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Theorem 4.27. Let \( {H}_{1},{H}_{2} \) be Hilbert spaces, and let \( T \) be an operator from \( {H}_{1} \) into \( {H}_{2} \) such that \( D\left( T\right) = {H}_{1} \) . Then the following assertions are equivalent: (i) \( T \) is bounded (i.e., \( {f}_{n} \rightarrow f \) implies \( T{f}_{n} \rightarrow {Tf} \) ), ... | Proof. (i) implies (ii): If \( {f}_{n}\overset{w}{ \rightarrow }f \), then for every \( g \in {H}_{2} \) (notice that \( {T}^{ * } \) exists and \( \left. {{T}^{ * } \in B\left( {{H}_{2},{H}_{1}}\right) }\right) \) we have \[ \left\langle {g, T{f}_{n}}\right\rangle = \left\langle {{T}^{ * }g,{f}_{n}}\right\rangle \righ... | Yes |
Theorem 4.28. Let \( H \) be a Hilbert space and let \( \left( {T}_{n}\right) \) be a bounded sequence of symmetric operators from \( B\left( H\right) \). (a) If \( {T}_{n}\overset{w}{ \rightarrow }T \) for some \( T \in B\left( H\right) \), then \( T \) is also symmetric. | Proof. (a) For all \( f, g \in H \) we have \[ \langle f,{Tg}\rangle = \lim \left\langle {f,{T}_{n}g}\right\rangle = \lim \left\langle {{T}_{n}f, g}\right\rangle = \langle {Tf}, g\rangle . \] Therefore, \( T \) is symmetric. | Yes |
Theorem 4.29. For an operator \( P \in B\left( H\right) \) the following statements are equivalent:\n\n(i) \( P \) is an orthogonal projection,\n\n(ii) \( I - P \) is an orthogonal projection,\n\n(iii) \( P \) is idempotent and \( R\left( P\right) = N{\left( P\right) }^{ \bot } \) ,\n\n(iv) \( P \) is idempotent and se... | Proof. (i) and (ii) are equivalent: From the definition of orthogonal projections it follows immediately that \( P \) is the orthogonal projection onto \( M \) if and only if \( I - P \) is the orthogonal projection onto \( {M}^{ \bot } \) . From this it also follows that \( R\left( P\right) = M = {M}^{ \bot \bot } = N... | Yes |
Theorem 4.30. Let \( M \) and \( N \) be closed subspaces of a Hilbert space \( H \), and let \( {P}_{M} \) and \( {P}_{N} \) be the orthogonal projections onto \( M \) and \( N \), respectively.\n\n(a) \( P = {P}_{M}{P}_{N} \) is an orthogonal projection if and only if \( {P}_{M}{P}_{N} = {P}_{N}{P}_{M} \) holds; then... | Proof.\n\n(a) If \( P = {P}_{M}{P}_{N} \) is an orthogonal projection, then \( P \) is self-adjoint. Consequently, \( {P}_{M}{P}_{N} = P = {P}^{ * } = {\left( {P}_{M}{P}_{N}\right) }^{ * } = {P}_{N}^{ * }{P}_{M}^{ * } = {P}_{N}{P}_{M} \) . Let \( {P}_{M}{P}_{N} = \) \( {P}_{N}{P}_{M} \) hold. Then it follows that \( {P... | Yes |
Theorem 4.31. Let \( M \) and \( N \) be closed subspaces of the Hilbert space \( H \), and let \( {P}_{M} \) and \( {P}_{N} \) be the orthogonal projections onto \( M \) and \( N \), respectively.\n\n(a) We have \( 0 \leq {P}_{M} \leq I \).\n\n(b) The following statements are equivalent:\n\n(i) \( {P}_{M} \leq {P}_{N}... | Proof.\n\n(a) For all \( f \in H \) we have \( \langle {Of}, f\rangle = 0 \leq {\begin{Vmatrix}{P}_{M}f\end{Vmatrix}}^{2} = \left\langle {{P}_{M}f, f}\right\rangle \leq \parallel f{\parallel }^{2} = \) \( \langle {If}, f\rangle \) .\n\n(b) (i) implies (ii): If \( {P}_{M} \leq {P}_{N} \), then \( {\begin{Vmatrix}{P}_{M}... | Yes |
Theorem 4.33. Let \( {P}_{1} \) and \( {P}_{2} \) be orthogonal projections acting on the Hilbert space \( H \) . Then we have\n\n\[ \n\begin{Vmatrix}{{P}_{1} - {P}_{2}}\end{Vmatrix} = \max \left\{ {{\rho }_{12},{\rho }_{21}}\right\} .\n\]\n\nwhere\n\n\[ \n{\rho }_{jk} = \sup \left\{ {\begin{Vmatrix}{{P}_{j}h}\end{Vmat... | Proof.\n\n(a) By the definition of the norm of an operator we have (notice that \( \left. {R{\left( {P}_{k}\right) }^{ \bot } = N\left( {P}_{k}\right) }\right) \)\n\n\[ \n\begin{Vmatrix}{{P}_{1} - {P}_{2}}\end{Vmatrix} = \sup \left\{ {\begin{Vmatrix}{\left( {{P}_{1} - {P}_{2}}\right) f}\end{Vmatrix} : f \in H,\parallel... | Yes |
Theorem 4.34. Let \( {H}_{1} \) and \( {H}_{2} \) be Hilbert spaces and let \( U \) be an operator from \( {H}_{1} \) into \( {H}_{2} \) such that \( D\left( U\right) = {H}_{1} \). (a) The following assertions are equivalent: (i) \( U \) is a partial isometry with initial domain \( M \) and final domain \( N \), (ii) \... | Proof. (a) The equivalence of (i) and (ii) follows by (1.8) in the real case and by (1.4) in the complex case. (i) implies (iv): We have \( N\left( {U}^{ * }\right) = R{\left( U\right) }^{ \bot } = {N}^{ \bot } \) and (because (i) implies (ii)) \( \begin{Vmatrix}{{U}^{ * }{Uf}}\end{Vmatrix} = \begin{Vmatrix}{{P}_{M}f}\... | Yes |
Theorem 4.35. If \( P \) and \( Q \) are orthogonal projections on the Hilbert space \( H \) such that \( \parallel P - Q\parallel < 1 \), then we have\n\n(a) \( \dim R\left( P\right) = \dim R\left( Q\right) ,\dim R\left( {I - P}\right) = \dim R\left( {I - Q}\right) \),\n\n(b) \( P \) and \( Q \) are unitarily equivale... | Proof.\n\n(a) We have \( R\left( P\right) \cap R{\left( Q\right) }^{ \bot } = \{ 0\} \), because for \( f \in R\left( P\right) \cap R{\left( Q\right) }^{ \bot }, f \neq 0 \) we would have \( \parallel \left( {P - Q}\right) f\parallel = \parallel {Pf}\parallel = \parallel f\parallel \), consequently \( \parallel P - Q\p... | Yes |
Theorem 5.1. Let \( T \) be an operator from \( {H}_{1} \) into \( {H}_{2} \) . On \( D\left( T\right) \) by\n\n\[ \langle f, g{\rangle }_{T} = \langle f, g\rangle + \langle {Tf},{Tg}\rangle ,\parallel f{\parallel }_{T} = {\left\{ \parallel f{\parallel }^{2} + \parallel Tf{\parallel }^{2}\right\} }^{1/2} \]\n\n a scala... | Proof. The properties of a semi-scalar product are obviously satisfied. Because of the inequality \( \langle f, f{\rangle }_{T} \geq \langle f, f\rangle \) the semi-scalar product \( \langle .,.{\rangle }_{T} \) is positive, thus it is a scalar product.\n\nIf \( T \) is closed and \( \left( {f}_{n}\right) \) is a \( T ... | Yes |
Every bounded operator is closable. A bounded operator \( T \) is closed if and only if \( D\left( T\right) \) is closed. If \( T \) is bounded, then we have \( D\left( \bar{T}\right) = \overline{D\left( T\right) } \) ; the closure \( \bar{T} \) is the bounded extension of \( T \) onto \( \overline{D\left( T\right) } \... | 1. Let \( \left( {f}_{n}\right) \) be a sequence in \( D\left( T\right) \) such that \( {f}_{n} \rightarrow 0 \) and \( T{f}_{n} \rightarrow g \) . Then we have \( \begin{Vmatrix}{T{f}_{n}}\end{Vmatrix} \leq \parallel T\parallel \begin{Vmatrix}{f}_{n}\end{Vmatrix} \rightarrow 0 \) . Therefore \( g = 0 \) .\n\n## 2. For... | Yes |
Theorem 5.3. Let \( T \) be a densely defined operator from \( {H}_{1} \) into \( {H}_{2} \) . (a) \( {T}^{ * } \) is closed. | Proof. (a) By Theorem 4.16 we have \( G\left( {T}^{ * }\right) = {\left( UG\left( T\right) \right) }^{ \bot } \) . Therefore \( G\left( {T}^{ * }\right) \) is closed. | Yes |
An operator \( T \) from \( {H}_{1} \) into \( {H}_{2} \) is closable if and only if there exists a closed extension of \( T \) . | If \( T \) is closable, then we have \( T \subset \bar{T} \) . Therefore \( \bar{T} \) is a closed extension of \( T \) . If \( S \) is a closed extension of \( T \), then we have \( G\left( T\right) \subset G\left( S\right) \) \( = \overline{G\left( S\right) } \), hence \( \overline{G\left( T\right) } \subset G\left( ... | Yes |
Theorem 5.5. Let \( T \) and \( S \) be operators from \( {H}_{1} \) into \( {H}_{2} \), and let \( S \) be \( T \) -bounded with \( T \) -bound less than 1 . Then \( T + S \) is closed (closable) if and only if \( T \) is closed (closable); we have \( D\left( \overline{T + S}\right) = D\left( \bar{T}\right) \) . | Proof. As the \( T \) -bound of \( S \) is less than 1, there exist a \( b < 1 \) and an \( a > 0 \) such that \( \parallel {Sf}\parallel \leq a\parallel f\parallel + b\parallel {Tf}\parallel \) for all \( f \in D\left( T\right) \) . Consequently, for all \( f \in D = D\left( T\right) = D\left( {T + S}\right) \) we hav... | Yes |
Theorem 5.6 (Banach; closed graph theorem). Let \( {H}_{1} \) and \( {H}_{2} \) be Hilbert spaces and let \( T \) be an operator from \( {H}_{1} \) into \( {H}_{2} \) . Then the following statements are equivalent:\n\n(a) \( T \) is closed and \( D\left( T\right) \) is closed,\n\n(b) \( T \) is bounded and \( D\left( T... | Proof. (a) implies (b): We have to show that \( T \) is bounded. Without loss of generality we may assume that \( D\left( T\right) = \overline{D\left( T\right) } = {H}_{1} \) (otherwise we could consider \( T \) as an operator from the Hilbert space \( D\left( T\right) \) into \( {H}_{2} \) ). Consequently, \( {T}^{ * ... | Yes |
Theorem 5.7. Let \( {H}_{1} \) and \( {H}_{2} \) be Hilbert spaces and let \( T \) be an operator from \( {H}_{1} \) into \( {H}_{2} \) such that \( D\left( T\right) = {H}_{1} \) and \( D\left( {T}^{ * }\right) \) is dense in \( {H}_{2} \) . Then \( T \) belongs to \( B\left( {{H}_{1},{H}_{2}}\right) \) . In particular... | Proof. By Theorem 5.3 the operator \( T \) is closable. Because \( D\left( T\right) = {H}_{1} \), we have \( D\left( T\right) = D\left( \bar{T}\right) \), i.e., \( T = \bar{T} \) . Therefore \( T \) is closed and \( D\left( T\right) = {H}_{1} \) . Then \( T \) is bounded by Theorem 5.6. | Yes |
Theorem 5.8. Let \( {H}_{1} \) and \( {H}_{2} \) be Hilbert spaces and let \( T \) be an injective operator from \( {H}_{1} \) into \( {H}_{2} \) such that \( R\left( T\right) = {H}_{2} \) . The operator \( T \) is closed if and only if \( {T}^{-1} \in B\left( {{H}_{1},{H}_{2}}\right) \) . | Proof. By Part (e) of the proposition preceding Theorem 5.1 the operator \( T \) is closed if and only if \( {T}^{-1} \) is closed. The assertion follows immediately from this and from Theorem 5.6. | No |
Theorem 5.9. Let \( {H}_{1},{H}_{2} \) and \( {H}_{3} \) be Hilbert spaces, let \( T \) be a closed operator from \( {H}_{1} \) into \( {H}_{2} \), and let \( S \) be a closable operator from \( {H}_{1} \) into \( {H}_{3} \) such that \( D\left( S\right) \supset D\left( T\right) \) . Then \( S \) is \( T \) -bounded. | Proof. On account of Theorem 5.6 it is enough to show that the operator \( {S}_{0} \) from \( \left( {D\left( T\right) ,\langle .,.{\rangle }_{T}}\right) \) into \( {H}_{3} \), defined by \( D\left( {S}_{0}\right) = D\left( T\right) \) and \( {S}_{0}f = {Sf} \) for \( f \in D\left( T\right) \), is closed. As \( D\left(... | Yes |
Theorem 5.10. Let \( S \) and \( T \) be bijective operators from \( {H}_{1} \) into \( {H}_{2} \) . If \( D\left( S\right) \subset D\left( T\right) \), then\n\n\[ \n{T}^{-1} - {S}^{-1} = {T}^{-1}\left( {S - T}\right) {S}^{-1}.\n\]\n\nIf \( D\left( S\right) = D\left( T\right) \), then\n\n\[ \n{T}^{-1} - {S}^{-1} = {T}^... | Proof. It is enough to prove the first assertion. We shall prove that \( {T}^{-1} = {S}^{-1} + {T}^{-1}\left( {S - T}\right) {S}^{-1} \) . Since \( {T}^{-1}{Tf} = f \) for all \( f \in D\left( T\right) \) and \( S{S}^{-1}g = g \) for all \( g \in {H}_{2} \), it follows that\n\n\[ \n{S}^{-1} + {T}^{-1}\left( {S - T}\rig... | Yes |
Theorem 5.11. Assume that \( {H}_{1} \) and \( {H}_{2} \) are Hilbert spaces, \( T \) is a closed bijective operator from \( {H}_{1} \) into \( {H}_{2}, S \) is an operator from \( {H}_{1} \) into \( {H}_{2} \) such that \( D\left( S\right) \supset D\left( T\right) \), and \( \begin{Vmatrix}{S{T}^{-1}}\end{Vmatrix} < 1... | Proof. For all \( f \in D\left( T\right) \) we have\n\n\[ \parallel {Sf}\parallel = \begin{Vmatrix}{S{T}^{-1}{Tf}}\end{Vmatrix} \leq \begin{Vmatrix}{S{T}^{-1}}\end{Vmatrix}\parallel {Tf}\parallel \]\n\ni.e., \( S \) is \( T \) -bounded with \( T \) -bound less than 1. By Theorem 5.5 the operator \( T + S \) is closed, ... | Yes |
Theorem 5.12. Let \( T \) be a densely defined operator on \( H \) . Then \( \sigma \left( {T}^{ * }\right) = \) \( \sigma {\left( T\right) }^{ * } \) and \( \rho \left( {T}^{ * }\right) = \rho {\left( T\right) }^{ * } \) (here for any subset \( M \) of the complex numbers \( \left. {{M}^{ * } = \left\{ {{z}^{ * } : z ... | Proof. Because of (5.11) it is enough to prove that \( \rho \left( T\right) = \rho {\left( {T}^{ * }\right) }^{ * } \) . To prove this it is enough to show that \( \rho \left( T\right) \subset \rho {\left( {T}^{ * }\right) }^{ * } \), since because of the equality \( {T}^{* * } = T \) we also have \( \rho \left( {T}^{ ... | Yes |
Theorem 5.13. Let \( S \) and \( T \) be closed operators on \( H \). (a) For all \( z,{z}^{\prime } \in \rho \left( T\right) \) we have the first resolvent identity \[ R\left( {z, T}\right) - R\left( {{z}^{\prime }, T}\right) = \left( {{z}^{\prime } - z}\right) R\left( {z, T}\right) R\left( {{z}^{\prime }, T}\right) \... | Proof. The first resolvent identity follows from Theorem 5.10 if in there we replace \( T \) by \( z - T \) and \( S \) by \( z - T \). | No |
Theorem 5.14. If \( T \) is a closed operator on the Hilbert space \( H \), then \( \rho \left( T\right) \) is open, consequently \( \sigma \left( T\right) \) is closed. More precisely, if \( {z}_{0} \in \rho \left( T\right) \), then \( z \in \rho \left( T\right) \) for all \( z \in \mathbb{K} \) such that \( \left| {z... | Proof. Let \( {z}_{0} \in \rho \left( T\right) \), and let \( \left| {z - {z}_{0}}\right| < {\begin{Vmatrix}R\left( {z}_{0}, T\right) \end{Vmatrix}}^{-1} \) . If in Theorem 5.11 we replace \( T \) by \( {z}_{0} - T \) and \( S \) by \( \left( {z - {z}_{0}}\right) I \), then it follows that \( z - T = {z}_{0} - T + \lef... | Yes |
Theorem 5.15. Let \( T \) be a closed operator on the Hilbert space \( H \) . The resolvent \( R\left( {., T}\right) : \rho \left( T\right) \rightarrow B\left( H\right) \) is a continuous function (i.e., for any \( {z}_{0} \in \) \( \rho \left( T\right) \) and any sequence \( \left( {z}_{n}\right) \) from \( \rho \left... | Proof. Let \( {z}_{0}, z \in \rho \left( T\right) \) such that \( \left| {z - {z}_{0}}\right| < {\begin{Vmatrix}R\left( {z}_{0}, T\right) \end{Vmatrix}}^{-1} \) . Then by Theorem 5.14 we have\n\n\[ \begin{Vmatrix}{R\left( {z, T}\right) - R\left( {{z}_{0}, T}\right) }\end{Vmatrix} \leq \mathop{\sum }\limits_{{n = 1}}^{\... | Yes |
Theorem 5.16. Let \( T \) be a closed operator on the complex Hilbert space \( H \) , and let \( f, g \in H \) . Then the functions\n\n\[ R\\left( {., T}\\right) : \\rho \\left( T\\right) \\rightarrow B\\left( H\\right) ,\\;z \\mapsto R\\left( {z, T}\\right) \]\n\n\[ R\\left( {., T}\\right) f : \\rho \\left( T\\right) ... | Proof. Let \( {z}_{0} \in \rho \left( T\right) \), and let \( r = {\\begin{Vmatrix}R\left( {z}_{0}, T\right) \\end{Vmatrix}}^{-1} \) . Then for all \( z \in \mathbb{C} \) such that \( \\left| {z - {z}_{0}}\\right| < r \) we have\n\n\[ R\\left( {z, T}\\right) = \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }{\\left( {z}_{... | Yes |
Assume that \( H \) is a Hilbert space and \( T \in B\left( H\right), r\left( T\right) = \) \( \limsup {\begin{Vmatrix}{T}^{n}\end{Vmatrix}}^{1/n} \). (a) We have \( r\left( T\right) \leq {\begin{Vmatrix}{T}^{m}\end{Vmatrix}}^{1/m} \) for all \( m \in \mathbb{N} \), and thus \( r\left( T\right) = \lim {\begin{Vmatrix}{... | (a) Let \( m \in \mathbb{N} \) . Every \( n \in \mathbb{N} \) can be uniquely represented in the form \( n = m{p}_{n} + {q}_{n} \) with \( {p}_{n},{q}_{n} \in \mathbb{N} \) and \( {q}_{n} < m \) . If we denote \( C = \) \( \max \left\{ {1,\parallel T\parallel ,\begin{Vmatrix}{T}^{2}\end{Vmatrix},\ldots ,\begin{Vmatrix}... | Yes |
Theorem 5.18. Let \( T \) be a Hermitian operator on the pre-Hilbert space \( H \) . Every eigenvalue of \( T \) is real; eigenvectors belonging to different eigenvalues are orthogonal. If \( H \) is complex, then for any \( z \in \mathbb{C} \smallsetminus \mathbb{R} \) the operator \( z - T \) is continuously invertib... | Proof. Let \( z \) be an eigenvalue of \( T \) and let \( f \in N\left( {z - T}\right), f \neq 0 \) . Then \( {z}^{ * }\parallel f{\parallel }^{2} = \langle {Tf}, f\rangle = \langle f,{Tf}\rangle = z\parallel f{\parallel }^{2} \), thus \( z = {z}^{ * } \), i.e., \( z \in \mathbb{R} \) . If \( {z}_{1},{z}_{2} \) are two... | Yes |
Theorem 5.19. Let \( T \) be a symmetric operator on the Hilbert space \( H \) . If \( H = N\left( {s - T}\right) + R\left( {s - T}\right) \) for some \( s \in \overline{\mathbb{R}} \), then \( T \) is self-adjoint and \( H = \) \( N\left( {s - T}\right) \oplus R\left( {s - T}\right) \) . Special case: If \( R\left( {s... | Proof. From \( \bar{T} = {T}^{* * } \) and \( T \subset {T}^{ * } \) it follows that \( N\left( {s - T}\right) \subset N\left( {s - \bar{T}}\right) \) \( = R{\left( s - {T}^{ * }\right) }^{ \bot } \subset R{\left( s - T\right) }^{ \bot } \), thus that \( N\left( {s - T}\right) \bot R\left( {s - T}\right) \) and \( H = ... | Yes |
A symmetric operator \( T \) on a Hilbert space is essentially self-adjoint if and only if \( {T}^{ * } \) is symmetric. We then have \( \bar{T} = {T}^{ * } \) . | Proof. If \( T \) is essentially self-adjoint, then \( {T}^{ * } = {\left( \bar{T}\right) }^{ * } = \bar{T} = {T}^{* * } \), consequently \( {T}^{ * } \) is self-adjoint (therefore symmetric) and we have \( \bar{T} = {T}^{ * } \) . If \( {T}^{ * } \) is symmetric, then since \( \bar{T} \) is symmetric by Theorem 5.4(b)... | Yes |
Theorem 5.22. If \( T \) is a symmetric operator on the complex Hilbert space \( H \) and for some \( n \in \mathbb{N}, n \geq 2 \) we have \( \overline{R\left( {i - {T}^{n}}\right) } = H \) or \( \overline{R\left( {-i - {T}^{n}}\right) } = H \) (respectively \( R\left( {i - {T}^{n}}\right) = H \) or \( \left. {R\left(... | Proof.\n\n(a) Let \( \overline{R\left( {i - {T}^{n}}\right) } = \overline{R\left( {{T}^{n} - i}\right) } = H \) . There are numbers \( {\gamma }_{ \pm } \in \mathbb{C} \) such that \( \operatorname{Im}{\gamma }_{ + } > 0,\operatorname{Im}{\gamma }_{ - } < 0 \) and \( {\gamma }_{ \pm }^{n} = i \) . We then have\n\n\[ \l... | Yes |
Theorem 5.23. (a) The symmetric operator \( T \) on the complex Hilbert space \( H \) is self-adjoint if and only if \( \sigma \left( T\right) \subset \mathbb{R} \) . | Proof. (a) By Theorems 5.18 and 5.21 the operator \( T \) is self-adjoint if and only if \( z - T \) is surjective and continuously invertible for all \( z \in \mathbb{C} \smallsetminus \mathbb{R} \), i.e., if and only if \( \mathbb{C} \smallsetminus \mathbb{R} \subset \rho \left( T\right) \), or, equivalently, \( \sig... | Yes |
Theorem 5.24. If \( T \) is self-adjoint, then the following statements are equivalent:\n\n(i) \( z \in \rho \left( T\right) \),\n\n(ii) there exists a \( c > 0 \) such that \( \parallel \left( {z - T}\right) f\parallel \geq c\parallel f\parallel \) for all \( f \in D\left( T\right) \) (i.e., \( \left( {z - T}\right) \... | Proof. If \( z \in \rho \left( T\right) \), then \( \left( {z - T}\right) \) is injective and \( R\left( {z, T}\right) \) is continuous. If \( \left( {z - T}\right) \) is injective and \( R\left( {z, T}\right) \) is continuous, then \( z \notin {\sigma }_{p}\left( T\right) \) and thus by Theorem 5.23(b) the set \( D\le... | Yes |
Theorem 5.25. Assume that \( {H}_{1} \) and \( {H}_{2} \) are Hilbert spaces, \( A \) and \( B \) are operators from \( {\mathrm{H}}_{1} \) into \( {\mathrm{H}}_{2} \) such that\n\n\[ D\left( A\right) \subset D\left( B\right) \text{ and }\parallel {Bf}\parallel \leq C\parallel {Af}\parallel \text{ for }f \in D\left( A\... | Proof. For \( \left| \kappa \right| < \left( {1/{2C}}\right) \) and for all \( f \in D\left( A\right) \) we have\n\n\[ \begin{aligned} \parallel {Bf}\parallel & \leq C\parallel {Af}\parallel \leq C\left\{ {\parallel \left( {A + {\kappa B}}\right) f\parallel + \left| \kappa \right| \parallel {Bf}\parallel }\right\} \\ &... | Yes |
Theorem 5.26. Assume that \( {H}_{1} \) and \( {H}_{2} \) are Hilbert spaces, \( T \) and \( S \) are operators from \( {H}_{1} \) into \( {H}_{2};T \) is closed, and \( S \) is \( T \) -bounded. Furthermore, denote by \( \mathbf{\Omega } \) the set\n\n\[ \Omega = \{ z \in \mathbb{K} : T + {zS}\text{ is closed }\} \]\n... | Proof. If \( {z}_{0} \in \Omega \), then \( T + {z}_{0}S \) is closed and \( D\left( {T + {z}_{0}S}\right) = D\left( T\right) \) . The operator \( S \) is therefore \( \left( {T + {z}_{0}S}\right) \) -bounded by Theorem 5.9. Hence \( T + {zS} \) is closed for \( z \) sufficiently near \( {z}_{0} \) . Thus \( \Omega \) ... | Yes |
Theorem 5.27. Assume that \( {H}_{1} \) and \( {H}_{2} \) are Hilbert spaces, \( T \) and \( S \) are operators from \( {H}_{1} \) into \( {H}_{2} \) . Let \( T \) be densely defined, let \( S \) be \( T \) -bounded, let \( {S}^{ * } \) be \( {T}^{ * } \) -bounded, and let\n\n\[ \n\Omega = \left\{ {z \in \mathbb{K} : T... | Proof. Let \( {Q}_{z} \) be the orthogonal projection (in \( {H}_{1} \times {H}_{2} \) ) onto \( G\left( {T + {zS}}\right) \) and let \( {Q}_{z}^{\prime } \) be the orthogonal projection onto \( {U}^{-1}G\left( {{T}^{ * } + {z}^{ * }{S}^{ * }}\right) \), where \( U \) is defined as in Section 4.4. By Theorem 5.26 the o... | Yes |
Theorem 5.28 (Rellich-Kato). If \( T \) is self-adjoint (essentially self-adjoint) on the Hilbert space \( H \), the operator \( S \) is symmetric and \( T \) -bounded with \( T \) -bound less than 1, then \( T + S \) is self-adjoint (essentially self-adjoint with \( \overline{T + S} = \bar{T} + \bar{S} \) and \( D\lef... | Proof. (a) If \( T \) is self-adjoint \( \left( {T = {T}^{ * }}\right) \), then because of the inclusion \( S \subset {S}^{ * } \) the operator \( {S}^{ * } \) is \( {T}^{ * } \) -bounded with \( {T}^{ * } \) -bound less than 1 . In Theorem 5.27 we therefore have \( \{ z \in \mathbb{K} : \left| z\right| \leq 1\} \subse... | Yes |
Theorem 5.29 (Wüst [58]). Let \( T \) be a self-adjoint operator on the Hilbert space \( \mathrm{H} \), let \( \mathrm{S} \) be symmetric and \( T \) -bounded, and let\n\n\[ \Omega = \left\{ {z \in \mathbb{K} : T + {zS}}\right. \text{and}\left. {T + {z}^{ * }S\text{are closed}}\right\} \text{,}\]\n\n\[ {\mathbf{\Omega ... | Proof. We have \( {\left( T + zS\right) }^{ * } = T + z{S}^{ * } = T + {zS} \) for every \( z \in \mathbb{R} \cap {\Omega }_{0} \) by Theorem 5.27. | Yes |
Theorem 5.34. Let \( S \) be a symmetric operator on the (real or complex) Hilbert space \( \mathrm{H} \) . (a) If \( \parallel {Sf}\parallel \geq \gamma \parallel f\parallel \) for all \( f \in D\left( S\right) \) with some \( \gamma > 0 \), then there exists a self-adjoint extension \( T \) of \( S \) such that \( \p... | Proof. (a) The operator \( A = {S}^{-1} \) is Hermitian \( (\langle {ASf},{Sg}\rangle = \langle f,{Sg}\rangle = \langle {Sf}, g\rangle = \left. {\langle {Sf},{ASg}\rangle \text{for all}{Sf},{Sg} \in D\left( A\right) = R\left( S\right) }\right) \) and bounded, \( \parallel A\parallel \leq {\gamma }^{-1};A \) is injectiv... | Yes |
Theorem 5.35. If \( t \) is a bounded sesquilinear form on \( H \), then there exists exactly one \( T \in B\left( H\right) \) such that \( t\left( {f, g}\right) = \langle {Tf}, g\rangle \) for all \( f, g \in H \) . We then have \( \parallel T\parallel = \parallel t\parallel \) . | Proof. For every \( f \in H \) the function \( g \mapsto t\left( {f, g}\right) \) is a continuous linear functional on \( H \), since we have \( \left| {t\left( {f, g}\right) }\right| \leq \parallel t\parallel \parallel f\parallel \parallel g\parallel \) . Therefore for each \( f \in H \) there exists exactly one \( \w... | Yes |
Theorem 5.36. Let \( \left( {H,\langle .,.\rangle }\right) \) be a Hilbert space and let \( {H}_{1} \) be a dense subspace of \( H \) . Assume that a scalar product \( \langle .,.{\rangle }_{1} \) is defined on \( {H}_{1} \) in such a way that \( \left( {{H}_{1},\langle .,.{\rangle }_{1}}\right) \) is a Hilbert space a... | Proof. Existence: The element \( \widehat{f} \) in (5.21) is uniquely determined, since \( {H}_{1} \) is dense. Since the mapping \( f \mapsto \widehat{f} \) is also linear,(5.21) defines a linear operator. We can also consider \( T \) as an operator from \( {H}_{1} = \left( {{H}_{1},\langle .,.{\rangle }_{1}}\right) \... | Yes |
Theorem 5.37. Assume that \( H \) is a Hilbert space, \( D \) is a dense subspace of \( H \) and \( s \) is a semi-bounded symmetric sesquilinear form on \( D \) with lower bound \( \gamma \) . Let \( \parallel \cdot {\parallel }_{s} \) be compatible with \( \parallel \cdot \parallel \) . There exists exactly one semi-... | Proof. If we replace \( \left( {{H}_{1},\langle .,.{\rangle }_{1}}\right) \) by \( \left( {{H}_{s},\langle .,.{\rangle }_{s}}\right) \) in Theorem 5.36, then we obtain exactly one self-adjoint operator \( {T}_{0} \) such that \( D\left( {T}_{0}\right) \subset {H}_{s} \) and\n\n\[ \left\langle {{T}_{0}f, g}\right\rangle... | Yes |
Theorem 5.38. Let \( S \) be a semi-bounded symmetric operator with lower bound \( \gamma \) . Then there exists a semi-bounded self-adjoint extension of \( S \) with lower bound \( \gamma \) . If we define \( s\left( {f, g}\right) = \langle {Sf}, g\rangle \) for \( f, g \in D\left( S\right) \), and \( {H}_{s} \) as ab... | Proof. By Theorem 5.37 there exists exactly one self-adjoint operator \( T \) with \( D\left( T\right) \subset {H}_{s} \) and \[ \langle {Tf}, g\rangle = s\left( {f, g}\right) = \langle {Sf}, g\rangle \text{ for }f \in D\left( S\right) \cap D\left( T\right), g \in D\left( S\right) . \] \( \gamma \) is a lower bound for... | Yes |
Let \( \left( {{H}_{1},\langle .,.{\rangle }_{1}}\right) \) and \( \left( {{H}_{2},\langle .,.{\rangle }_{2}}\right) \) be Hilbert spaces and let \( A \) be a densely defined closed operator from \( {H}_{1} \) into \( {H}_{2} \) . Then \( {A}^{ * }A \) is a self-adjoint operator on \( {H}_{1} \) with lower bound \( 0\l... | As \( A \) is closed, \( D\left( A\right) \) is a Hilbert space with the scalar product \( \langle f, g{\rangle }_{A} = \langle {Af},{Ag}{\rangle }_{2} + \langle f, g{\rangle }_{1} \), and \( \parallel f{\parallel }_{A} \geq \parallel f{\parallel }_{1} \) for all \( f \in D\left( A\right) \) . Therefore by Theorem 5.36... | Yes |
Theorem 5.40. Let \( {A}_{1} \) and \( {A}_{2} \) be densely defined closed operators from \( H \) into \( {\mathrm{H}}_{1} \) and from \( \mathrm{H} \) into \( {\mathrm{H}}_{2} \), respectively. Then \( {A}_{1}^{ * }{A}_{1} = {A}_{2}^{ * }{A}_{2} \) if and only if \( D\left( {A}_{1}\right) = D\left( {A}_{2}\right) \) ... | Proof. Assume that \( D\left( {A}_{1}\right) = D\left( {A}_{2}\right) \) and \( \begin{Vmatrix}{{A}_{1}f}\end{Vmatrix} = \begin{Vmatrix}{{A}_{2}f}\end{Vmatrix} \) for all \( f \in D\left( {A}_{1}\right) \) . It follows from (1.4) in the complex case and from (1.8) in the real case that\n\n\[ \left\langle {{A}_{1}f,{A}_... | Yes |
(1) Every normal operator \( T \) is closed and maximal normal (i.e., for every normal operator \( N \) the inclusion \( T \subset N \) implies \( T = N \) ). | (1) The \( T \) -norm and the \( {T}^{ * } \) -norm coincide on \( D\left( T\right) \) . Since \( {T}^{ * } \) is closed, that \( T \) is closed follows from Theorem 5.1. If \( N \) is normal and \( T \subset N \) , then \( D\left( T\right) \subset D\left( N\right) = D\left( {N}^{ * }\right) \subset D\left( {T}^{ * }\r... | Yes |
Theorem 5.41. Let \( T \) be a normal operator.\n\n(a) For every \( z \in \mathbb{K} \) we have \( N\left( {z - T}\right) = N\left( {{z}^{ * } - {T}^{ * }}\right) \) .\n\n(b) If \( {z}_{1},{z}_{2} \) are distinct eigenvalues of \( T \) and \( {f}_{1},{f}_{2} \) are corresponding eigenvectors, then \( {f}_{1} \bot {f}_{... | Proof.\n\n(a) The statement is evident, as \( z - T \) is normal.\n\n(b) By part (a) we have\n\n\[ \left( {{z}_{1} - {z}_{2}}\right) \left\langle {{f}_{1},{f}_{2}}\right\rangle = \left\langle {{z}_{1}^{ * }{f}_{1},{f}_{2}}\right\rangle - \left\langle {{f}_{1},{z}_{2}{f}_{2}}\right\rangle \]\n\n\[ = \left\langle {{T}^{ ... | Yes |
Theorem 5.42. Let \( T \) be normal and injective. Then we have:\n\n(i) \( R\left( T\right) \) is dense,\n\n(ii) \( {T}^{ * } \) is injective,\n\n(iii) \( {T}^{-1} \) is normal,\n\n(iv) \( R\left( T\right) = R\left( {T}^{ * }\right) \) . | Proof. \( R\left( T\right) \) is dense because of the equalities \( R{\left( T\right) }^{ \bot } = N\left( {T}^{ * }\right) = N\left( T\right) = \) \( \{ 0\} \) . Consequently, \( {T}^{ * } \) is injective, and we have \( {\left( {T}^{ * }\right) }^{-1} = {\left( {T}^{-1}\right) }^{ * } \) . Therefore it follows that\n... | Yes |
Theorem 5.43. If \( T \) is normal, then\n\n\[ \rho \left( T\right) = \{ z \in \mathbb{K} : \left( {z - T}\right) \text{ is continuously invertible }\} \]\n\n\[ = \{ z \in \mathbb{K} : R\left( {z - T}\right) = H\} \]\n\n\[ {\sigma }_{p}\left( T\right) = \{ z \in \mathbb{K} : \overline{R\left( {z - T}\right) } \neq H\} ... | Proof. If \( z \in \rho \left( T\right) \), then \( {\left( z - T\right) }^{-1} \in B\left( H\right) \) (in particular, \( z - T \) is continuously invertible). If \( z - T \) is continuously invertible, then (as \( T \) is closed)\n\n\[ R\left( {z - T}\right) = D\left( {\left( z - T\right) }^{-1}\right) = \overline{D\... | Yes |
Theorem 5.44. If \( T \in B\\left( H\\right) \) is normal, then the spectral radius \( r\\left( T\\right) \) equals \( \\parallel T\\parallel \) . | Proof. By Theorem 4.46 we have for all \( T \in B\\left( H\\right) \)\n\n\[ \n\\begin{Vmatrix}{{T}^{ * }T}\\end{Vmatrix} = \\sup \\left\\{ {\\left| {\\left\\langle {{T}^{ * }{Tf}, f}\\right\\rangle : f \in H,\\parallel f}\\right| \leq 1}\\right\\} \n\]\n\n\[ \n= \\sup \\left\\{ {\\parallel {Tf}{\\parallel }^{2} : f \in... | Yes |
Theorem 6.1. Let \( T \) be an operator from \( {H}_{1} \) into \( {H}_{2} \) such that \( D\left( T\right) = {H}_{1} \) . The operator \( T \) is a bounded operator of rank \( m \) if and only if there are linearly independent elements \( {f}_{1},\ldots ,{f}_{m} \) from \( {H}_{1} \) and linearly independent elements ... | Proof. If \( T \) has the form (6.1), then \( R\left( T\right) \subset L\left( {{g}_{1},\ldots ,{g}_{m}}\right) \) . For every \( {j}_{0} \in \{ 1,\ldots, m\} \) there exists an \( {h}_{{j}_{0}} \in L\left( {{f}_{1},\ldots ,{f}_{m}}\right) \) such that \( {h}_{{j}_{0}} \neq 0 \), and \( {h}_{{j}_{0}} \bot {f}_{j} \) fo... | Yes |
Every compact operator is bounded. If \( T \) is compact, then \( \bar{T} \) is also compact. | Proof. Assume that \( T \) is not bounded. Then there exists a sequence \( \left( {f}_{n}\right) \) from \( D\left( T\right) \) with the properties \( \begin{Vmatrix}{f}_{n}\end{Vmatrix} \leq 1 \) and \( \begin{Vmatrix}{T{f}_{n}}\end{Vmatrix} \geq n \) for all \( n \in \mathbb{N} \) . Therefore, no subsequence \( \left... | Yes |
An operator \( T \in B\left( {{H}_{1},{H}_{2}}\right) \) is compact if and only if there exists a sequence \( \left( {T}_{n}\right) \) of finite rank operators from \( B\left( {{H}_{1},{H}_{2}}\right) \) for which \( \begin{Vmatrix}{{T}_{n} - T}\end{Vmatrix} \rightarrow 0 \) . For every compact operator \( T \) the sub... | One direction has already been proved. Let \( T \) now be compact. First we show that \( N{\left( T\right) }^{ \bot } \) is separable. Let \( \left\{ {{e}_{\alpha } : \alpha \in A}\right\} \) be an ONB of \( N{\left( T\right) }^{ \bot } \) . As \( T \) is compact, \( T{e}_{{\alpha }_{n}} \rightarrow 0 \) for every sequ... | Yes |
Theorem 6.6. Assume that \( H \) is a Hilbert space, \( T \in {B}_{\infty }\left( H\right) ,\lambda \in \mathbb{K},\lambda \neq 0 \) . Then \( R\left( {\lambda - T}\right) \) is closed. | Proof. Let \( g \in \overline{R\left( {\lambda - T}\right) } \) . Then there exists a sequence \( \left( {f}_{n}^{\prime }\right) \) from \( H \) such that \( {g}_{n} = \left( {\lambda - T}\right) {f}_{n}^{\prime } \rightarrow g \) . If \( {f}_{n} \) is the orthogonal projection of \( {f}_{n}^{\prime } \) onto \( N{\le... | Yes |
Theorem 6.8. Let \( H \) be a Hilbert space and let \( T \in {B}_{\infty }\left( H\right) \) . If \( \lambda \neq 0 \) is an eigenvalue of \( T \) (hence \( {\lambda }^{ * } \) is an eigenvalue of \( {T}^{ * } \) ) then \( N\left( {\lambda - T}\right) \) and \( N\left( {{\lambda }^{ * } - {T}^{ * }}\right) = R{\left( \... | Proof. We have \( \dim N\left( {\lambda - T}\right) \leq \dim N\left( {{\lambda }^{ * } - {T}^{ * }}\right) \) or \( \dim N\left( {{\lambda }^{ * } - {T}^{ * }}\right) \leq \) \( \dim N\left( {\lambda - T}\right) \) . We treat the first case. There exists then an isometric mapping \( V \) of \( N\left( {\lambda - T}\ri... | Yes |
An operator \( T \in B\left( {{H}_{1},{H}_{2}}\right) \) is a Hilbert-Schmidt operator if and only if \( {T}^{ * } \) is a Hilbert-Schmidt operator. | Proof. Let \( T \in B\left( {{H}_{1},{H}_{2}}\right) \) be a Hilbert-Schmidt operator and let \( \left\{ {e}_{\alpha }\right. \) : \( \alpha \in A\} \) be an ONB of \( {H}_{1} \) such that \( \sum {\begin{Vmatrix}T{e}_{\alpha }\end{Vmatrix}}^{2} < \infty \) . If \( \left\{ {{e}_{\beta }^{\prime } : \beta \in B}\right\}... | Yes |
Theorem 6.11. Let \( {M}_{1} \) and \( {M}_{2} \) be measurable subsets of \( {\mathbb{R}}^{p} \) and \( {\mathbb{R}}^{q} \) , respectively \( {}^{2} \) . The operator \( T \in B\left( {{L}_{2}\left( {M}_{1}\right) ,{L}_{2}\left( {M}_{2}\right) }\right) \) is a Hilbert-Schmidt operator if and only if there exists a ker... | Proof. If \( T \) is of this form, then \( T \in B\left( {{L}_{2}\left( {M}_{1}\right) ,{L}_{2}\left( {M}_{2}\right) }\right) \), by Section 4.1, Example 3. If \( \left\{ {{e}_{n} : n \in \mathbb{N}}\right\} \) and \( \left\{ {{f}_{m} : m \in \mathbb{N}}\right\} \) are orthonormal bases of \( {L}_{2}\left( {M}_{1}\righ... | Yes |
Every Carleman operator is closable. The closure of a Carleman operator is a Carleman operator. Every maximal Carleman operator is closed. | Since an operator is a Carleman operator if and only if it is a restriction of a maximal Carleman operator, it is sufficient to show that every maximal Carleman operator is closed. Let \( k : M \rightarrow H \) be measurable, and let \( {T}_{k} \) be defined by (6.6). Take a sequence \( \left( {f}_{n}\right) \) from \(... | Yes |
Theorem 6.14 (Korotkov [46]). An operator \( T \) from \( H \) into \( {L}_{2}\left( M\right) \) is a Carleman operator if and only if there exists a measurable function \( \kappa : M \rightarrow \mathbb{R} \) such that for all \( f \in D\left( T\right) \)\n\n\[ \left| {{Tf}\left( x\right) }\right| \leq \parallel f\par... | Proof. If \( T \) is a Carleman operator induced by \( k \), then (6.7) holds with \( \kappa \left( x\right) = \parallel k\left( x\right) \parallel \) . Let (6.7) now be satisfied. Then there exists a bounded function \( g : M \rightarrow \left( {0,\infty }\right) \) for which \( {g\kappa } \in {L}_{2}\left( M\right) \... | Yes |
Theorem 6.17. An operator \( T \) from \( {L}_{2}\left( {M}_{1}\right) \) into \( {L}_{2}\left( {M}_{2}\right) \) is a Carleman operator if and only if there exists a measurable function \( K : {M}_{2} \times {M}_{1} \rightarrow \mathbb{C} \) such that \( K\left( {x, \cdot }\right) \in {L}_{2}\left( {M}_{1}\right) \) a... | Proof. If \( T \) is induced by a Carleman kernel \( K \) in the sense of (6.8), then the assumption of Theorem 6.14 (Korotkov) is fulfilled with \( \kappa \left( x\right) = \) \( \parallel K\left( {x,}\right) \parallel \) ; so \( T \) is a Carleman operator. If \( T \) is a Carleman operator, then we proceed as in the... | Yes |
Theorem 6.18. We have \( {\left( {T}_{k,0}\right) }^{ * } = {T}_{k} \) . (In what follows we write \( {T}_{k,0}^{ * } \) for \( \left. {{\left( {T}_{k,0}\right) }^{ * } \cdot }\right) | Proof. By the definition of \( {T}_{k,0} \) we have for all \( f \in D\left( {T}_{k}\right) \) and \( g \in D\left( {T}_{k,0}\right) \)\n\n\[ \left\langle {g,{T}_{k}f}\right\rangle = \langle g,\langle k\left( \cdot \right), f\rangle \rangle = {\int }_{M}g{\left( x\right) }^{ * }\langle k\left( x\right), f\rangle \mathr... | Yes |
Theorem 6.19. Let \( T \) be a densely defined Carleman operator from \( {L}_{2}\left( {M}_{1}\right) \) into \( {L}_{2}\left( {M}_{2}\right) \) that is induced by the Carleman kernel \( K \) . The adjoint \( {T}^{ * } \) is a Carleman operator if and only if \( {K}^{ + } \) is a Carleman kernel \( \left( {{K}^{ + }\le... | Proof. By assumption, \( T \subset {T}_{K} \) . As \( T \) is closable, \( D\left( {T}^{ * }\right) \) is dense. If \( {T}^{ * } \) is defined by the Carleman kernel \( H : {M}_{1} \times {M}_{2} \rightarrow \mathbb{C} \), then \( {T}^{ * } \subset {T}_{H} \) . Consequently, \( \bar{T} = {T}^{* * } \supset {T}_{H}^{ * ... | Yes |
Theorem 6.20. Let \( {H}_{1},{H}_{2},\left\{ {{e}_{n} : n \in \mathbb{N}}\right\} \), and \( \left\{ {{e}_{n}^{\prime } : n \in \mathbb{N}}\right\} \) be as above. If \( \left( {a}_{jk}\right) \) is a matrix such that \( \mathop{\sum }\limits_{{j = 1}}^{\infty }{\left| {a}_{jk}\right| }^{2} < \infty \) for all \( k \in... | Proof. For all \( n \in \mathbb{N} \) we have\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{jk}\left\langle {{e}_{k},{e}_{n}}\right\rangle = {a}_{jn}\text{ and }\mathop{\sum }\limits_{{j = 1}}^{\infty }{\left| \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{jk}\left\langle {e}_{k},{e}_{n}\right\rangle \right| }^{2}... | Yes |
Theorem 6.21. Let \( {H}_{1} \) and \( {H}_{2} \) be separable Hilbert spaces, and let \( T \) be a densely defined operator from \( {H}_{1} \) into \( {H}_{2} \) . The operator \( T \) is closable if and only if there exist orthonormal bases \( \left\{ {{e}_{n} : n \in \mathbb{N}}\right\} \) of \( {H}_{1} \) and \( \l... | Proof. If \( T \) has this form, then it is closable by Theorem 6.20. If \( T \) is closable, then there are orthonormal bases \( \left\{ {{e}_{n} : n \in \mathbb{N}}\right\} \) of \( {H}_{1} \) in \( D\left( T\right) \) and (since \( D\left( {T}^{ * }\right) \) is dense) \( \left\{ {{e}_{n}^{\prime } : n \in \mathbb{N... | Yes |
Theorem 6.22. If \( {\sum }_{j, k}{\left| {a}_{jk}\right| }^{2} = {C}^{2} < \infty \), then the operator \( A \) defined by (6.9) is a Hilbert-Schmidt operator, and \( \parallel A\parallel \mid = C \) . | Proof. By Theorem \( {6.20}, A \) is densely defined and closed. For the basis \( \left\{ {{e}_{n} : n \in \mathbb{N}}\right\} \) we have\n\n\[ \mathop{\sum }\limits_{n}{\begin{Vmatrix}A{e}_{n}\end{Vmatrix}}^{2} = \mathop{\sum }\limits_{n}{\begin{Vmatrix}\mathop{\sum }\limits_{j}\left( \mathop{\sum }\limits_{k}{a}_{jk}... | Yes |
Theorem 6.23. Let \( {a}_{jk} = {b}_{jk}{c}_{jk} \) ; furthermore, let\n\n\[ \mathop{\sum }\limits_{k}{\left| {b}_{jk}\right| }^{2} \leq {C}_{1}^{2}\;\text{ for all }\;j \in \mathbb{N}, \]\n\nand\n\n\[ \mathop{\sum }\limits_{j}{\left| {c}_{jk}\right| }^{2} \leq {C}_{2}^{2}\;\text{ for all }\;k \in \mathbb{N}, \]\n\nthe... | Proof. For every \( f \in {H}_{1} \) we have\n\n\[ \mathop{\sum }\limits_{j}{\left| \mathop{\sum }\limits_{k}{a}_{jk}\left\langle {e}_{k}, f\right\rangle \right| }^{2} = \mathop{\sum }\limits_{j}{\left| \mathop{\sum }\limits_{k}{b}_{jk}{c}_{jk}\left\langle {e}_{k}, f\right\rangle \right| }^{2} \]\n\n\[ \leq \mathop{\su... | Yes |
Theorem 6.25. Let \( K : {M}_{2} \times {M}_{1} \rightarrow \mathbb{C} \) be measurable, and let \( {K}_{1},{K}_{2} \) be measurable such that \( K\left( {x, y}\right) = {K}_{1}\left( {x, y}\right) {K}_{2}\left( {x, y}\right) \). Let \( \left( {M}_{1j}\right) \) and \( \left( {M}_{2j}\right) \) be increasing sequences ... | Proof. Let us set\n\n\[{H}_{j}\left( {x, y}\right) = \left\{ \begin{array}{l} K\left( {x, y}\right) \;\text{ for }\;\left( {x, y}\right) \in {M}_{2j} \times {M}_{1j}, \\ 0\;\text{ otherwise } \end{array}\right.\]\n\nand \( {L}_{j} = K - {H}_{j} \) . Then \( {T}_{{H}_{l}} \) is a Hilbert-Schmidt (hence compact) operator... | Yes |
Theorem 6.27. Let \( \left( {a, b}\right) \) be an arbitrary open interval in \( \mathbb{R} \), and let \( f \in {W}_{2, n}\left( {a, b}\right) \) . If \( - \infty < a \), then \( {f}^{\left( j\right) } \) can be extended continuously to a for all \( j \in \{ 0,1,\ldots, n - 1\} \) ; if \( a = - \infty \), then \( \mat... | Proof. Let \( c \in \left( {a, b}\right) \) . If \( a > - \infty \), then\n\n\[ \n{\int }_{a}^{c}\left| {{f}^{\left( j + 1\right) }\left( x\right) }\right| \mathrm{d}x \leq {\left\{ \left( c - a\right) {\int }_{a}^{c}{\left| {f}^{\left( j + 1\right) }\left( x\right) \right| }^{2}\mathrm{\;d}x\right\} }^{1/2} < \infty ,... | Yes |
Theorem 6.28. We have \( k \in R\left( {T}_{n,0}\right) \) if and only if \( k \in {C}_{0}^{\infty }\left( {a, b}\right) \) and\n\n\[{\int }_{a}^{b}{x}^{j}k\left( x\right) \mathrm{d}x = 0\;\text{ for all }\;j \in \{ 0,1,\ldots, n - 1\} .\] | Proof. If \( k = {T}_{n,0}g \) for some \( g \in D\left( {T}_{n,0}\right) = {C}_{0}^{\infty }\left( {a, b}\right) \), then we obviously have \( k \in {C}_{0}^{\infty }\left( {a, b}\right) \), and for \( j \in \{ 0,1,\ldots, n - 1\} \)\n\n\[{\int }_{a}^{b}{x}^{j}k\left( x\right) \mathrm{d}x = {\int }_{a}^{b}{x}^{j}\left... | Yes |
Theorem 6.30. In the case \( \left( {a, b}\right) = \mathbb{R} \) we have \( \overline{{T}_{n,0}} = {T}_{n} = {T}_{n}^{ * } \) ; thus \( {T}_{n,0} \) is essentially self-adjoint and \( {T}_{n} \) is self-adjoint. | Proof. The relations \( {T}_{n,0}^{ * } = {T}_{n} \) and \( {T}_{n,0} \subset {T}_{n} \) imply that \( {T}_{n}^{ * } \subset {T}_{n,0}^{ * } = {T}_{n} \) . We show that \( {T}_{n} \) is symmetric; then it will follow that \( {T}_{n}^{ * } \subset {T}_{n} \subset {T}_{n}^{ * } \) ; hence \( {T}_{n} = {T}_{n}^{ * } = {T}... | Yes |
Theorem 6.31. In case \( \left( {a, b}\right) \neq \mathbb{R} \) we have\n\n\[ D\left( \overline{{T}_{n,0}}\right) = \left\{ {f \in {W}_{2, n}\left( {a, b}\right) : \begin{array}{l} f\left( a\right) = {f}^{\prime }\left( a\right) = \cdots = {f}^{\left( n - 1\right) }\left( a\right) = 0\text{ if }a > - \infty , \\ f\lef... | Proof. We write \( {W}_{2, n}^{0}\left( {a, b}\right) \) for the subspace given in the theorem. Let \( {S}_{n} \) be the operator induced on \( {W}_{2, n}^{0}\left( {a, b}\right) \) by \( {\tau }_{n} \) . Then one verifies easily that \( {S}_{n} \) and \( {T}_{n} \) are formal adjoints of each other. Therefore \( {S}_{... | Yes |
Theorem 6.32. The operator \( {T}_{{2n},0} \) is non-negative. The Friedrichs extension \( {T}_{{2n}, F} \) of \( {T}_{{2n},0} \) is given by the formulae\n\n\[ D\left( {T}_{{2n}, F}\right) = \left\{ {f \in D\left( {T}_{2n}\right) : \begin{array}{l} f\left( a\right) = {f}^{\prime }\left( a\right) = \cdots = {f}^{\left(... | Proof. For all \( f \in D\left( {T}_{{2n},0}\right) = {C}_{0}^{\infty }\left( {a, b}\right) \) we obviously have\n\n\[ \left\langle {f,{T}_{{2n},0}f}\right\rangle = {\left( -1\right) }^{n}{\int }_{a}^{b}f{\left( x\right) }^{ * }{f}^{\left( 2n\right) }\left( x\right) \mathrm{d}x = {\int }_{a}^{b}{\left| {f}^{\left( n\ri... | Yes |
If \( \left( {a, b}\right) = \mathbb{R} \), then \[ \sigma \left( {T}_{2n}\right) = \lbrack 0,\infty )\;\text{ and }\;\sigma \left( {T}_{{2n} - 1}\right) = \mathbb{R}\;\text{ for all }\;n \in \mathbb{N}. \] | Proof. (a) \( {T}_{2n} = \overline{{T}_{{2n},0}} \) is non-negative. Hence for every \( s < 0 \) we have \[ {\begin{Vmatrix}\left( s - {T}_{2n}\right) f\end{Vmatrix}}^{2} = {\left| s\right| }^{2}\parallel f{\parallel }^{2} - 2\operatorname{Re}s\left\langle {f,{T}_{2n}f}\right\rangle + {\begin{Vmatrix}{T}_{2n}f\end{Vmat... | Yes |
Theorem 6.34. Let \( T \) be a self-adjoint operator on \( {L}_{2}\left( {a, b}\right) \) induced by \( {\tau }_{n} \) (i.e., \( {T}_{n,0} \subset T = {T}^{ * } \subset {T}_{n} \) ). Let\n\n\[ \left( {\sigma f}\right) \left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{n}{a}_{j}\left( x\right) \frac{{\mathrm{d}}^{j}f\le... | Proof. If \( c = \sup \left\{ {\left| {{a}_{n}\left( x\right) }\right| : x \in \left( {a, b}\right) }\right\} \), then by Theorem 6.26, the operator \( S \) is obviously \( T \) -bounded with \( T \) -bound \( c < 1 \) . The assertion follows from this by Theorem 5.28. | Yes |
Theorem 7.2. (The expansion theorem for compact normal operators). If \( T \) is a compact normal operator on a complex Hilbert space, then there exists a zero-sequence (or a finite sequence) \( \left( {\mu }_{i}\right) \) from \( \mathbb{C} \) and an orthonormal sequence \( \left( {f}_{j}\right) \) from \( \mathrm{H} ... | Proof. Let \( T = \mathop{\sum }\limits_{j}{\lambda }_{j}{P}_{j} \) be the representation from Theorem 7.1. For every \( j \) let \( \left\{ {{g}_{j,1},{g}_{j,2},\ldots ,{g}_{j,{k}_{i}}}\right\} \) be an orthonormal basis of \( R\left( {P}_{j}\right) \) ; furthermore, let \( {\mu }_{j, k} = {\lambda }_{j} \) for \( k =... | No |
Theorem 7.3. Let \( T \) be a compact normal operator on a complex Hilbert space, and let \( \left( {\lambda }_{i}\right) \) be the sequence of non-zero eigenvalues of \( T \) ; every eigenvalue counted according to its multiplicity, and \( \left| {\lambda }_{n}\right| \geq \left| {\lambda }_{n + 1}\right| \) for all \... | Proof. It follows from the relations \( r\left( T\right) = \parallel T\parallel \) and \( \sigma \left( T\right) \smallsetminus \{ 0\} \subset {\sigma }_{p}\left( T\right) \) that \( \left| {\lambda }_{1}\right| = \parallel T\parallel \) (cf. Theorem 5.17(c) and (d)). Let \( \left( {f}_{j}\right) \) be an orthonormal s... | Yes |
Let \( n \in \mathbb{N}, n \geq 2 \) . Every normal compact operator \( T \) on a complex Hilbert space has exactly one normal compact \( {}^{1}n \) th root whose eigenvalues all lie in \( \{ z \in \mathbb{C} : 0 \leq \arg z < {2\pi }/n\} \) . Every non-negative selfadjoint compact operator has exactly one non-negative... | The operator \( {A}_{n} = \mathop{\sum }\limits_{j}{\lambda }_{j}^{1/n}{P}_{j} \) with \( 0 \leq \arg {\lambda }_{j}^{1/n} < {2\pi }/n \) has the required property. Let \( B = \mathop{\sum }\limits_{k}{\mu }_{k}{Q}_{k} \) be an operator having the same property. Then we have in particular that\n\n\[ \mathop{\sum }\limi... | Yes |
Theorem 7.5. Let \( T \) be compact. Then \( \parallel \left| T\right| f\parallel = \parallel {Tf}\parallel \) for all \( f \in H \) . There exists an isometric operator \( U \) from \( \overline{R\left( \left| T\right| \right) } \) onto \( \overline{R\left( T\right) } \) such that \( T = U\left| T\right| \) , and \( \... | Proof. For all \( f \in H \) we have\n\n\[ \parallel \left| T\right| f{\parallel }^{2} = \langle \left| T\right| f,\left| T\right| f\rangle = \left\langle {{\left| T\right| }^{2}f, f}\right\rangle = \left\langle {{T}^{ * }{Tf}, f}\right\rangle = \langle {Tf},{Tf}\rangle = \parallel {Tf}{\parallel }^{2}. \]\n\nIf for ev... | Yes |
Theorem 7.6. Let \( T \in {B}_{\infty }\left( {{H}_{1},{H}_{2}}\right) ,{s}_{j} = {s}_{j}\left( T\right) \) . Then there exist orthonormal sequences \( \left( {f}_{j}\right) \) from \( {H}_{1} \) and \( \left( {g}_{j}\right) \) from \( {H}_{2} \) (these sequences can be finite) for which\n\n\[{Tf} = \mathop{\sum }\limi... | Proof. By Theorem 7.2 the compact operator \( \left| T\right| \) has a representation of the above form. Since \( \left| T\right| \) is non-negative, all the \( {s}_{j} \) are positive. Consequently, with the operator \( U \) from Theorem 7.5 it follows for all \( f \in {H}_{1} \) that\n\n\[{Tf} = U\left| T\right| f = ... | Yes |
Theorem 7.7. Let \( S \) and \( T \) be from \( {B}_{\infty }\left( {H,{H}_{1}}\right) \) . Then\n\n\[ \n{s}_{1}\left( T\right) = \parallel T\parallel \]\n\n\[ \n{s}_{j + 1}\left( T\right) = \mathop{\inf }\limits_{{{g}_{1},\ldots ,{g}_{j} \in H}}\sup \left\{ {\parallel {Tf}\parallel : f \in H, f \bot {g}_{1},\ldots ,{g... | Proof. The formulae (7.4) follow from (7.3), since the \( {s}_{j}\left( T\right) \) are the eigenvalues of \( \left| T\right| \) and since \( \parallel {Tf}\parallel = \parallel \left| T\right| f\parallel \) for all \( f \in H \) . For \( S, T \in \) \( {B}_{\infty }\left( {H,{H}_{1}}\right) \) and \( j, k \in {\mathbb... | Yes |
Theorem 7.8.\n\n(a) If \( S, T \in {B}_{p}\left( {H,{H}_{1}}\right) \left( {0 < p < \infty }\right) \), then \( S + T \) also belongs to \( {B}_{p}\left( {H,{H}_{1}}\right) \), and\n\n\[ \parallel S + T{\parallel }_{p} \leq {2}^{1/p}\left( {\parallel S{\parallel }_{p} + \parallel T{\parallel }_{p}}\right) \;\text{ for ... | Proof.\n\n(a) By virtue of (7.5) we have\n\n\[ \mathop{\sum }\limits_{j}{s}_{j}{\left( S + T\right) }^{p} = \mathop{\sum }\limits_{j}\left\{ {{s}_{{2j} - 1}{\left( S + T\right) }^{p} + {s}_{2j}{\left( S + T\right) }^{p}}\right\} \]\n\n\[ \leq \mathop{\sum }\limits_{j}\left\{ {{\left( {s}_{j}\left( S\right) + {s}_{j}\le... | Yes |
Theorem 7.9. Let \( p, q, r > 0 \) with \( \left( {1/p}\right) + \left( {1/q}\right) = \left( {1/r}\right) \) . We have \( T \in \) \( {B}_{r}\left( {H,{H}_{1}}\right) \) if and only if there exist operators \( {T}_{1} \in {B}_{p}\left( {H,{H}_{2}}\right) \) and \( {T}_{2} \in \) \( {B}_{q}\left( {{H}_{2},{H}_{1}}\righ... | Proof. By Theorem 7.8(b) we have \( {T}_{2}{T}_{1} \in {B}_{r}\left( {H,{H}_{1}}\right) \) for \( {T}_{1} \in {B}_{p}\left( {H,{H}_{2}}\right) \) and \( {T}_{2} \in {B}_{q}\left( {{H}_{2},{H}_{1}}\right) \) . Now let \( T \in {B}_{r}\left( {H,{H}_{1}}\right) \) and let (cf. Theorem 7.6)\n\n\[ \n{Tf} = \mathop{\sum }\li... | Yes |
(a) The set \( {B}_{2}\left( {H,{H}_{1}}\right) \) coincides with the set of Hilbert-Schmidt operators. For \( T \in {B}_{2}\left( {H,{H}_{1}}\right) \) we have \( \parallel T{\parallel }_{2} = \parallel \mid T\parallel \mid \) . | (a) Let \( T \in {B}_{2}\left( {H,{H}_{1}}\right) \) . If \( {f}_{1},{f}_{2},\ldots \) are the orthonormalized eigenelements of \( \left| T\right| \) that belong to the non-zero eigenvalues \( {s}_{j}\left( T\right) \) and if \( \left\{ {{g}_{\alpha } : \alpha \in }\right. \) \( A\} \) is an ONB of \( N\left( \left| T\... | Yes |
(a) Definition (7.8) of the trace of an operator \( T \in {B}_{1}\left( H\right) \) is independent of the choice of the orthonormal basis. | (a) By Theorem 7.10(b) there exist operators \( {T}_{1} \in {B}_{2}\left( {H,{H}_{2}}\right) \) and \( {T}_{2} \in \) \( {B}_{2}\left( {{H}_{2}, H}\right) \) such that \( T = {T}_{2}{T}_{1} \) . If \( \left\{ {{e}_{\alpha } : \alpha \in A}\right\} \) and \( \left\{ {{f}_{\beta } : \beta \in B}\right\} \) are arbitrary ... | Yes |
Theorem 7.12. An operator \( T \) from \( {H}_{1} \) into \( {H}_{2} \) such that \( D\left( T\right) = {H}_{1} \) is in \( {B}_{1}\left( {{H}_{1},{H}_{2}}\right) \) if and only if there are sequences \( \left( {\varphi }_{n}\right) \) from \( {H}_{1} \) and \( \left( {\psi }_{m}\right) \) from \( {H}_{2} \) such that ... | Proof. If \( T \in {B}_{1}\left( {{H}_{1},{H}_{2}}\right) \), then Theorem 7.6 gives the required representation (7.10); moreover, it follows that \( \parallel T{\parallel }_{1} \) is greater than or equal to the given infimum. If (7.10) holds, then \( T \) is compact, since \( T \) is the limit of the finite rank oper... | Yes |
Theorem 7.14. Let \( E \) be a spectral family on the Hilbert space \( H \), and let \( u : \mathbb{R} \rightarrow \mathbb{K} \) be an \( E \) -measurable function. Then the formulae\n\n\[ D\left( {\widehat{E}\left( u\right) }\right) = \left\{ {f \in H : u \in {L}_{2}\left( {\mathbb{R},{\rho }_{f}}\right) }\right\} \]\... | Proof. The mapping \( \widehat{E}\left( u\right) : D\left( {\widehat{E}\left( u\right) }\right) \rightarrow H \) is well-defined, because of (7.17)-(7.21). We show that this mapping is linear. It is clear that \( f \in \) \( D\left( {\widehat{E}\left( u\right) }\right) \) and \( a \in \mathbb{K} \) imply that \( {af} \... | Yes |
Let \( E \) be the spectral family of Example 1. If \( u : \mathbb{R} \rightarrow \mathbb{C} \) is a step function, \( u\left( t\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}{\chi }_{{J}_{j}}\left( t\right) \), then | \[ \left( {\int u\left( t\right) \mathrm{d}E\left( t\right) f}\right) \left( x\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}{\chi }_{{J}_{j}}\left( {g\left( x\right) }\right) f\left( x\right) \] \[ = u\left( {g\left( x\right) }\right) f\left( x\right) ,\;f \in {L}_{2}\left( M\right) . \] | Yes |
If \( E \) is the spectral family of Example 2 on \( { \oplus }_{\alpha \in A}{L}_{2}\left( {\mathbb{R},{\rho }_{\alpha }}\right) \) and \( u : \mathbb{R} \rightarrow \mathbb{C} \) is an \( E \) -measurable function (cf. Exercise 7.18), then\n\n\[ D\left( {\widehat{E}\left( u\right) }\right) = \left\{ {\left( {f}_{\alp... | We call this operator the maximal multiplication operator induced by \( u \) on \( { \oplus }_{\alpha \in A}{L}_{2}\left( {\mathbb{R},{\rho }_{\alpha }}\right) \) . The proof is along the same lines as in Example 1. | No |
Let \( E \) be the spectral family of Example 3. Then every function \( u : \mathbb{R} \rightarrow \mathbb{K} \) is \( E \) -measurable. | since for all \( t \in \left\{ {{\lambda }_{j} : j = }\right. \) \( 1,2,\ldots \} \)\n\n\[ u\left( t\right) = \mathop{\sum }\limits_{n}u\left( {\lambda }_{n}\right) {\chi }_{\left\{ {\lambda }_{n}\right\} }\left( t\right) = \mathop{\lim }\limits_{{m \rightarrow \infty }}\mathop{\sum }\limits_{{n = 1}}^{m}u\left( {\lamb... | No |
Theorem 7.15. Let \( {H}_{1} \) and \( {H}_{2} \) be Hilbert spaces, let \( U \) be a unitary operator from \( {H}_{1} \) onto \( {H}_{2} \), and let \( E \) be a spectral family on \( {H}_{1} \) . Then by the formula\n\n\[ F\left( t\right) = {UE}\left( t\right) {U}^{-1},\;t \in \mathbb{R} \]\n\na spectral family is de... | Proof. It is clear that \( F \) is a spectral family on \( {H}_{2} \) . If \( {\rho }_{f}\left( t\right) = \parallel E\left( t\right) f{\parallel }^{2} \) and \( {\sigma }_{g}\left( t\right) = \parallel F\left( t\right) g{\parallel }^{2} \), then \( {\rho }_{f}\left( t\right) = {\sigma }_{Uf}\left( t\right) \) obviousl... | Yes |
Theorem 7.16. Let \( E \) be a spectral family on the Hilbert space \( H \) . Then there exists a family \( \left\{ {{\rho }_{\alpha } : \alpha \in A}\right\} \) of right continuous non-decreasing functions (the cardinality of \( A \) is at most the dimension of \( H \) ) and a unitary operator\n\n\( U : H \rightarrow ... | Proof. For any \( f \in H, f \neq 0 \) let \( {H}_{f} = \overline{L\{ E\left( t\right) f : t \in \mathbb{R}\} } \) and let \( {\rho }_{f}\left( t\right) = \) \( \parallel E\left( t\right) f{\parallel }^{2} \) . Then the formula\n\n\[ {U}_{f,0}\left( {\mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}E\left( {t}_{j}\right) f}\r... | Yes |
Assume that \( T \) is a compact self-adjoint operator on \( H,\left( {\lambda }_{j}\right) \) is the sequence of non-zero eigenvalues of \( T,\left( {P}_{i}\right) \) is the sequence of the orthogonal projections onto the eigenspaces \( N\left( {{\lambda }_{j} - T}\right) \) and \( {P}_{0} \) is the orthogonal project... | defines the spectral family of \( T \), since\n\n\[ \widehat{E}\left( \mathrm{{id}}\right) f = \mathop{\sum }\limits_{j}{\lambda }_{j}{P}_{j}f + 0{P}_{0}f = {Tf}\text{ for all }f \in H \]\n\nby Theorem 7.1. | Yes |
Theorem 7.18 (Spectral representation theorem). Let \( T \) be a self-adjoint operator on \( H \) . Then there exist a family \( \left\{ {{\rho }_{\alpha } : \alpha \in A}\right\} \) of right continuous non-decreasing functions and a unitary operator \( U \) from \( H \) onto \( { \oplus }_{\alpha \in A}{L}_{2}\left( {... | Proof. If \( E \) is the spectral family of \( T \), and \( { \oplus }_{\alpha \in A}{L}_{2}\left( {\mathbb{R},{\rho }_{\alpha }}\right), U \), and \( F \) are constructed as in Theorem 7.16, then by Theorem 7.16 and Theorem 7.17 \[ T = \widehat{E}\left( \mathrm{{id}}\right) = {U}^{-1}\widehat{F}\left( \mathrm{{id}}\ri... | Yes |
Theorem 7.19. If \( u\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{n}{c}_{j}{t}^{j} \), then \( u\left( T\right) = \mathop{\sum }\limits_{{j = 0}}^{n}{c}_{j}{T}^{j} \), where we set \( {T}^{0} = I \) . | Proof. The assertion obviously holds for \( n = 0 \) (cf. Theorem 7.14(d)). Let us assume that it holds for polynomials of degree \( \leq n - 1 \) . Then \( v\left( T\right) = \) \( \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}{T}^{j - 1} \) for \( v\left( t\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}{t}^{j - 1} \... | Yes |
Theorem 7.21. A self-adjoint \( T \) on \( H \) is bounded if and only if there exist real numbers \( {\gamma }_{1} \) and \( {\gamma }_{2} \) for which\n\n\[ E\left( t\right) = \left\{ \begin{array}{lll} 0 & \text{ for } & t < {\gamma }_{1}, \\ I & \text{ for } & t \geq {\gamma }_{2}. \end{array}\right. \] | We can then choose\n\n\[ {\gamma }_{1} = m = \inf \{ \langle f,{Tf}\rangle : f \in D\left( T\right) ,\parallel f\parallel = 1\} ,\]\n\n\[ {\gamma }_{2} = M = \sup \{ \langle f,{Tf}\rangle : f \in D\left( T\right) ,\parallel f\parallel = 1\} . \]\n\nFor \( m < t < M \) we have \( E\left( t\right) \neq 0 \) and \( E\left... | Yes |
Theorem 7.22. Let \( T \) be a self-adjoint operator on \( H \), let \( E \) be the spectral family of \( T \), and let \( {T}_{0} \) be a restriction of \( T \) for which \( {\bar{T}}_{0} = T \) . Then the following statements are equivalent:\n\n(i) \( s \in \sigma \left( T\right) \) ;\n\n(ii) there exists a sequence ... | Proof. The equivalence of (i) and (ii) immediately follows from Theorem 5.24.\n\n(ii) implies (iii): Since \( D\left( {T}_{0}\right) \) is a core of \( T \), for every \( n \in \mathbb{N} \) there exists a \( {g}_{n} \in D\left( {T}_{0}\right) \) for which \( \begin{Vmatrix}{{g}_{n} - {f}_{n}}\end{Vmatrix} \leq {n}^{-1... | Yes |
Corollary 1. Let \( a < b \) . If \( E\left( b\right) - E\left( a\right) \neq 0 \) then \( (a, b\rbrack \cap \sigma \left( T\right) \neq \varnothing \) . We have \( \left( {a, b}\right) \cap \sigma \left( T\right) \neq \varnothing \) if and only if \( E\left( {b - }\right) - E\left( a\right) \neq 0 \) . | Proof.\n\n(a) Assume that \( (a, b\rbrack \subset \rho \left( T\right) \) . Then by Theorem 7.22 the spectral family \( E \) is constant in some neighborhood of \( s \) for every \( s \in (a, b\rbrack \) . Consequently, \( E \) is constant in \( (a, b\rbrack \), and thus \( E\left( b\right) - E\left( a\right) = s - \) ... | Yes |
Corollary 2. A self-adjoint operator \( T \) is bounded from below if and only if its spectrum is bounded from below. The greatest lower bound of \( T \) is equal to \( \min \sigma \left( T\right) \) . | Proof. By Part 3 of the proposition preceding Theorem 7.20, we have \( \langle f,{Tf}\rangle \geq \gamma \parallel f{\parallel }^{2} \) for all \( f \in D\left( T\right) \) if and only if \( E\left( t\right) = 0 \) for \( t < \gamma \) . If \( E\left( t\right) = 0 \) for \( t < \gamma \), then by Theorem 7.22 no spectr... | Yes |
Theorem 7.23. Let \( T,{T}_{0} \) and \( E \) be as in Theorem 7.22. Then the following assertions are equivalent:\n\n(i) \( s \in {\sigma }_{p}\left( T\right) \) ;\n\n(ii) there exists a Cauchy sequence \( \left( {f}_{n}\right) \) from \( D\left( T\right) \) for which \( \lim \begin{Vmatrix}{f}_{n}\end{Vmatrix} > 0 \)... | Proof. (i) implies (ii): If \( f \) is an eigenelement of \( T \) belonging to the eigenvalue \( s \), then we can choose the constant sequence \( {f}_{n} = f \).\n\n(ii) implies (iii): Since \( D\left( {T}_{0}\right) \) is a core of \( T \), for every \( n \in \mathbb{N} \) there exists a \( {g}_{n} \in D\left( {T}_{0... | Yes |
Theorem 7.25. Let \( T \) be a self-adjoint operator on \( H \), and let \( H = {H}_{1} \oplus {H}_{2} \oplus {H}_{3} \) with \( \dim {H}_{3} = m < \infty \) . Assume that the orthogonal projection \( {P}_{j} \) onto \( {H}_{j} \) maps \( D\left( T\right) \) into itself for \( j = 1,2.{}^{5} \) If\n\n\[ \langle f,{Tf}\... | Proof. Let us assume that \( \dim R\left( {E\left( {b - }\right) - E\left( a\right) }\right) \geq m + 1 \) . Then there exists an \( f \in R\left( {E\left( {b - }\right) - E\left( a\right) }\right) \cap {H}_{3}^{ \bot } \) such that \( f \neq 0 \) . Hence, \( f = {P}_{1}f + {P}_{2}f \) , and, putting \( c = \left( {a +... | Yes |
Theorem 7.26. (a) Let \( T \) be a self-adjoint operator on \( H \), and let \( {H}_{1} \) be a closed subspace of H such that \( \dim {H}_{1}^{ \bot } = m < \infty \) . Assume that \( {P}_{1}D\left( T\right) \subset D\left( T\right) \) for the orthogonal projection \( {P}_{1} \) onto \( {H}_{1} \), and that \[ \langle... | (a) With \( {H}_{2} = \{ 0\} \) and \( {H}_{3} = {H}_{1}^{ \bot } \) the assumptions of Theorem 7.25 are fulfilled for every \( a < b \) . Consequently, \( \left( {-\infty, b}\right) \cap \sigma \left( T\right) \) contains only isolated eigenvalues of finite multiplicity, with total multiplicity less than or equal to \... | Yes |
Theorem 7.27. Let \( T \) be a self-adjoint operator on \( H \) with spectral family \( E \) . Denote, for every \( f \in H \), by \( {\rho }_{f} \) the measure induced on \( \mathbb{R} \) by means of \( \parallel E\left( \text{.}\right) f{\parallel }^{2} \) . (a) \( {\mathrm{H}}_{p} \) equals the set of those \( f \in... | ## Proof. (a) If \( {f}_{j} \) is an eigenelement for the eigenvalue \( {\lambda }_{j} \), and \( f = \mathop{\sum }\limits_{{j = 1}}^{\infty }{c}_{j}{f}_{j} \), then we obviously have \( E\left( \left\{ {{\lambda }_{j} : j \in \mathbb{N}}\right\} \right) f = f \) . If \( A = \left\{ {{\lambda }_{j} : j \in \mathbb{N}}... | Yes |
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