Ψ : C ( σ ( A ) ) → B ( H ) \Psi\colon C(\sigma(A))\to B(H) Ψ : C ( σ ( A )) → B ( H ) Ψ ( ι ) = A \Psi(\iota)=A Ψ ( ι ) = A Ψ ( 1 ) = I \Psi(1)=I Ψ ( 1 ) = I p p p f f f Ψ ( f ) = p ( A ) = φ ( f ) \Psi(f)=p(A)=\varphi(f) Ψ ( f ) = p ( A ) = φ ( f ) 補題 1.3 (1) の一意性によりΨ = Φ A \Psi=\Phi_A Ψ = Φ A
主張 (2) を示す。S A = A S SA=AS S A = A S p p p S p ( A ) = p ( A ) S Sp(A)=p(A)S S p ( A ) = p ( A ) S 補題 1.3 (2) により主張を得る。▨
定理 1.5. H ≠ { 0 } H\ne\{0\} H = { 0 } U ∈ B ( H ) U\in B(H) U ∈ B ( H ) ι : σ ( U ) → C \iota\colon\sigma(U)\to\C ι : σ ( U ) → C
単位的 *-準同型Φ : C ( σ ( U ) ) → B ( H ) \Phi\colon C(\sigma(U))\to B(H) Φ : C ( σ ( U )) → B ( H ) Φ ( ι ) = U \Phi(\iota)=U Φ ( ι ) = U Φ \Phi Φ Φ U \Phi_U Φ U f ∈ C ( σ ( U ) ) f\in C(\sigma(U)) f ∈ C ( σ ( U )) f ( U ) = Φ U ( f ) f(U)=\Phi_U(f) f ( U ) = Φ U ( f ) f ↦ f ( U ) f\mapsto f(U) f ↦ f ( U ) U U U f ∈ C ( σ ( U ) ) f\in C(\sigma(U)) f ∈ C ( σ ( U )) ∥ f ( U ) ∥ = ∥ f ∥ ∞ \|f(U)\|=\|f\|_\infty ∥ f ( U ) ∥ = ∥ f ∥ ∞ q q q σ ( U ) \sigma(U) σ ( U ) f f f f ( U ) = q ( U ) f(U)=q(U) f ( U ) = q ( U )
S ∈ B ( H ) S\in B(H) S ∈ B ( H ) S U = U S SU=US S U = U S f ∈ C ( σ ( U ) ) f\in C(\sigma(U)) f ∈ C ( σ ( U )) S f ( U ) = f ( U ) S Sf(U)=f(U)S S f ( U ) = f ( U ) S 証明. §E12.1 定理 4.1 と§E12.1 定理 3.5 によりσ ( U ) \sigma(U) σ ( U ) §E12.1 命題 4.5 によりσ ( U ) ⊆ { λ ∈ C : ∣ λ ∣ = 1 } \sigma(U)\subseteq\{\lambda\in\C:|\lambda|=1\} σ ( U ) ⊆ { λ ∈ C : ∣ λ ∣ = 1 } 補題 1.2 により0 ∈ ρ ( U ) 0\in\rho(U) 0 ∈ ρ ( U ) U − 1 = U ∗ U^{-1}=U^* U − 1 = U ∗ σ ( U ) \sigma(U) σ ( U ) A \mathcal A A
Laurent 多項式q , q ′ q,q' q , q ′ σ ( U ) \sigma(U) σ ( U ) 補題 1.2 により∥ q ( U ) − q ′ ( U ) ∥ = max { ∣ q ( λ ) − q ′ ( λ ) ∣ : λ ∈ σ ( U ) } = 0 \|q(U)-q'(U)\|=\max\{|q(\lambda)-q'(\lambda)|:\lambda\in\sigma(U)\}=0 ∥ q ( U ) − q ′ ( U ) ∥ = max { ∣ q ( λ ) − q ′ ( λ ) ∣ : λ ∈ σ ( U )} = 0 q ( U ) = q ′ ( U ) q(U)=q'(U) q ( U ) = q ′ ( U ) q q q f f f q ( U ) q(U) q ( U ) φ : A → B ( H ) \varphi\colon\mathcal A\to B(H) φ : A → B ( H ) 補題 1.2 によりすべてのf ∈ A f\in\mathcal A f ∈ A ∥ φ ( f ) ∥ = ∥ f ∥ ∞ \|\varphi(f)\|=\|f\|_\infty ∥ φ ( f ) ∥ = ∥ f ∥ ∞ A \mathcal A A 1 1 1 φ \varphi φ φ ( 1 ) = I \varphi(1)=I φ ( 1 ) = I q ( z ) = ∑ k = − n n c k z k q(z)=\sum_{k=-n}^nc_kz^k q ( z ) = ∑ k = − n n c k z k q ♯ ( z ) = ∑ k = − n n c k ‾ z − k q^\sharp(z)=\sum_{k=-n}^n\overline{c_k}z^{-k} q ♯ ( z ) = ∑ k = − n n c k z − k ∣ λ ∣ = 1 |\lambda|=1 ∣ λ ∣ = 1 λ \lambda λ λ ‾ = λ − 1 \overline\lambda=\lambda^{-1} λ = λ − 1 λ ∈ σ ( U ) \lambda\in\sigma(U) λ ∈ σ ( U ) q ( λ ) ‾ = q ♯ ( λ ) \overline{q(\lambda)}=q^\sharp(\lambda) q ( λ ) = q ♯ ( λ ) A \mathcal A A U ∗ = U − 1 U^*=U^{-1} U ∗ = U − 1 §E12.1 命題 1.3 (1) 、§E12.1 命題 1.3 (2) 、§E12.1 命題 1.3 (3) により、すべてのk ∈ Z k\in\Z k ∈ Z ( U k ) ∗ = U − k (U^k)^*=U^{-k} ( U k ) ∗ = U − k q ( U ) ∗ = q ♯ ( U ) q(U)^*=q^\sharp(U) q ( U ) ∗ = q ♯ ( U ) φ \varphi φ A \mathcal A A ι \iota ι σ ( U ) \sigma(U) σ ( U ) §E2.32 系 3.5 によりC ( σ ( U ) ) C(\sigma(U)) C ( σ ( U ))
主張 (1) を示す。補題 1.3 (1) により、φ \varphi φ Φ U : C ( σ ( U ) ) → B ( H ) \Phi_U\colon C(\sigma(U))\to B(H) Φ U : C ( σ ( U )) → B ( H ) f f f ∥ Φ U ( f ) ∥ = ∥ f ∥ ∞ \|\Phi_U(f)\|=\|f\|_\infty ∥ Φ U ( f ) ∥ = ∥ f ∥ ∞ Φ U ( ι ) = φ ( ι ) = U \Phi_U(\iota)=\varphi(\iota)=U Φ U ( ι ) = φ ( ι ) = U Ψ : C ( σ ( U ) ) → B ( H ) \Psi\colon C(\sigma(U))\to B(H) Ψ : C ( σ ( U )) → B ( H ) Ψ ( ι ) = U \Psi(\iota)=U Ψ ( ι ) = U ι ⋅ ι ‾ = 1 \iota\cdot\overline\iota=1 ι ⋅ ι = 1 Ψ ( ι ‾ ) \Psi(\overline\iota) Ψ ( ι ) U U U Ψ ( ι ‾ ) = U − 1 \Psi(\overline\iota)=U^{-1} Ψ ( ι ) = U − 1 Ψ ( 1 ) = I \Psi(1)=I Ψ ( 1 ) = I q q q f f f Ψ ( f ) = q ( U ) = φ ( f ) \Psi(f)=q(U)=\varphi(f) Ψ ( f ) = q ( U ) = φ ( f ) 補題 1.3 (1) の一意性によりΨ = Φ U \Psi=\Phi_U Ψ = Φ U
主張 (2) を示す。S U = U S SU=US S U = U S U − 1 S = U − 1 S U U − 1 = U − 1 U S U − 1 = S U − 1 U^{-1}S=U^{-1}SUU^{-1}=U^{-1}USU^{-1}=SU^{-1} U − 1 S = U − 1 S U U − 1 = U − 1 U S U − 1 = S U − 1 q q q S q ( U ) = q ( U ) S Sq(U)=q(U)S S q ( U ) = q ( U ) S 補題 1.3 (2) により主張を得る。▨
2 射影値測度による表現 補題 2.1. K ⊆ C K\subseteq\C K ⊆ C H H H ι : K → C \iota\colon K\to\C ι : K → C
単位的 *-準同型Φ , Ψ : C ( K ) → B ( H ) \Phi,\Psi\colon C(K)\to B(H) Φ , Ψ : C ( K ) → B ( H ) Φ ( ι ) = Ψ ( ι ) \Phi(\iota)=\Psi(\iota) Φ ( ι ) = Ψ ( ι ) Φ = Ψ \Phi=\Psi Φ = Ψ
E E E K K K x , y ∈ H x,y\in H x , y ∈ H E x , y E_{x,y} E x , y E , F E,F E , F K K K Φ E ( ι ) = Φ F ( ι ) \Phi_E(\iota)=\Phi_F(\iota) Φ E ( ι ) = Φ F ( ι ) E = F E=F E = F E E E K K K Φ E \Phi_E Φ E C ( K ) C(K) C ( K ) K K K V V V E ( V ) ≠ 0 E(V)\ne0 E ( V ) = 0 証明. 主張 (1) を示す。Φ ( ι ) = Ψ ( ι ) \Phi(\iota)=\Psi(\iota) Φ ( ι ) = Ψ ( ι ) Φ , Ψ : C ( K ) → B ( H ) \Phi,\Psi\colon C(K)\to B(H) Φ , Ψ : C ( K ) → B ( H ) Φ ( ι ‾ ) = Φ ( ι ) ∗ = Ψ ( ι ) ∗ = Ψ ( ι ‾ ) \Phi(\overline\iota)=\Phi(\iota)^*=\Psi(\iota)^*=\Psi(\overline\iota) Φ ( ι ) = Φ ( ι ) ∗ = Ψ ( ι ) ∗ = Ψ ( ι ) Φ \Phi Φ Ψ \Psi Ψ ι , ι ‾ \iota,\overline\iota ι , ι A \mathcal A A A \mathcal A A ι ∈ A \iota\in\mathcal A ι ∈ A K K K §E2.32 系 3.5 によりA \mathcal A A C ( K ) C(K) C ( K ) f ∈ C ( K ) f\in C(K) f ∈ C ( K ) A \mathcal A A f n f_n f n ∥ f n − f ∥ ∞ → 0 \|f_n-f\|_\infty\to0 ∥ f n − f ∥ ∞ → 0 §E12.7 補題 6.1 (2) により
∥ Φ ( f ) − Ψ ( f ) ∥ ≤ ∥ Φ ( f − f n ) ∥ + ∥ Ψ ( f n − f ) ∥ ≤ 2 ∥ f − f n ∥ ∞ ⟶ 0 \|\Phi(f)-\Psi(f)\|\le\|\Phi(f-f_n)\|+\|\Psi(f_n-f)\|\le2\|f-f_n\|_\infty\longrightarrow0 ∥Φ ( f ) − Ψ ( f ) ∥ ≤ ∥Φ ( f − f n ) ∥ + ∥Ψ ( f n − f ) ∥ ≤ 2∥ f − f n ∥ ∞ ⟶ 0 であるから、Φ ( f ) = Ψ ( f ) \Phi(f)=\Psi(f) Φ ( f ) = Ψ ( f )
主張 (2) を示す。x , y ∈ H x,y\in H x , y ∈ H ν = E x , y \nu=E_{x,y} ν = E x , y ν \nu ν B ( K ) \mathcal B(K) B ( K ) §E10.27 定理 2.3 により∣ ν ∣ |\nu| ∣ ν ∣ K K K d ( z , w ) = ∣ z − w ∣ d(z,w)=|z-w| d ( z , w ) = ∣ z − w ∣ §E10.39 定理 1.1 により、すべてのB ∈ B ( K ) B\in\mathcal B(K) B ∈ B ( K )
∣ ν ∣ ( B ) = sup { ∣ ν ∣ ( F ) : F ⊆ B , F は K の閉集合 } |\nu|(B)=\sup\{|\nu|(F):F\subseteq B,\ F\text{ は }K\text{ の閉集合}\} ∣ ν ∣ ( B ) = sup { ∣ ν ∣ ( F ) : F ⊆ B , F は K の閉集合 } である。K K K §E2.19 命題 4.1 によりコンパクトであるから、∣ ν ∣ |\nu| ∣ ν ∣ ∣ ν ∣ ( K ) < ∞ |\nu|(K)<\infty ∣ ν ∣ ( K ) < ∞ ∣ ν ∣ |\nu| ∣ ν ∣ ν \nu ν
主張 (3) を示す。Φ E ( ι ) = Φ F ( ι ) \Phi_E(\iota)=\Phi_F(\iota) Φ E ( ι ) = Φ F ( ι ) E , F E,F E , F K K K C ( K ) ⊆ B b ( K ) C(K)\subseteq B_b(K) C ( K ) ⊆ B b ( K ) §E12.7 定理 4.5 (2) によりΦ E \Phi_E Φ E Φ F \Phi_F Φ F C ( K ) C(K) C ( K ) 主張 (1) により一致する。その共通の制限をΦ \Phi Φ 主張 (2) によりE , F E,F E , F §E12.7 定理 6.2 (1) をΦ \Phi Φ E = F E=F E = F
主張 (4) を示す。Φ E \Phi_E Φ E C ( K ) C(K) C ( K ) V V V K K K z 0 ∈ V z_0\in V z 0 ∈ V { z ∈ K : ∣ z − z 0 ∣ < r } ⊆ V \{z\in K:|z-z_0|<r\}\subseteq V { z ∈ K : ∣ z − z 0 ∣ < r } ⊆ V r > 0 r>0 r > 0 f ( z ) = max { 0 , 1 − ∣ z − z 0 ∣ / r } f(z)=\max\{0,1-|z-z_0|/r\} f ( z ) = max { 0 , 1 − ∣ z − z 0 ∣/ r } z ∈ K z\in K z ∈ K f ∈ C ( K ) f\in C(K) f ∈ C ( K ) f ( z 0 ) = 1 f(z_0)=1 f ( z 0 ) = 1 f = f 1 V f=f\mathbf 1_V f = f 1 V E ( V ) = 0 E(V)=0 E ( V ) = 0 §E12.7 定理 4.5 (2) によりΦ E ( f ) = Φ E ( f ) E ( V ) = 0 \Phi_E(f)=\Phi_E(f)E(V)=0 Φ E ( f ) = Φ E ( f ) E ( V ) = 0 f = 0 f=0 f = 0 f ( z 0 ) = 1 f(z_0)=1 f ( z 0 ) = 1 E ( V ) ≠ 0 E(V)\ne0 E ( V ) = 0
K K K V V V E ( V ) ≠ 0 E(V)\ne0 E ( V ) = 0 f ∈ C ( K ) f\in C(K) f ∈ C ( K ) M = ∥ f ∥ ∞ M=\|f\|_\infty M = ∥ f ∥ ∞ 0 ≤ t < M 0\le t<M 0 ≤ t < M { z ∈ K : ∣ f ( z ) ∣ > t } \{z\in K:|f(z)|>t\} { z ∈ K : ∣ f ( z ) ∣ > t } t ≥ M t\ge M t ≥ M
inf { t ≥ 0 : E ( { z ∈ K : ∣ f ( z ) ∣ > t } ) = 0 } = M \inf\{t\ge0:E(\{z\in K:|f(z)|>t\})=0\}=M inf { t ≥ 0 : E ({ z ∈ K : ∣ f ( z ) ∣ > t }) = 0 } = M であり、§E12.7 命題 4.6 (1) により∥ Φ E ( f ) ∥ = ∥ f ∥ ∞ \|\Phi_E(f)\|=\|f\|_\infty ∥ Φ E ( f ) ∥ = ∥ f ∥ ∞ Φ E \Phi_E Φ E C ( K ) C(K) C ( K ) Φ E ( f ) = 0 \Phi_E(f)=0 Φ E ( f ) = 0 ∥ f ∥ ∞ = 0 \|f\|_\infty=0 ∥ f ∥ ∞ = 0 f = 0 f=0 f = 0 ▨
定理 2.2 (スペクトル定理). H ≠ { 0 } H\ne\{0\} H = { 0 } T ∈ B ( H ) T\in B(H) T ∈ B ( H ) ι : σ ( T ) → C \iota\colon\sigma(T)\to\C ι : σ ( T ) → C Φ T \Phi_T Φ T C ( σ ( T ) ) → B ( H ) C(\sigma(T))\to B(H) C ( σ ( T )) → B ( H ) ι \iota ι T T T 定理 1.4 (1) 、定理 1.5 (1) ）とし、f ∈ C ( σ ( T ) ) f\in C(\sigma(T)) f ∈ C ( σ ( T )) f ( T ) = Φ T ( f ) f(T)=\Phi_T(f) f ( T ) = Φ T ( f )
σ ( T ) \sigma(T) σ ( T ) E E E Φ E ( ι ) = T \Phi_E(\iota)=T Φ E ( ι ) = T E E E E T E_T E T x , y ∈ H x,y\in H x , y ∈ H ( E T ) x , y (E_T)_{x,y} ( E T ) x , y f ∈ C ( σ ( T ) ) f\in C(\sigma(T)) f ∈ C ( σ ( T )) Φ E T ( f ) = f ( T ) \Phi_{E_T}(f)=f(T) Φ E T ( f ) = f ( T ) g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) g ( T ) = Φ E T ( g ) g(T)=\Phi_{E_T}(g) g ( T ) = Φ E T ( g ) g ↦ g ( T ) g\mapsto g(T) g ↦ g ( T ) f ↦ f ( T ) f\mapsto f(T) f ↦ f ( T ) B b ( σ ( T ) ) B_b(\sigma(T)) B b ( σ ( T )) B ( H ) B(H) B ( H ) g ↦ g ( T ) g\mapsto g(T) g ↦ g ( T ) T T T 有界 Borel 汎関数計算 (bounded Borel functional calculus )σ ( T ) \sigma(T) σ ( T ) V V V E T ( V ) ≠ 0 E_T(V)\ne0 E T ( V ) = 0 S ∈ B ( H ) S\in B(H) S ∈ B ( H ) S T = T S ST=TS S T = T S B ⊆ σ ( T ) B\subseteq\sigma(T) B ⊆ σ ( T ) S E T ( B ) = E T ( B ) S SE_T(B)=E_T(B)S S E T ( B ) = E T ( B ) S g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) S g ( T ) = g ( T ) S Sg(T)=g(T)S S g ( T ) = g ( T ) S 証明. 主張 (1) を示す。σ ( T ) \sigma(T) σ ( T ) C \C C §E12.7 定理 6.2 (1) をΦ T \Phi_T Φ T σ ( T ) \sigma(T) σ ( T ) E T E_T E T x , y ∈ H x,y\in H x , y ∈ H ( E T ) x , y (E_T)_{x,y} ( E T ) x , y f ∈ C ( σ ( T ) ) f\in C(\sigma(T)) f ∈ C ( σ ( T )) Φ E T ( f ) = f ( T ) \Phi_{E_T}(f)=f(T) Φ E T ( f ) = f ( T ) Φ E T ( ι ) = ι ( T ) = T \Phi_{E_T}(\iota)=\iota(T)=T Φ E T ( ι ) = ι ( T ) = T §E12.7 定理 4.5 (2) によりg ↦ Φ E T ( g ) g\mapsto\Phi_{E_T}(g) g ↦ Φ E T ( g )
E E E σ ( T ) \sigma(T) σ ( T ) Φ E ( ι ) = T \Phi_E(\iota)=T Φ E ( ι ) = T Φ E ( ι ) = T = Φ E T ( ι ) \Phi_E(\iota)=T=\Phi_{E_T}(\iota) Φ E ( ι ) = T = Φ E T ( ι ) 補題 2.1 (3) によりE = E T E=E_T E = E T
主張 (2) を示す。Φ E T \Phi_{E_T} Φ E T C ( σ ( T ) ) C(\sigma(T)) C ( σ ( T )) Φ T \Phi_T Φ T 定理 1.4 (1) または定理 1.5 (1) により等長である。したがって、補題 2.1 (4) により、σ ( T ) \sigma(T) σ ( T ) V V V E T ( V ) ≠ 0 E_T(V)\ne0 E T ( V ) = 0
主張 (3) を示す。定理 1.4 (2) または定理 1.5 (2) により、すべてのf ∈ C ( σ ( T ) ) f\in C(\sigma(T)) f ∈ C ( σ ( T )) S f ( T ) = f ( T ) S Sf(T)=f(T)S S f ( T ) = f ( T ) S §E12.7 定理 6.2 (2) をΦ T \Phi_T Φ T E T E_T E T ▨
命題 2.3. H ≠ { 0 } H\ne\{0\} H = { 0 } T ∈ B ( H ) T\in B(H) T ∈ B ( H ) E T E_T E T 定理 2.2 (1) の射影値測度とする。このとき、すべてのλ ∈ σ ( T ) \lambda\in\sigma(T) λ ∈ σ ( T ) E T ( { λ } ) E_T(\{\lambda\}) E T ({ λ }) ker ( λ I − T ) \ker(\lambda I-T) ker ( λ I − T )
証明. λ ∈ σ ( T ) \lambda\in\sigma(T) λ ∈ σ ( T ) P = E T ( { λ } ) P=E_T(\{\lambda\}) P = E T ({ λ }) ι \iota ι σ ( T ) → C \sigma(T)\to\C σ ( T ) → C 定理 2.2 (1) と§E12.7 定理 4.5 (2) によりT − λ I = Φ E T ( ι − λ ) T-\lambda I=\Phi_{E_T}(\iota-\lambda) T − λ I = Φ E T ( ι − λ )
( T − λ I ) P = Φ E T ( ( ι − λ ) 1 { λ } ) = Φ E T ( 0 ) = 0 (T-\lambda I)P=\Phi_{E_T}\bigl((\iota-\lambda)\mathbf 1_{\{\lambda\}}\bigr)=\Phi_{E_T}(0)=0 ( T − λ I ) P = Φ E T ( ( ι − λ ) 1 { λ } ) = Φ E T ( 0 ) = 0 である。したがってim P ⊆ ker ( λ I − T ) \im P\subseteq\ker(\lambda I-T) im P ⊆ ker ( λ I − T )
x ∈ ker ( λ I − T ) x\in\ker(\lambda I-T) x ∈ ker ( λ I − T ) μ x = ( E T ) x , x \mu_x=(E_T)_{x,x} μ x = ( E T ) x , x §E12.7 定理 4.5 (3) により
∫ σ ( T ) ∣ ι − λ ∣ 2 d μ x = ∥ ( T − λ I ) x ∥ 2 = 0 \int_{\sigma(T)}|\iota-\lambda|^2\,d\mu_x=\|(T-\lambda I)x\|^2=0 ∫ σ ( T ) ∣ ι − λ ∣ 2 d μ x = ∥ ( T − λ I ) x ∥ 2 = 0 である。k ∈ N ≥ 1 k\in\NN k ∈ N ≥ 1 N k = { t ∈ σ ( T ) : ∣ t − λ ∣ ≥ 1 / k } N_k=\{t\in\sigma(T):|t-\lambda|\ge1/k\} N k = { t ∈ σ ( T ) : ∣ t − λ ∣ ≥ 1/ k } k − 2 μ x ( N k ) ≤ ∫ σ ( T ) ∣ ι − λ ∣ 2 d μ x = 0 k^{-2}\mu_x(N_k)\le\int_{\sigma(T)}|\iota-\lambda|^2\,d\mu_x=0 k − 2 μ x ( N k ) ≤ ∫ σ ( T ) ∣ ι − λ ∣ 2 d μ x = 0 σ ( T ) ∖ { λ } = ⋃ k N k \sigma(T)\setminus\{\lambda\}=\bigcup_kN_k σ ( T ) ∖ { λ } = ⋃ k N k μ x ( σ ( T ) ∖ { λ } ) = 0 \mu_x(\sigma(T)\setminus\{\lambda\})=0 μ x ( σ ( T ) ∖ { λ }) = 0 §E12.7 命題 3.2 (1) と§E12.7 命題 3.2 (2) により
∥ x − P x ∥ 2 = ∥ E T ( σ ( T ) ∖ { λ } ) x ∥ 2 = μ x ( σ ( T ) ∖ { λ } ) = 0 \|x-Px\|^2=\|E_T(\sigma(T)\setminus\{\lambda\})x\|^2=\mu_x(\sigma(T)\setminus\{\lambda\})=0 ∥ x − P x ∥ 2 = ∥ E T ( σ ( T ) ∖ { λ }) x ∥ 2 = μ x ( σ ( T ) ∖ { λ }) = 0 であるから、x = P x ∈ im P x=Px\in\im P x = P x ∈ im P im P = ker ( λ I − T ) \im P=\ker(\lambda I-T) im P = ker ( λ I − T ) §E12.7 定義 3.1 条件 (a) と§E12.7 補題 1.5 によりP P P ker ( λ I − T ) \ker(\lambda I-T) ker ( λ I − T ) ▨
3 乗算作用素形 補題 3.1. H ≠ { 0 } H\ne\{0\} H = { 0 } T ∈ B ( H ) T\in B(H) T ∈ B ( H ) E T E_T E T g ( T ) g(T) g ( T ) g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) 定理 2.2 (1) のとおりとする。x ∈ H x\in H x ∈ H μ x = ( E T ) x , x \mu_x=(E_T)_{x,x} μ x = ( E T ) x , x L 2 ( μ x ) = L 2 ( σ ( T ) , B ( σ ( T ) ) , μ x ; C ) L^2(\mu_x)=L^2(\sigma(T),\mathcal B(\sigma(T)),\mu_x;\C) L 2 ( μ x ) = L 2 ( σ ( T ) , B ( σ ( T )) , μ x ; C ) { f ( T ) x : f ∈ C ( σ ( T ) ) } \{f(T)x:f\in C(\sigma(T))\} { f ( T ) x : f ∈ C ( σ ( T ))} H H H H x H_x H x H x H_x H x x x x 巡回部分空間 (cyclic subspace )g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) u ∈ L 2 ( μ x ) u\in L^2(\mu_x) u ∈ L 2 ( μ x ) g u gu g u L 2 ( μ x ) L^2(\mu_x) L 2 ( μ x )
μ x \mu_x μ x σ ( T ) \sigma(T) σ ( T ) μ x ( σ ( T ) ) = ∥ x ∥ 2 \mu_x(\sigma(T))=\|x\|^2 μ x ( σ ( T )) = ∥ x ∥ 2 H x H_x H x x x x H H H W x : L 2 ( μ x ) → H x W_x\colon L^2(\mu_x)\to H_x W x : L 2 ( μ x ) → H x f ∈ C ( σ ( T ) ) f\in C(\sigma(T)) f ∈ C ( σ ( T )) W x [ f ] = f ( T ) x W_x[f]=f(T)x W x [ f ] = f ( T ) x W x W_x W x g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) W x [ g ] = g ( T ) x W_x[g]=g(T)x W x [ g ] = g ( T ) x すべてのg ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) u ∈ L 2 ( μ x ) u\in L^2(\mu_x) u ∈ L 2 ( μ x ) g ( T ) W x u = W x ( g u ) g(T)W_xu=W_x(gu) g ( T ) W x u = W x ( g u ) T ( H x ) ⊆ H x T(H_x)\subseteq H_x T ( H x ) ⊆ H x T ∗ ( H x ) ⊆ H x T^*(H_x)\subseteq H_x T ∗ ( H x ) ⊆ H x T W x u = W x ( ι u ) TW_xu=W_x(\iota u) T W x u = W x ( ι u ) ι : σ ( T ) → C \iota\colon\sigma(T)\to\C ι : σ ( T ) → C
証明. 主張 (1) を示す。定理 2.2 (1) によりμ x \mu_x μ x §E12.7 命題 3.2 (2) によりμ x \mu_x μ x μ x ( σ ( T ) ) = ∥ x ∥ 2 \mu_x(\sigma(T))=\|x\|^2 μ x ( σ ( T )) = ∥ x ∥ 2 §E12.7 補題 4.3 (4) により∣ μ x ∣ = μ x |\mu_x|=\mu_x ∣ μ x ∣ = μ x §E10.34 定義 3.1 によりμ x \mu_x μ x
主張 (2) を示す。H x H_x H x x = 1 ( T ) x ∈ H x x=1(T)x\in H_x x = 1 ( T ) x ∈ H x §E10.1 命題 1.4 (1) によりH x H_x H x H H H g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) §E12.7 定理 4.5 (3) により
∥ g ( T ) x ∥ 2 = ∫ σ ( T ) ∣ g ∣ 2 d μ x = ∥ [ g ] ∥ L 2 ( μ x ) 2 \|g(T)x\|^2=\int_{\sigma(T)}|g|^2\,d\mu_x=\|[g]\|_{L^2(\mu_x)}^2 ∥ g ( T ) x ∥ 2 = ∫ σ ( T ) ∣ g ∣ 2 d μ x = ∥ [ g ] ∥ L 2 ( μ x ) 2 である。定理 2.2 (1) により写像g ↦ g ( T ) x g\mapsto g(T)x g ↦ g ( T ) x [ g ] = [ g ′ ] [g]=[g'] [ g ] = [ g ′ ] ∥ g ( T ) x − g ′ ( T ) x ∥ = ∥ [ g − g ′ ] ∥ = 0 \|g(T)x-g'(T)x\|=\|[g-g']\|=0 ∥ g ( T ) x − g ′ ( T ) x ∥ = ∥ [ g − g ′ ] ∥ = 0 D = { [ g ] : g ∈ B b ( σ ( T ) ) } D=\{[g]:g\in B_b(\sigma(T))\} D = {[ g ] : g ∈ B b ( σ ( T ))} L 2 ( μ x ) L^2(\mu_x) L 2 ( μ x ) [ g ] ↦ g ( T ) x [g]\mapsto g(T)x [ g ] ↦ g ( T ) x D D D H H H σ ( T ) \sigma(T) σ ( T ) C c ( σ ( T ) , C ) = C ( σ ( T ) ) C_c(\sigma(T),\C)=C(\sigma(T)) C c ( σ ( T ) , C ) = C ( σ ( T )) §E10.34 定理 1.2 により{ [ f ] : f ∈ C ( σ ( T ) ) } \{[f]:f\in C(\sigma(T))\} {[ f ] : f ∈ C ( σ ( T ))} L 2 ( μ x ) L^2(\mu_x) L 2 ( μ x ) D D D §E10.4 系 2.2 により、有界線形写像W : L 2 ( μ x ) → H W\colon L^2(\mu_x)\to H W : L 2 ( μ x ) → H D D D [ g ] ↦ g ( T ) x [g]\mapsto g(T)x [ g ] ↦ g ( T ) x D D D d n → u d_n\to u d n → u W u = lim n W d n Wu=\lim_nW d_n W u = lim n W d n ∥ W u ∥ = lim n ∥ d n ∥ = ∥ u ∥ \|Wu\|=\lim_n\|d_n\|=\|u\| ∥ W u ∥ = lim n ∥ d n ∥ = ∥ u ∥ W W W
u ∈ L 2 ( μ x ) u\in L^2(\mu_x) u ∈ L 2 ( μ x ) { [ f ] : f ∈ C ( σ ( T ) ) } \{[f]:f\in C(\sigma(T))\} {[ f ] : f ∈ C ( σ ( T ))} [ f n ] → u [f_n]\to u [ f n ] → u W u = lim n f n ( T ) x ∈ H x Wu=\lim_nf_n(T)x\in H_x W u = lim n f n ( T ) x ∈ H x W W W W u n → w Wu_n\to w W u n → w ∥ u n − u m ∥ = ∥ W u n − W u m ∥ \|u_n-u_m\|=\|Wu_n-Wu_m\| ∥ u n − u m ∥ = ∥ W u n − W u m ∥ ( u n ) (u_n) ( u n ) L 2 ( μ x ) L^2(\mu_x) L 2 ( μ x ) u u u W u = w Wu=w W u = w { f ( T ) x : f ∈ C ( σ ( T ) ) } \{f(T)x:f\in C(\sigma(T))\} { f ( T ) x : f ∈ C ( σ ( T ))} H x H_x H x W W W H x H_x H x ( u , v ) ↦ ⟨ W u , W v ⟩ (u,v)\mapsto\langle Wu,Wv\rangle ( u , v ) ↦ ⟨ W u , W v ⟩ ( u , v ) ↦ ⟨ u , v ⟩ (u,v)\mapsto\langle u,v\rangle ( u , v ) ↦ ⟨ u , v ⟩ §E12.1 補題 2.3 (1) により一致する。§E12.1 命題 2.4 (3) により、W W W H x H_x H x W x W_x W x W x W_x W x W x [ g ] = g ( T ) x W_x[g]=g(T)x W x [ g ] = g ( T ) x g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) W ′ : L 2 ( μ x ) → H x W'\colon L^2(\mu_x)\to H_x W ′ : L 2 ( μ x ) → H x f ∈ C ( σ ( T ) ) f\in C(\sigma(T)) f ∈ C ( σ ( T )) W ′ [ f ] = f ( T ) x W'[f]=f(T)x W ′ [ f ] = f ( T ) x W ′ W' W ′ W x W_x W x { [ f ] : f ∈ C ( σ ( T ) ) } \{[f]:f\in C(\sigma(T))\} {[ f ] : f ∈ C ( σ ( T ))} W ′ = W x W'=W_x W ′ = W x
主張 (3) を示す。g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) ∣ g u ∣ ≤ ∥ g ∥ ∞ ∣ u ∣ |gu|\le\|g\|_\infty|u| ∣ g u ∣ ≤ ∥ g ∥ ∞ ∣ u ∣ u ↦ g u u\mapsto gu u ↦ g u L 2 ( μ x ) L^2(\mu_x) L 2 ( μ x ) ∥ g ∥ ∞ \|g\|_\infty ∥ g ∥ ∞ h ∈ B b ( σ ( T ) ) h\in B_b(\sigma(T)) h ∈ B b ( σ ( T )) 定理 2.2 (1) の乗法性により
g ( T ) W x [ h ] = g ( T ) h ( T ) x = ( g h ) ( T ) x = W x [ g h ] g(T)W_x[h]=g(T)h(T)x=(gh)(T)x=W_x[gh] g ( T ) W x [ h ] = g ( T ) h ( T ) x = ( g h ) ( T ) x = W x [ g h ] である。g ( T ) W x g(T)W_x g ( T ) W x u ↦ W x ( g u ) u\mapsto W_x(gu) u ↦ W x ( g u ) L 2 ( μ x ) L^2(\mu_x) L 2 ( μ x ) H H H D D D u u u g ( T ) W x u = W x ( g u ) g(T)W_xu=W_x(gu) g ( T ) W x u = W x ( g u ) g ( T ) ( H x ) = g ( T ) W x ( L 2 ( μ x ) ) ⊆ H x g(T)(H_x)=g(T)W_x(L^2(\mu_x))\subseteq H_x g ( T ) ( H x ) = g ( T ) W x ( L 2 ( μ x )) ⊆ H x ι ∈ C ( σ ( T ) ) \iota\in C(\sigma(T)) ι ∈ C ( σ ( T )) ι ( T ) = T \iota(T)=T ι ( T ) = T 定理 2.2 (1) によりι ‾ ( T ) = T ∗ \overline\iota(T)=T^* ι ( T ) = T ∗ T ( H x ) ⊆ H x T(H_x)\subseteq H_x T ( H x ) ⊆ H x T ∗ ( H x ) ⊆ H x T^*(H_x)\subseteq H_x T ∗ ( H x ) ⊆ H x T W x u = W x ( ι u ) TW_xu=W_x(\iota u) T W x u = W x ( ι u ) ▨
証明. x ∈ H x\in H x ∈ H H x H_x H x μ x \mu_x μ x L 2 ( μ x ) L^2(\mu_x) L 2 ( μ x ) W x W_x W x 補題 3.1 のとおりとする。
H ∖ { 0 } H\setminus\{0\} H ∖ { 0 } S S S x , x ′ ∈ S x,x'\in S x , x ′ ∈ S H x ⊆ H x ′ ⊥ H_x\subseteq H_{x'}^\perp H x ⊆ H x ′ ⊥ O \mathcal O O ∅ ∈ O \emptyset\in\mathcal O ∅ ∈ O C \mathcal C C O \mathcal O O ⋃ C \bigcup\mathcal C ⋃ C C \mathcal C C ⋃ C ∈ O \bigcup\mathcal C\in\mathcal O ⋃ C ∈ O C \mathcal C C §E1.20 定理 2.1 (3) により、O \mathcal O O S S S y ∈ H y\in H y ∈ H x ∈ S x\in S x ∈ S H x H_x H x y ≠ 0 y\ne0 y = 0 x ∈ S x\in S x ∈ S f , f ′ ∈ C ( σ ( T ) ) f,f'\in C(\sigma(T)) f , f ′ ∈ C ( σ ( T )) f ( T ) ∗ = f ‾ ( T ) f(T)^*=\overline f(T) f ( T ) ∗ = f ( T ) ( f ‾ f ′ ) ( T ) x ∈ H x (\overline ff')(T)x\in H_x ( f f ′ ) ( T ) x ∈ H x
⟨ f ( T ) y , f ′ ( T ) x ⟩ = ⟨ y , ( f ‾ f ′ ) ( T ) x ⟩ = 0 \langle f(T)y,f'(T)x\rangle=\langle y,(\overline ff')(T)x\rangle=0 ⟨ f ( T ) y , f ′ ( T ) x ⟩ = ⟨ y , ( f f ′ ) ( T ) x ⟩ = 0 であり、§E10.3 補題 1.2 による内積の連続性により、すべてのx ∈ S x\in S x ∈ S H y ⊆ H x ⊥ H_y\subseteq H_x^\perp H y ⊆ H x ⊥ y ∈ S y\in S y ∈ S y ∈ H y ⊆ H y ⊥ y\in H_y\subseteq H_y^\perp y ∈ H y ⊆ H y ⊥ y = 0 y=0 y = 0 y ∉ S y\notin S y ∈ / S S ∪ { y } ∈ O S\cup\{y\}\in\mathcal O S ∪ { y } ∈ O S S S S S S ⋃ x ∈ S H x \bigcup_{x\in S}H_x ⋃ x ∈ S H x M M M M ⊥ = { 0 } M^\perp=\{0\} M ⊥ = { 0 } §E10.3 命題 2.3 によりM M M H H H
x ∈ S x\in S x ∈ S P x P_x P x H x H_x H x §E12.7 命題 1.6 (3) により、V y = ( P x y ) x ∈ S Vy=(P_xy)_{x\in S} V y = ( P x y ) x ∈ S V : H → ⨁ x ∈ S H x V\colon H\to\bigoplus_{x\in S}H_x V : H → ⨁ x ∈ S H x I = S I=S I = S μ i = μ x \mu_i=\mu_x μ i = μ x i = x i=x i = x ( u x ) x ∈ S ∈ ⨁ x ∈ S L 2 ( μ x ) (u_x)_{x\in S}\in\bigoplus_{x\in S}L^2(\mu_x) ( u x ) x ∈ S ∈ ⨁ x ∈ S L 2 ( μ x ) ( W x u x ) x ∈ S (W_xu_x)_{x\in S} ( W x u x ) x ∈ S D D D D D D ⟨ D u , D v ⟩ = ∑ x ⟨ W x u x , W x v x ⟩ = ∑ x ⟨ u x , v x ⟩ = ⟨ u , v ⟩ \langle Du,Dv\rangle=\sum_x\langle W_xu_x,W_xv_x\rangle=\sum_x\langle u_x,v_x\rangle=\langle u,v\rangle ⟨ D u , D v ⟩ = ∑ x ⟨ W x u x , W x v x ⟩ = ∑ x ⟨ u x , v x ⟩ = ⟨ u , v ⟩ ⨁ x ∈ S H x \bigoplus_{x\in S}H_x ⨁ x ∈ S H x ( h x ) x ∈ S (h_x)_{x\in S} ( h x ) x ∈ S ( W x ∗ h x ) x ∈ S (W_x^*h_x)_{x\in S} ( W x ∗ h x ) x ∈ S D D D D D D §E12.1 命題 2.4 (3) によりD D D W = D ∗ V W=D^*V W = D ∗ V §E12.1 命題 1.3 (3) と§E12.1 命題 1.3 (4) によりW ∗ W = V ∗ D D ∗ V = I W^*W=V^*DD^*V=I W ∗ W = V ∗ D D ∗ V = I W W ∗ = D ∗ V V ∗ D = I WW^*=D^*VV^*D=I W W ∗ = D ∗ V V ∗ D = I W W W
g ∈ B b ( σ ( T ) ) g\in B_b(\sigma(T)) g ∈ B b ( σ ( T )) 補題 3.1 (3) によりg ( T ) ( H x ) ⊆ H x g(T)(H_x)\subseteq H_x g ( T ) ( H x ) ⊆ H x 定理 2.2 (1) によりg ( T ) ∗ = g ‾ ( T ) g(T)^*=\overline g(T) g ( T ) ∗ = g ( T ) 補題 3.1 (3) をg ‾ \overline g g g ‾ ( T ) ( H x ) ⊆ H x \overline g(T)(H_x)\subseteq H_x g ( T ) ( H x ) ⊆ H x §E12.6 補題 1.3 (1) によりg ( T ) ( H x ⊥ ) ⊆ H x ⊥ g(T)(H_x^\perp)\subseteq H_x^\perp g ( T ) ( H x ⊥ ) ⊆ H x ⊥ y ∈ H y\in H y ∈ H y = P x y + ( y − P x y ) y=P_xy+(y-P_xy) y = P x y + ( y − P x y ) g ( T ) y = g ( T ) P x y + g ( T ) ( y − P x y ) g(T)y=g(T)P_xy+g(T)(y-P_xy) g ( T ) y = g ( T ) P x y + g ( T ) ( y − P x y ) H x H_x H x H x ⊥ H_x^\perp H x ⊥ P x g ( T ) y = g ( T ) P x y P_xg(T)y=g(T)P_xy P x g ( T ) y = g ( T ) P x y u ∈ ⨁ x L 2 ( μ x ) u\in\bigoplus_xL^2(\mu_x) u ∈ ⨁ x L 2 ( μ x ) y = V ∗ D u y=V^*Du y = V ∗ D u x ∈ S x\in S x ∈ S P x y = W x u x P_xy=W_xu_x P x y = W x u x 補題 3.1 (3) により
V g ( T ) V ∗ D u = ( P x g ( T ) y ) x ∈ S = ( g ( T ) W x u x ) x ∈ S = ( W x ( g u x ) ) x ∈ S = D ( ( g u x ) x ∈ S ) Vg(T)V^*Du=\bigl(P_xg(T)y\bigr)_{x\in S}=\bigl(g(T)W_xu_x\bigr)_{x\in S}=\bigl(W_x(gu_x)\bigr)_{x\in S}=D\bigl((gu_x)_{x\in S}\bigr) V g ( T ) V ∗ D u = ( P x g ( T ) y ) x ∈ S = ( g ( T ) W x u x ) x ∈ S = ( W x ( g u x ) ) x ∈ S = D ( ( g u x ) x ∈ S ) であるから、W g ( T ) W ∗ u = D ∗ V g ( T ) V ∗ D u = ( g u x ) x ∈ S Wg(T)W^*u=D^*Vg(T)V^*Du=(gu_x)_{x\in S} W g ( T ) W ∗ u = D ∗ V g ( T ) V ∗ D u = ( g u x ) x ∈ S g = ι g=\iota g = ι ι ( T ) = T \iota(T)=T ι ( T ) = T
H H H C ⊆ H C\subseteq H C ⊆ H x ∈ S x\in S x ∈ S x ∈ H x x\in H_x x ∈ H x x , x ′ ∈ S x,x'\in S x , x ′ ∈ S x / ∥ x ∥ x/\|x\| x /∥ x ∥ x ′ / ∥ x ′ ∥ x'/\|x'\| x ′ /∥ x ′ ∥ 2 \sqrt2 2 x ∈ S x\in S x ∈ S ∥ c x − x / ∥ x ∥ ∥ < 2 / 2 \|c_x-x/\|x\|\|<\sqrt2/2 ∥ c x − x /∥ x ∥∥ < 2 /2 c x ∈ C c_x\in C c x ∈ C x ↦ c x x\mapsto c_x x ↦ c x S S S ▨
例 3.3. H = C 2 H=\C^2 H = C 2 T ∈ B ( H ) T\in B(H) T ∈ B ( H ) x ∈ H x\in H x ∈ H 補題 3.1 のH x H_x H x H H H 定理 3.2 をT T T I I I
証明. T ∗ = T T^*=T T ∗ = T T T T λ ∈ C \lambda\in\C λ ∈ C λ T − T = ( λ − 1 ) T \lambda T-T=(\lambda-1)T λ T − T = ( λ − 1 ) T H ≠ { 0 } H\ne\{0\} H = { 0 } λ ≠ 1 \lambda\ne1 λ = 1 σ ( T ) = { 1 } \sigma(T)=\{1\} σ ( T ) = { 1 } f ∈ C ( σ ( T ) ) f\in C(\sigma(T)) f ∈ C ( σ ( T )) f ( 1 ) 1 f(1)\mathbf 1 f ( 1 ) 1 f ↦ f ( T ) f\mapsto f(T) f ↦ f ( T ) f ( T ) = f ( 1 ) T f(T)=f(1)T f ( T ) = f ( 1 ) T x ∈ H x\in H x ∈ H { f ( T ) x : f ∈ C ( σ ( T ) ) } = { c x : c ∈ C } \{f(T)x:f\in C(\sigma(T))\}=\{cx:c\in\C\} { f ( T ) x : f ∈ C ( σ ( T ))} = { c x : c ∈ C } 1 1 1 H x = { c x : c ∈ C } H_x=\{cx:c\in\C\} H x = { c x : c ∈ C } dim H = 2 \dim H=2 dim H = 2 H x ≠ H H_x\ne H H x = H
I I I { i } \{i\} { i } σ ( T ) = { 1 } \sigma(T)=\{1\} σ ( T ) = { 1 } u u u u ( 1 ) 1 u(1)\mathbf 1 u ( 1 ) 1 L 2 ( σ ( T ) , B ( σ ( T ) ) , μ i ; C ) L^2(\sigma(T),\mathcal B(\sigma(T)),\mu_i;\C) L 2 ( σ ( T ) , B ( σ ( T )) , μ i ; C ) [ 1 ] [\mathbf 1] [ 1 ] 1 1 1 W W W H H H dim H = 2 \dim H=2 dim H = 2 ▨
4 点スペクトルと連続スペクトル 命題 4.1. H ≠ { 0 } H\ne\{0\} H = { 0 }
正規作用素T ∈ B ( H ) T\in B(H) T ∈ B ( H ) σ r ( T ) = ∅ \sigma_r(T)=\emptyset σ r ( T ) = ∅ σ ( T ) \sigma(T) σ ( T ) σ p ( T ) \sigma_p(T) σ p ( T ) σ c ( T ) \sigma_c(T) σ c ( T )
T ∈ B ( H ) T\in B(H) T ∈ B ( H ) E T E_T E T 定理 2.2 (1) の射影値測度とする。このとき
σ p ( T ) = { λ ∈ σ ( T ) : E T ( { λ } ) ≠ 0 } , σ c ( T ) = { λ ∈ σ ( T ) : E T ( { λ } ) = 0 } \sigma_p(T)=\{\lambda\in\sigma(T):E_T(\{\lambda\})\ne0\},\qquad\sigma_c(T)=\{\lambda\in\sigma(T):E_T(\{\lambda\})=0\} σ p ( T ) = { λ ∈ σ ( T ) : E T ({ λ }) = 0 } , σ c ( T ) = { λ ∈ σ ( T ) : E T ({ λ }) = 0 } 主張 (2) のT T T σ ( T ) \sigma(T) σ ( T ) T T T 証明. 主張 (1) を示す。λ ∈ σ ( T ) \lambda\in\sigma(T) λ ∈ σ ( T ) λ I − T \lambda I-T λ I − T §E12.1 命題 2.4 (2) によりker ( T ∗ − λ ‾ I ) = ker ( T − λ I ) = { 0 } \ker(T^*-\overline\lambda I)=\ker(T-\lambda I)=\{0\} ker ( T ∗ − λ I ) = ker ( T − λ I ) = { 0 } §E12.1 命題 1.3 (1) 、§E12.1 命題 1.3 (2) 、§E12.1 命題 1.3 (5) により( λ I − T ) ∗ = λ ‾ I − T ∗ (\lambda I-T)^*=\overline\lambda I-T^* ( λ I − T ) ∗ = λ I − T ∗ §E12.1 命題 1.4 (2) によりim ( λ I − T ) ‾ = { 0 } ⊥ = H \tpcl{\im(\lambda I-T)}=\{0\}^\perp=H im ( λ I − T ) = { 0 } ⊥ = H λ ∈ σ c ( T ) \lambda\in\sigma_c(T) λ ∈ σ c ( T ) §E12.1 定義 3.3 により主張を得る。
主張 (2) を示す。λ ∈ σ ( T ) \lambda\in\sigma(T) λ ∈ σ ( T ) 命題 2.3 によりker ( λ I − T ) = im E T ( { λ } ) \ker(\lambda I-T)=\im E_T(\{\lambda\}) ker ( λ I − T ) = im E T ({ λ }) λ ∈ σ p ( T ) \lambda\in\sigma_p(T) λ ∈ σ p ( T ) E T ( { λ } ) ≠ 0 E_T(\{\lambda\})\ne0 E T ({ λ }) = 0 T T T 主張 (1) により第二の等式を得る。
主張 (3) を示す。λ \lambda λ σ ( T ) \sigma(T) σ ( T ) { λ } \{\lambda\} { λ } σ ( T ) \sigma(T) σ ( T ) 定理 2.2 (2) によりE T ( { λ } ) ≠ 0 E_T(\{\lambda\})\ne0 E T ({ λ }) = 0 主張 (2) によりλ ∈ σ p ( T ) \lambda\in\sigma_p(T) λ ∈ σ p ( T ) ▨
命題 4.2. H ≠ { 0 } H\ne\{0\} H = { 0 } T ∈ B ( H ) T\in B(H) T ∈ B ( H ) E T E_T E T 定理 2.2 (1) の射影値測度とする。Λ \Lambda Λ T T T λ ∈ Λ \lambda\in\Lambda λ ∈ Λ P λ P_\lambda P λ ker ( λ I − T ) \ker(\lambda I-T) ker ( λ I − T ) Q Q Q ker T \ker T ker T
σ ( T ) ∖ { 0 } = Λ \sigma(T)\setminus\{0\}=\Lambda σ ( T ) ∖ { 0 } = Λ σ ( T ) \sigma(T) σ ( T ) Borel 集合B ⊆ σ ( T ) B\subseteq\sigma(T) B ⊆ σ ( T ) B ∩ Λ B\cap\Lambda B ∩ Λ ( λ j ) (\lambda_j) ( λ j ) x ∈ H x\in H x ∈ H E T ( B ) x = ∑ j P λ j x + 1 B ( 0 ) Q x E_T(B)x=\sum_jP_{\lambda_j}x+\mathbf 1_B(0)Qx E T ( B ) x = j ∑ P λ j x + 1 B ( 0 ) Q x λ ∈ Λ \lambda\in\Lambda λ ∈ Λ E T ( { λ } ) = P λ E_T(\{\lambda\})=P_\lambda E T ({ λ }) = P λ 0 ∈ σ ( T ) 0\in\sigma(T) 0 ∈ σ ( T ) E T ( { 0 } ) = Q E_T(\{0\})=Q E T ({ 0 }) = Q
証明. 主張 (1) を示す。§E12.1 定義 3.3 (1) によりΛ ⊆ σ p ( T ) ∖ { 0 } ⊆ σ ( T ) ∖ { 0 } \Lambda\subseteq\sigma_p(T)\setminus\{0\}\subseteq\sigma(T)\setminus\{0\} Λ ⊆ σ p ( T ) ∖ { 0 } ⊆ σ ( T ) ∖ { 0 } §E12.2 定理 6.1 (1) によりσ ( T ) ∖ { 0 } ⊆ Λ \sigma(T)\setminus\{0\}\subseteq\Lambda σ ( T ) ∖ { 0 } ⊆ Λ §E12.6 定理 4.1 (1) によりΛ \Lambda Λ σ ( T ) \sigma(T) σ ( T )
主張 (2) を示す。命題 2.3 により、λ ∈ Λ \lambda\in\Lambda λ ∈ Λ E T ( { λ } ) = P λ E_T(\{\lambda\})=P_\lambda E T ({ λ }) = P λ 0 ∈ σ ( T ) 0\in\sigma(T) 0 ∈ σ ( T ) E T ( { 0 } ) E_T(\{0\}) E T ({ 0 }) ker ( 0 I − T ) = ker T \ker(0I-T)=\ker T ker ( 0 I − T ) = ker T Q Q Q 0 ∉ σ ( T ) 0\notin\sigma(T) 0 ∈ / σ ( T ) 0 ∉ B 0\notin B 0 ∈ / B 主張 (1) によりB B B B ∩ Λ B\cap\Lambda B ∩ Λ 0 ∈ B 0\in B 0 ∈ B { 0 } \{0\} { 0 } 0 ∈ B 0\in B 0 ∈ B { 0 } \{0\} { 0 } §E12.7 命題 3.2 (5) を適用し、§E12.7 命題 3.2 (1) のE T ( ∅ ) = 0 E_T(\emptyset)=0 E T ( ∅ ) = 0 ▨
例 4.3. m m m [ 0 , 1 ] [0,1] [ 0 , 1 ] H = L 2 ( [ 0 , 1 ] , B ( [ 0 , 1 ] ) , m ; C ) H=L^2([0,1],\mathcal B([0,1]),m;\C) H = L 2 ([ 0 , 1 ] , B ([ 0 , 1 ]) , m ; C ) ( A u ) ( t ) = t u ( t ) (Au)(t)=tu(t) ( A u ) ( t ) = t u ( t ) A A A σ ( A ) = σ c ( A ) = [ 0 , 1 ] \sigma(A)=\sigma_c(A)=[0,1] σ ( A ) = σ c ( A ) = [ 0 , 1 ] σ p ( A ) = ∅ \sigma_p(A)=\emptyset σ p ( A ) = ∅ 定理 2.2 (1) の射影値測度E A E_A E A B ⊆ [ 0 , 1 ] B\subseteq[0,1] B ⊆ [ 0 , 1 ] E A ( B ) u = 1 B u E_A(B)u=\mathbf 1_Bu E A ( B ) u = 1 B u A A A A A A ker A = { 0 } \ker A=\{0\} ker A = { 0 } 命題 4.2 (2) の右辺はB = [ 0 , 1 ] B=[0,1] B = [ 0 , 1 ] 0 0 0 E A ( [ 0 , 1 ] ) = I ≠ 0 E_A([0,1])=I\ne0 E A ([ 0 , 1 ]) = I = 0 命題 4.2 からコンパクト性の仮定を外すことはできない。
証明. F ( B ) u = 1 B u F(B)u=\mathbf 1_Bu F ( B ) u = 1 B u B ∈ B ( [ 0 , 1 ] ) B\in\mathcal B([0,1]) B ∈ B ([ 0 , 1 ]) §E12.7 例 7.2 (1) によりF F F [ 0 , 1 ] [0,1] [ 0 , 1 ] ι : [ 0 , 1 ] → C \iota\colon[0,1]\to\C ι : [ 0 , 1 ] → C Φ F ( ι ) u = ι u = A u \Phi_F(\iota)u=\iota u=Au Φ F ( ι ) u = ι u = A u ι \iota ι §E12.7 定理 4.5 (2) によりA ∗ = Φ F ( ι ‾ ) = A A^*=\Phi_F(\overline\iota)=A A ∗ = Φ F ( ι ) = A A A A u ∈ H u\in H u ∈ H ⟨ A u , u ⟩ = ∫ 0 1 t ∣ u ( t ) ∣ 2 d t \langle Au,u\rangle=\int_0^1t|u(t)|^2\,dt ⟨ A u , u ⟩ = ∫ 0 1 t ∣ u ( t ) ∣ 2 d t 0 0 0 ∥ u ∥ 2 \|u\|^2 ∥ u ∥ 2 A A A §E12.1 命題 4.4 (1) によりσ ( A ) ⊆ [ 0 , 1 ] \sigma(A)\subseteq[0,1] σ ( A ) ⊆ [ 0 , 1 ]
λ ∈ [ 0 , 1 ] \lambda\in[0,1] λ ∈ [ 0 , 1 ] λ ∈ ρ ( A ) \lambda\in\rho(A) λ ∈ ρ ( A ) 0 < δ < 1 / ( ∥ R ( λ , A ) ∥ + 1 ) 0<\delta<1/(\|R(\lambda,A)\|+1) 0 < δ < 1/ ( ∥ R ( λ , A ) ∥ + 1 ) δ \delta δ J = [ 0 , 1 ] ∩ [ λ − δ , λ + δ ] J=[0,1]\cap[\lambda-\delta,\lambda+\delta] J = [ 0 , 1 ] ∩ [ λ − δ , λ + δ ] u = 1 J u=\mathbf 1_J u = 1 J m ( J ) ≥ δ > 0 m(J)\ge\delta>0 m ( J ) ≥ δ > 0 ∥ u ∥ > 0 \|u\|>0 ∥ u ∥ > 0 J J J ∣ λ − t ∣ ≤ δ |\lambda-t|\le\delta ∣ λ − t ∣ ≤ δ ∥ ( λ I − A ) u ∥ ≤ δ ∥ u ∥ \|(\lambda I-A)u\|\le\delta\|u\| ∥ ( λ I − A ) u ∥ ≤ δ ∥ u ∥
∥ u ∥ = ∥ R ( λ , A ) ( λ I − A ) u ∥ ≤ ∥ R ( λ , A ) ∥ δ ∥ u ∥ \|u\|=\|R(\lambda,A)(\lambda I-A)u\|\le\|R(\lambda,A)\|\,\delta\,\|u\| ∥ u ∥ = ∥ R ( λ , A ) ( λ I − A ) u ∥ ≤ ∥ R ( λ , A ) ∥ δ ∥ u ∥ であり、( 1 − ∥ R ( λ , A ) ∥ δ ) ∥ u ∥ ≤ 0 (1-\|R(\lambda,A)\|\delta)\|u\|\le0 ( 1 − ∥ R ( λ , A ) ∥ δ ) ∥ u ∥ ≤ 0 ∥ R ( λ , A ) ∥ δ < 1 \|R(\lambda,A)\|\delta<1 ∥ R ( λ , A ) ∥ δ < 1 ∥ u ∥ > 0 \|u\|>0 ∥ u ∥ > 0 [ 0 , 1 ] ⊆ σ ( A ) [0,1]\subseteq\sigma(A) [ 0 , 1 ] ⊆ σ ( A ) σ ( A ) = [ 0 , 1 ] \sigma(A)=[0,1] σ ( A ) = [ 0 , 1 ]
F F F σ ( A ) \sigma(A) σ ( A ) Φ F ( ι ) = A \Phi_F(\iota)=A Φ F ( ι ) = A 定理 2.2 (1) の一意性によりE A = F E_A=F E A = F λ ∈ [ 0 , 1 ] \lambda\in[0,1] λ ∈ [ 0 , 1 ] m ( { λ } ) = 0 m(\{\lambda\})=0 m ({ λ }) = 0 E A ( { λ } ) u = 1 { λ } u = 0 E_A(\{\lambda\})u=\mathbf 1_{\{\lambda\}}u=0 E A ({ λ }) u = 1 { λ } u = 0 命題 4.1 (2) によりσ p ( A ) = ∅ \sigma_p(A)=\emptyset σ p ( A ) = ∅ σ c ( A ) = [ 0 , 1 ] \sigma_c(A)=[0,1] σ c ( A ) = [ 0 , 1 ] A A A ker A = ker ( 0 I − A ) = { 0 } \ker A=\ker(0I-A)=\{0\} ker A = ker ( 0 I − A ) = { 0 } A A A 命題 4.2 (2) によりE A ( [ 0 , 1 ] ) = 0 E_A([0,1])=0 E A ([ 0 , 1 ]) = 0 E A ( [ 0 , 1 ] ) = F ( [ 0 , 1 ] ) = I E_A([0,1])=F([0,1])=I E A ([ 0 , 1 ]) = F ([ 0 , 1 ]) = I H ≠ { 0 } H\ne\{0\} H = { 0 } A A A ▨
5 正作用素の平方根と極分解 定理 5.1. H H H A ∈ B ( H ) A\in B(H) A ∈ B ( H ) B ∈ B ( H ) B\in B(H) B ∈ B ( H ) B 2 = A B^2=A B 2 = A B B B A 1 / 2 A^{1/2} A 1/2 S ∈ B ( H ) S\in B(H) S ∈ B ( H ) S A = A S SA=AS S A = A S S A 1 / 2 = A 1 / 2 S SA^{1/2}=A^{1/2}S S A 1/2 = A 1/2 S H ≠ { 0 } H\ne\{0\} H = { 0 } s ( t ) = t 1 / 2 s(t)=t^{1/2} s ( t ) = t 1/2 t ∈ σ ( A ) t\in\sigma(A) t ∈ σ ( A ) s ∈ C ( σ ( A ) ) s\in C(\sigma(A)) s ∈ C ( σ ( A )) A 1 / 2 = s ( A ) A^{1/2}=s(A) A 1/2 = s ( A )
証明. H = { 0 } H=\{0\} H = { 0 } B ( H ) = { 0 } B(H)=\{0\} B ( H ) = { 0 } B = 0 B=0 B = 0 H ≠ { 0 } H\ne\{0\} H = { 0 }
§E12.1 命題 2.7 (1) によりA A A §E12.1 命題 4.4 (3) によりσ ( A ) ⊆ [ 0 , ∞ ) \sigma(A)\subseteq[0,\infty) σ ( A ) ⊆ [ 0 , ∞ ) s s s σ ( A ) \sigma(A) σ ( A ) 0 0 0 B = s ( A ) B=s(A) B = s ( A ) §E12.7 補題 6.1 (1) をΦ A \Phi_A Φ A B B B s 2 = ι s^2=\iota s 2 = ι B 2 = ι ( A ) = A B^2=\iota(A)=A B 2 = ι ( A ) = A S A = A S SA=AS S A = A S 定理 1.4 (2) によりS B = B S SB=BS S B = B S
C ∈ B ( H ) C\in B(H) C ∈ B ( H ) C 2 = A C^2=A C 2 = A C A = C 3 = A C CA=C^3=AC C A = C 3 = A C B = s ( A ) B=s(A) B = s ( A ) 定理 1.4 (2) によりC B = B C CB=BC C B = B C B B B C C C §E12.1 命題 2.7 (1) により自己共役作用素である。x ∈ H x\in H x ∈ H y = ( B − C ) x y=(B-C)x y = ( B − C ) x ( B + C ) ( B − C ) = B 2 − C 2 = 0 (B+C)(B-C)=B^2-C^2=0 ( B + C ) ( B − C ) = B 2 − C 2 = 0
⟨ B y , y ⟩ + ⟨ C y , y ⟩ = ⟨ ( B + C ) ( B − C ) x , y ⟩ = 0 \langle By,y\rangle+\langle Cy,y\rangle=\langle(B+C)(B-C)x,y\rangle=0 ⟨ B y , y ⟩ + ⟨ C y , y ⟩ = ⟨( B + C ) ( B − C ) x , y ⟩ = 0 であり、二つの項はともに0 0 0 0 0 0 §E12.1 命題 2.7 (3) により∥ B y ∥ 2 ≤ ∥ B ∥ ⟨ B y , y ⟩ = 0 \|By\|^2\le\|B\|\langle By,y\rangle=0 ∥ B y ∥ 2 ≤ ∥ B ∥ ⟨ B y , y ⟩ = 0 ∥ C y ∥ 2 ≤ ∥ C ∥ ⟨ C y , y ⟩ = 0 \|Cy\|^2\le\|C\|\langle Cy,y\rangle=0 ∥ C y ∥ 2 ≤ ∥ C ∥ ⟨ C y , y ⟩ = 0 ( B − C ) y = 0 (B-C)y=0 ( B − C ) y = 0 B − C B-C B − C
∥ ( B − C ) x ∥ 2 = ⟨ ( B − C ) 2 x , x ⟩ = ⟨ ( B − C ) y , x ⟩ = 0 \|(B-C)x\|^2=\langle(B-C)^2x,x\rangle=\langle(B-C)y,x\rangle=0 ∥ ( B − C ) x ∥ 2 = ⟨( B − C ) 2 x , x ⟩ = ⟨( B − C ) y , x ⟩ = 0 であり、C x = B x Cx=Bx C x = B x C = B C=B C = B ▨
定義 5.2. H 1 , H 2 H_1,H_2 H 1 , H 2 K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } V ∈ B ( H 1 , H 2 ) V\in B(H_1,H_2) V ∈ B ( H 1 , H 2 ) x ∈ ( ker V ) ⊥ x\in(\ker V)^\perp x ∈ ( ker V ) ⊥ ∥ V x ∥ = ∥ x ∥ \|Vx\|=\|x\| ∥ V x ∥ = ∥ x ∥ V V V 部分等長作用素 (partial isometry )
補題 5.3. H 1 , H 2 H_1,H_2 H 1 , H 2 K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } V ∈ B ( H 1 , H 2 ) V\in B(H_1,H_2) V ∈ B ( H 1 , H 2 ) M = ( ker V ) ⊥ M=(\ker V)^\perp M = ( ker V ) ⊥
すべてのx ∈ H 1 x\in H_1 x ∈ H 1 ∥ V x ∥ ≤ ∥ x ∥ \|Vx\|\le\|x\| ∥ V x ∥ ≤ ∥ x ∥
V ∗ V V^*V V ∗ V M M M im V = V ( M ) \im V=V(M) im V = V ( M ) im V \im V im V H 2 H_2 H 2 定理 5.4. H 1 , H 2 H_1,H_2 H 1 , H 2 T ∈ B ( H 1 , H 2 ) T\in B(H_1,H_2) T ∈ B ( H 1 , H 2 ) §E12.1 命題 2.7 (4) によりT ∗ T T^*T T ∗ T ∣ T ∣ = ( T ∗ T ) 1 / 2 |T|=(T^*T)^{1/2} ∣ T ∣ = ( T ∗ T ) 1/2
すべてのx ∈ H 1 x\in H_1 x ∈ H 1 ∥ ∣ T ∣ x ∥ = ∥ T x ∥ \||T|x\|=\|Tx\| ∥∣ T ∣ x ∥ = ∥ T x ∥ ker ∣ T ∣ = ker T \ker|T|=\ker T ker ∣ T ∣ = ker T im ∣ T ∣ ‾ = ( ker T ) ⊥ \tpcl{\im|T|}=(\ker T)^\perp im ∣ T ∣ = ( ker T ) ⊥
部分等長作用素V ∈ B ( H 1 , H 2 ) V\in B(H_1,H_2) V ∈ B ( H 1 , H 2 ) ker V = ker T \ker V=\ker T ker V = ker T T = V ∣ T ∣ T=V|T| T = V ∣ T ∣ T = V ∣ T ∣ T=V|T| T = V ∣ T ∣ T T T 極分解 (polar decomposition )
主張 (2) のV V V V ∗ V V^*V V ∗ V ( ker T ) ⊥ (\ker T)^\perp ( ker T ) ⊥ V ∗ T = ∣ T ∣ V^*T=|T| V ∗ T = ∣ T ∣ im V = im T ‾ \im V=\tpcl{\im T} im V = im T 証明. 主張 (1) を示す。§E12.1 命題 2.7 (1) により∣ T ∣ |T| ∣ T ∣ x ∈ H 1 x\in H_1 x ∈ H 1
∥ ∣ T ∣ x ∥ 2 = ⟨ ∣ T ∣ 2 x , x ⟩ = ⟨ T ∗ T x , x ⟩ = ∥ T x ∥ 2 \||T|x\|^2=\langle|T|^2x,x\rangle=\langle T^*Tx,x\rangle=\|Tx\|^2 ∥∣ T ∣ x ∥ 2 = ⟨ ∣ T ∣ 2 x , x ⟩ = ⟨ T ∗ T x , x ⟩ = ∥ T x ∥ 2 である。したがってker ∣ T ∣ = ker T \ker|T|=\ker T ker ∣ T ∣ = ker T §E12.1 命題 1.4 (2) を∣ T ∣ = ∣ T ∣ ∗ |T|=|T|^* ∣ T ∣ = ∣ T ∣ ∗ im ∣ T ∣ ‾ = ( ker ∣ T ∣ ) ⊥ = ( ker T ) ⊥ \tpcl{\im|T|}=(\ker|T|)^\perp=(\ker T)^\perp im ∣ T ∣ = ( ker ∣ T ∣ ) ⊥ = ( ker T ) ⊥
主張 (2) を示す。R = im ∣ T ∣ R=\im|T| R = im ∣ T ∣ M = R ‾ M=\tpcl R M = R ∣ T ∣ x = ∣ T ∣ x ′ |T|x=|T|x' ∣ T ∣ x = ∣ T ∣ x ′ 主張 (1) により∥ T x − T x ′ ∥ = ∥ ∣ T ∣ x − ∣ T ∣ x ′ ∥ = 0 \|Tx-Tx'\|=\||T|x-|T|x'\|=0 ∥ T x − T x ′ ∥ = ∥∣ T ∣ x − ∣ T ∣ x ′ ∥ = 0 V 0 ( ∣ T ∣ x ) = T x V_0(|T|x)=Tx V 0 ( ∣ T ∣ x ) = T x V 0 : R → H 2 V_0\colon R\to H_2 V 0 : R → H 2 主張 (1) によりV 0 V_0 V 0 §E10.1 命題 1.4 (1) によりM M M R R R M M M §E10.4 系 2.2 によりV 0 V_0 V 0 V 1 : M → H 2 V_1\colon M\to H_2 V 1 : M → H 2 V 1 V_1 V 1 P M P_M P M M M M V = V 1 P M V=V_1P_M V = V 1 P M 主張 (1) と§E10.3 命題 2.3 によりM ⊥ = ker T M^\perp=\ker T M ⊥ = ker T x ∈ H 1 x\in H_1 x ∈ H 1 ∥ V x ∥ = ∥ P M x ∥ \|Vx\|=\|P_Mx\| ∥ V x ∥ = ∥ P M x ∥ V x = 0 Vx=0 V x = 0 x ∈ M ⊥ x\in M^\perp x ∈ M ⊥ ker V = M ⊥ = ker T \ker V=M^\perp=\ker T ker V = M ⊥ = ker T ( ker V ) ⊥ = M (\ker V)^\perp=M ( ker V ) ⊥ = M V = V 1 V=V_1 V = V 1 V V V x ∈ H 1 x\in H_1 x ∈ H 1 V ∣ T ∣ x = V 1 ∣ T ∣ x = T x V|T|x=V_1|T|x=Tx V ∣ T ∣ x = V 1 ∣ T ∣ x = T x
V ′ ∈ B ( H 1 , H 2 ) V'\in B(H_1,H_2) V ′ ∈ B ( H 1 , H 2 ) ker V ′ = ker T \ker V'=\ker T ker V ′ = ker T T = V ′ ∣ T ∣ T=V'|T| T = V ′ ∣ T ∣ V ′ V' V ′ V V V R R R M M M M ⊥ = ker T M^\perp=\ker T M ⊥ = ker T 0 0 0 §E10.3 定理 2.2 によりH 1 = M ⊕ M ⊥ H_1=M\oplus M^\perp H 1 = M ⊕ M ⊥ V ′ = V V'=V V ′ = V
主張 (3) を示す。( ker V ) ⊥ = M = ( ker T ) ⊥ (\ker V)^\perp=M=(\ker T)^\perp ( ker V ) ⊥ = M = ( ker T ) ⊥ 補題 5.3 (2) によりV ∗ V = P M V^*V=P_M V ∗ V = P M im ∣ T ∣ ⊆ M \im|T|\subseteq M im ∣ T ∣ ⊆ M V ∗ T = V ∗ V ∣ T ∣ = P M ∣ T ∣ = ∣ T ∣ V^*T=V^*V|T|=P_M|T|=|T| V ∗ T = V ∗ V ∣ T ∣ = P M ∣ T ∣ = ∣ T ∣ 補題 5.3 (3) によりim V = V ( M ) \im V=V(M) im V = V ( M ) im T = V ( R ) \im T=V(R) im T = V ( R ) im T ‾ ⊆ im V \tpcl{\im T}\subseteq\im V im T ⊆ im V V V V V ( M ) = V ( R ‾ ) ⊆ V ( R ) ‾ = im T ‾ V(M)=V(\tpcl R)\subseteq\tpcl{V(R)}=\tpcl{\im T} V ( M ) = V ( R ) ⊆ V ( R ) = im T im V = im T ‾ \im V=\tpcl{\im T} im V = im T ▨
例 5.5.
ℓ 2 \ell^2 ℓ 2 S S S §E12.1 例 2.9 の右シフトとする。このとき∣ S ∣ = I |S|=I ∣ S ∣ = I 定理 5.4 (2) の部分等長作用素はS S S W ∈ B ( ℓ 2 ) W\in B(\ell^2) W ∈ B ( ℓ 2 ) S = W ∣ S ∣ S=W|S| S = W ∣ S ∣ W = S W=S W = S W W W S = W ∣ S ∣ S=W|S| S = W ∣ S ∣ H ≠ { 0 } H\ne\{0\} H = { 0 } T = 0 ∈ B ( H ) T=0\in B(H) T = 0 ∈ B ( H ) ∣ T ∣ = 0 |T|=0 ∣ T ∣ = 0 定理 5.4 (2) の部分等長作用素は0 0 0 I I I T = I ∣ T ∣ T=I|T| T = I ∣ T ∣ ker I = { 0 } ≠ H = ker T \ker I=\{0\}\ne H=\ker T ker I = { 0 } = H = ker T 定理 5.4 (2) の一意性から条件ker V = ker T \ker V=\ker T ker V = ker T 証明. (1) を示す。§E12.1 例 2.9 によりS ∗ S = I S^*S=I S ∗ S = I I I I I 2 = I I^2=I I 2 = I 定理 5.1 の一意性により∣ S ∣ = I |S|=I ∣ S ∣ = I S S S x x x ∥ S x ∥ = ∥ x ∥ \|Sx\|=\|x\| ∥ S x ∥ = ∥ x ∥ ker S = { 0 } \ker S=\{0\} ker S = { 0 } S S S S = S ∣ S ∣ S=S|S| S = S ∣ S ∣ 定理 5.4 (2) の一意性により、その部分等長作用素はS S S S = W ∣ S ∣ = W I S=W|S|=WI S = W ∣ S ∣ = W I W = S W=S W = S §E12.1 例 2.9 によりS S S
(2) を示す。0 0 0 0 2 = 0 0^2=0 0 2 = 0 定理 5.1 の一意性により∣ T ∣ = 0 |T|=0 ∣ T ∣ = 0 ker V = H \ker V=H ker V = H V ∈ B ( H ) V\in B(H) V ∈ B ( H ) V = 0 V=0 V = 0 I I I x x x ∥ I x ∥ = ∥ x ∥ \|Ix\|=\|x\| ∥ I x ∥ = ∥ x ∥ I ∣ T ∣ = 0 = T I|T|=0=T I ∣ T ∣ = 0 = T ▨
6 演習 解答. ker V \ker V ker V V V V §E10.3 命題 2.3 によりM ⊥ = ker V M^\perp=\ker V M ⊥ = ker V P M P_M P M M M M §E10.3 定理 2.2 により、x ∈ H 1 x\in H_1 x ∈ H 1 x = P M x + k x=P_Mx+k x = P M x + k k ∈ ker V k\in\ker V k ∈ ker V ∥ x ∥ 2 = ∥ P M x ∥ 2 + ∥ k ∥ 2 \|x\|^2=\|P_Mx\|^2+\|k\|^2 ∥ x ∥ 2 = ∥ P M x ∥ 2 + ∥ k ∥ 2
補題 5.3 (1) を示す。V x = V P M x Vx=VP_Mx V x = V P M x V V V M M M ∥ V x ∥ = ∥ P M x ∥ ≤ ∥ x ∥ \|Vx\|=\|P_Mx\|\le\|x\| ∥ V x ∥ = ∥ P M x ∥ ≤ ∥ x ∥
補題 5.3 (2) を示す。C = I − V ∗ V C=I-V^*V C = I − V ∗ V §E12.1 命題 2.7 (4) によりV ∗ V V^*V V ∗ V C C C 補題 5.3 (1) によりすべてのx ∈ H 1 x\in H_1 x ∈ H 1 ⟨ C x , x ⟩ = ∥ x ∥ 2 − ∥ V x ∥ 2 ≥ 0 \langle Cx,x\rangle=\|x\|^2-\|Vx\|^2\ge0 ⟨ C x , x ⟩ = ∥ x ∥ 2 − ∥ V x ∥ 2 ≥ 0 C C C m ∈ M m\in M m ∈ M ⟨ C m , m ⟩ = ∥ m ∥ 2 − ∥ V m ∥ 2 = 0 \langle Cm,m\rangle=\|m\|^2-\|Vm\|^2=0 ⟨ C m , m ⟩ = ∥ m ∥ 2 − ∥ V m ∥ 2 = 0 §E12.1 命題 2.7 (3) により∥ C m ∥ 2 ≤ ∥ C ∥ ⟨ C m , m ⟩ = 0 \|Cm\|^2\le\|C\|\langle Cm,m\rangle=0 ∥ C m ∥ 2 ≤ ∥ C ∥ ⟨ C m , m ⟩ = 0 V ∗ V m = m V^*Vm=m V ∗ V m = m k ∈ ker V k\in\ker V k ∈ ker V V ∗ V k = 0 V^*Vk=0 V ∗ V k = 0 x ∈ H 1 x\in H_1 x ∈ H 1 V ∗ V x = V ∗ V P M x + V ∗ V k = P M x V^*Vx=V^*VP_Mx+V^*Vk=P_Mx V ∗ V x = V ∗ V P M x + V ∗ V k = P M x V ∗ V = P M V^*V=P_M V ∗ V = P M
補題 5.3 (3) を示す。V x = V P M x Vx=VP_Mx V x = V P M x im V = V ( M ) \im V=V(M) im V = V ( M ) V ( M ) V(M) V ( M ) m n ∈ M m_n\in M m n ∈ M V m n → z Vm_n\to z V m n → z ∥ m n − m n ′ ∥ = ∥ V m n − V m n ′ ∥ \|m_n-m_{n'}\|=\|Vm_n-Vm_{n'}\| ∥ m n − m n ′ ∥ = ∥ V m n − V m n ′ ∥ ( m n ) (m_n) ( m n ) H 1 H_1 H 1 m ∈ H 1 m\in H_1 m ∈ H 1 m n → m m_n\to m m n → m M M M m ∈ M m\in M m ∈ M V V V z = V m ∈ V ( M ) z=Vm\in V(M) z = V m ∈ V ( M ) im V \im V im V ▨
問題 6.2. ℓ 2 \ell^2 ℓ 2 e n e_n e n n ∈ N ≥ 1 n\in\NN n ∈ N ≥ 1 ( λ n ) n ∈ N ≥ 1 (\lambda_n)_{n\in\NN} ( λ n ) n ∈ N ≥ 1 Q ∩ [ 0 , 1 ] \Q\cap[0,1] Q ∩ [ 0 , 1 ] D x = ( λ n x n ) n ≥ 1 Dx=(\lambda_nx_n)_{n\ge1} D x = ( λ n x n ) n ≥ 1 E D E_D E D 定理 2.2 (1) の射影値測度とする。
D D D σ ( D ) = [ 0 , 1 ] \sigma(D)=[0,1] σ ( D ) = [ 0 , 1 ] すべての Borel 集合B ⊆ [ 0 , 1 ] B\subseteq[0,1] B ⊆ [ 0 , 1 ] x ∈ ℓ 2 x\in\ell^2 x ∈ ℓ 2 E D ( B ) x = ( 1 B ( λ n ) x n ) n ≥ 1 E_D(B)x=(\mathbf 1_B(\lambda_n)x_n)_{n\ge1} E D ( B ) x = ( 1 B ( λ n ) x n ) n ≥ 1
σ p ( D ) = Q ∩ [ 0 , 1 ] \sigma_p(D)=\Q\cap[0,1] σ p ( D ) = Q ∩ [ 0 , 1 ] σ c ( D ) = [ 0 , 1 ] ∖ Q \sigma_c(D)=[0,1]\setminus\Q σ c ( D ) = [ 0 , 1 ] ∖ Q D D D ( e n ) (e_n) ( e n ) σ c ( D ) \sigma_c(D) σ c ( D ) E D ( [ 0 , 1 ] ∖ Q ) = 0 E_D([0,1]\setminus\Q)=0 E D ([ 0 , 1 ] ∖ Q ) = 0 σ c ( D ) \sigma_c(D) σ c ( D ) σ ( D ) \sigma(D) σ ( D )
解答. (1) を示す。( λ n ) (\lambda_n) ( λ n ) §E12.1 例 2.8 によりD ∈ B ( ℓ 2 ) D\in B(\ell^2) D ∈ B ( ℓ 2 ) x ∈ ℓ 2 x\in\ell^2 x ∈ ℓ 2 ⟨ D x , x ⟩ = ∑ n λ n ∣ x n ∣ 2 \langle Dx,x\rangle=\sum_n\lambda_n|x_n|^2 ⟨ D x , x ⟩ = ∑ n λ n ∣ x n ∣ 2 0 0 0 ∥ x ∥ 2 \|x\|^2 ∥ x ∥ 2 §E12.1 命題 4.4 (1) によりσ ( D ) ⊆ [ 0 , 1 ] \sigma(D)\subseteq[0,1] σ ( D ) ⊆ [ 0 , 1 ] D e n = λ n e n De_n=\lambda_ne_n D e n = λ n e n §E12.1 定義 3.3 (1) によりQ ∩ [ 0 , 1 ] ⊆ σ p ( D ) ⊆ σ ( D ) \Q\cap[0,1]\subseteq\sigma_p(D)\subseteq\sigma(D) Q ∩ [ 0 , 1 ] ⊆ σ p ( D ) ⊆ σ ( D ) §E12.1 定理 3.5 によりσ ( D ) \sigma(D) σ ( D ) Q ∩ [ 0 , 1 ] \Q\cap[0,1] Q ∩ [ 0 , 1 ] [ 0 , 1 ] [0,1] [ 0 , 1 ] [ 0 , 1 ] ⊆ σ ( D ) [0,1]\subseteq\sigma(D) [ 0 , 1 ] ⊆ σ ( D ) σ ( D ) = [ 0 , 1 ] \sigma(D)=[0,1] σ ( D ) = [ 0 , 1 ]
(2) を示す。N ≥ 1 \NN N ≥ 1 P ( N ≥ 1 ) \mathcal P(\NN) P ( N ≥ 1 ) E E E C \C C F F F §E12.7 例 7.2 (2) のとおりとする。λ : N ≥ 1 → [ 0 , 1 ] \lambda\colon\NN\to[0,1] λ : N ≥ 1 → [ 0 , 1 ] n ↦ λ n n\mapsto\lambda_n n ↦ λ n §E12.7 命題 7.1 (1) によりG = E ∘ λ − 1 G=E\circ\lambda^{-1} G = E ∘ λ − 1 [ 0 , 1 ] [0,1] [ 0 , 1 ] G ( B ) x = ( 1 λ − 1 ( B ) ( n ) x n ) n ≥ 1 = ( 1 B ( λ n ) x n ) n ≥ 1 G(B)x=(\mathbf 1_{\lambda^{-1}(B)}(n)x_n)_{n\ge1}=(\mathbf 1_B(\lambda_n)x_n)_{n\ge1} G ( B ) x = ( 1 λ − 1 ( B ) ( n ) x n ) n ≥ 1 = ( 1 B ( λ n ) x n ) n ≥ 1 g ( z ) = z 1 [ 0 , 1 ] ( z ) g(z)=z\mathbf 1_{[0,1]}(z) g ( z ) = z 1 [ 0 , 1 ] ( z ) g ∈ B b ( C ) g\in B_b(\C) g ∈ B b ( C ) ι : [ 0 , 1 ] → C \iota\colon[0,1]\to\C ι : [ 0 , 1 ] → C n n n g ( λ n ) = ι ( λ n ) = λ n g(\lambda_n)=\iota(\lambda_n)=\lambda_n g ( λ n ) = ι ( λ n ) = λ n §E12.7 命題 7.1 (2) をG G G F F F Φ G ( ι ) = Φ E ( ι ∘ λ ) = Φ E ( g ∘ λ ) = Φ F ( g ) \Phi_G(\iota)=\Phi_E(\iota\circ\lambda)=\Phi_E(g\circ\lambda)=\Phi_F(g) Φ G ( ι ) = Φ E ( ι ∘ λ ) = Φ E ( g ∘ λ ) = Φ F ( g ) §E12.7 例 7.2 (2) によりΦ F ( g ) x = ( g ( λ n ) x n ) n ≥ 1 = D x \Phi_F(g)x=(g(\lambda_n)x_n)_{n\ge1}=Dx Φ F ( g ) x = ( g ( λ n ) x n ) n ≥ 1 = D x Φ G ( ι ) = D \Phi_G(\iota)=D Φ G ( ι ) = D G G G σ ( D ) = [ 0 , 1 ] \sigma(D)=[0,1] σ ( D ) = [ 0 , 1 ] Φ G ( ι ) = D \Phi_G(\iota)=D Φ G ( ι ) = D 定理 2.2 (1) の一意性によりE D = G E_D=G E D = G
(3) を示す。λ ∈ [ 0 , 1 ] \lambda\in[0,1] λ ∈ [ 0 , 1 ] (2) によりE D ( { λ } ) x = ( 1 { λ } ( λ n ) x n ) n ≥ 1 E_D(\{\lambda\})x=(\mathbf 1_{\{\lambda\}}(\lambda_n)x_n)_{n\ge1} E D ({ λ }) x = ( 1 { λ } ( λ n ) x n ) n ≥ 1 λ ∈ Q \lambda\in\Q λ ∈ Q λ = λ n \lambda=\lambda_n λ = λ n n n n E D ( { λ } ) e n = e n ≠ 0 E_D(\{\lambda\})e_n=e_n\ne0 E D ({ λ }) e n = e n = 0 λ ∉ Q \lambda\notin\Q λ ∈ / Q n n n 1 { λ } ( λ n ) = 0 \mathbf 1_{\{\lambda\}}(\lambda_n)=0 1 { λ } ( λ n ) = 0 E D ( { λ } ) = 0 E_D(\{\lambda\})=0 E D ({ λ }) = 0 命題 4.1 (2) により主張を得る。
[ 0 , 1 ] ∖ Q [0,1]\setminus\Q [ 0 , 1 ] ∖ Q n n n 1 [ 0 , 1 ] ∖ Q ( λ n ) = 0 \mathbf 1_{[0,1]\setminus\Q}(\lambda_n)=0 1 [ 0 , 1 ] ∖ Q ( λ n ) = 0 (2) によりE D ( [ 0 , 1 ] ∖ Q ) = 0 E_D([0,1]\setminus\Q)=0 E D ([ 0 , 1 ] ∖ Q ) = 0 [ 0 , 1 ] [0,1] [ 0 , 1 ] Q \Q Q σ c ( D ) \sigma_c(D) σ c ( D ) σ c ( D ) \sigma_c(D) σ c ( D ) λ \lambda λ Q ∩ [ 0 , 1 ] \Q\cap[0,1] Q ∩ [ 0 , 1 ] λ \lambda λ σ ( D ) \sigma(D) σ ( D ) ▨