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Proposition 6. The homomorphism \( {\mathrm{{inv}}}_{\mathrm{K}} : {\mathrm{B}}_{\mathrm{K}} \rightarrow \mathbf{Q}/\mathbf{Z} \) is an isomorphism. | This follows from the exact sequence and the fact that \( {\mathrm{B}}_{\bar{\mathrm{K}}} = 0 \) (prop. 5). | No |
Proposition 7. Let \( \mathrm{L} \) be a finite extension of \( \mathrm{K} \) of degree \( n \), and let \( {\operatorname{Res}}_{\mathrm{K}/\mathrm{L}} : {\mathrm{B}}_{\mathrm{K}} \rightarrow \) \( {\mathrm{B}}_{\mathrm{L}} \) be the canonical homomorphism of \( {\mathrm{B}}_{\mathrm{K}} \) into \( {\mathrm{B}}_{\math... | Proof. We begin with two special cases. a) \( \mathrm{L}/\mathrm{K} \) is unramified Then \( {\mathrm{K}}_{nr} = {\mathrm{L}}_{nr} \) ; put \( {\mathrm{g}}_{n} = \mathrm{G}\left( {{\mathrm{K}}_{nr}/\mathrm{L}}\right) = \mathrm{G}\left( {{\overline{\mathrm{K}}}_{nr}/\overline{\mathrm{L}}}\right) \) : it is the unique su... | Yes |
Corollary 1. Keeping the hypotheses and notation of prop. 7, a necessary and sufficient condition for an element \( a \in {\mathbf{B}}_{\mathbf{K}} \) to be split by \( \mathbf{L} \) is \( {na} = 0 \) . | Indeed, to say that \( a \) is split by \( \mathrm{L} \) means that \( {\operatorname{Res}}_{\mathrm{K}/\mathrm{L}}\left( a\right) = 0 \), equivalently \( 0 = {\operatorname{inv}}_{\mathrm{L}} \circ {\operatorname{Res}}_{\mathrm{K}/\mathrm{L}}\left( a\right) = {\operatorname{inv}}_{\mathrm{K}}\left( {na}\right) \), i.e... | Yes |
Corollary 2. Suppose \( \mathrm{L}/\mathrm{K} \) is Galois. Then the isomorphism \( {\operatorname{inv}}_{\mathrm{K}} : {\mathrm{B}}_{\mathrm{K}} \rightarrow \mathbf{Q}/\mathbf{Z} \) maps the subgroup \( {\mathrm{H}}^{2}\left( {\mathrm{\;L}/\mathrm{K}}\right) \) of \( {\mathrm{B}}_{\mathrm{K}} \) onto the subgroup \( \... | Clear. | No |
Corollary 3. Let \( \mathrm{D} \) be a central division algebra over \( \mathrm{K} \) of rank \( {n}^{2} \), and let \( a \in {\mathrm{B}}_{\mathrm{K}} \) be the corresponding element of the Brauer group. Then a has order \( n \) in \( {\mathrm{B}}_{\mathrm{K}} \), and every extension of \( \mathrm{K} \) of degree \( n... | We know that every maximal subfield of \( \mathrm{D} \) splits \( \mathrm{D} \) ; by cor. \( 1,{na} = 0 \) (this is a general fact-cf. Chap. X,§5, exer. 3). If \( d \) is the order of \( a, d \) divides \( n \) . Corollary 1 shows that \( \mathrm{D} \) is split by an extension \( \mathrm{L}/\mathrm{K} \) of degree \( d... | Yes |
For every integer \( n \), the cup product with \( {u}_{\mathrm{F}/\mathrm{E}} \) defines an isomorphism\n\n\[ \n{\widehat{\mathrm{H}}}^{n}\left( {{\mathrm{G}}_{\mathrm{F}/\mathrm{E}},\mathbf{Z}}\right) \rightarrow {\widehat{\mathrm{H}}}^{n + 2}\left( {{\mathrm{G}}_{\mathrm{F}/\mathrm{E}},{\mathrm{F}}^{ * }}\right) \n\... | This is Tate's theorem (th. 1 of Chap. XI). | No |
Proposition 13. Let \( \mathrm{L}/\mathrm{K} \) be an unramified extension. If we identify the groups \( {\mathrm{G}}_{\mathrm{L}/\mathrm{K}} \) and \( {\mathrm{G}}_{\bar{\mathrm{L}}/\bar{\mathrm{K}}} \), then\n\n\[ \left( {x,\mathrm{\;L}/\mathrm{K}}\right) = {\mathrm{F}}_{\mathrm{K}}^{v\left( x\right) } \]\n\nwhere \(... | Set \( g = {G}_{L/K} \) and \( F = {F}_{K} \) for brevity. Let \( \chi \) be a character of \( g \) . We must\ncheck that\n\n\[ \chi \left( \left( {x,\mathrm{\;L}/\mathrm{K}}\right) \right) = \chi \left( {\mathrm{F}}^{v\left( x\right) }\right) \]\n\nBy prop. 2 of Chap. XI, the left side is equal to \( {\operatorname{in... | Yes |
Proposition 14. For every \( q \in \mathbf{Z},{\widehat{\mathbf{H}}}^{q}\left( {\mathbf{G},{\widehat{\mathbf{L}}}_{nr}^{ * }}\right) = 0 \) . | This follows from prop. 11 of Chap. X combined with prop. 7 of Chap. V. | No |
Corollary 1. The exact sequence \( 0 \rightarrow {\widehat{\mathrm{U}}}_{nr} \rightarrow {\widehat{\mathrm{L}}}_{nr}^{ * } \rightarrow \mathbf{Z} \rightarrow 0 \) defines an isomorphism of \( {\mathrm{G}}^{a} = {\widehat{\mathrm{H}}}^{-2}\left( {\mathrm{G},\mathbf{Z}}\right) \) onto \( {\widehat{\mathrm{H}}}^{-1}\left(... | Obvious. | No |
Corollary 2. Let \( {s}_{i} \in \mathrm{G} \) and \( {z}_{i} \in {\widehat{\mathrm{L}}}_{nr}^{ * } \) be elements such that\n\n\[ \prod {z}_{i}^{{s}_{i} - 1} = 1 \]\n\nThen \( \prod {s}_{i}^{w\left( {z}_{i}\right) } = 1 \) in \( {\mathrm{G}}^{a} \) . | Let \( \pi \) uniformize \( {\widehat{\mathrm{L}}}_{nr} \) . Then \( {z}_{i} = {u}_{i}{\pi }^{{n}_{i}} \), with \( {n}_{i} = w\left( {z}_{i}\right) \) . We have\n\n\[ {\pi }^{\sum {n}_{i}\left( {{s}_{i} - 1}\right) }\prod {u}_{i}^{{s}_{i} - 1} = 1 \]\n\nLet \( \mathrm{I} \) be the augmentation ideal of \( \mathbf{Z}\le... | Yes |
Lemma 1. In order that an element \( x \) of \( {\widehat{\mathrm{L}}}_{nr} \) (resp. of \( {\widehat{\mathrm{K}}}_{nr} \) ) belong to \( \mathrm{L} \) (resp. to \( \mathrm{K} \) ), it is necessary and sufficient that \( \mathrm{F}x = x \) . | Necessity is obvious. Let us prove sufficiency for \( \mathrm{L} \) (that for \( \mathrm{K} \) is similar). Suppose \( x \in {\widehat{\mathrm{L}}}_{nr} \) is such that \( \mathrm{F}x = x \) . Let \( \pi \) be a uniformizer of \( \mathrm{L} \) . Write \( x = {\pi }^{n}u \), with \( w\left( u\right) = 0 \), so that \( \... | Yes |
Theorem 2 (Dwork). Let \( \mathrm{L}/\mathrm{K} \) be a totally ramified Galois extension, having abelian Galois group \( \mathrm{G} \) . Suppose that the residue field \( \overline{\mathrm{K}} = \mathrm{L} \) is quasi-finite. Let \( x \in {\mathrm{K}}^{ * } \) ; let \( y \in {\widehat{\mathrm{L}}}_{nr}^{ * } \) be suc... | [Given \( x \), there always exists \( y \) such that \( \mathrm{N}y = x \) as above, because \( \mathrm{N}\left( {\widetilde{\mathrm{L}}}_{nr}^{ * }\right) = {\widehat{\mathrm{K}}}_{nr}^{ * } \) (one could even choose \( y \) in \( {\mathrm{L}}_{nr}^{ * } \) if desired). Then \( \mathrm{N}\left( {y}^{\mathrm{F} - 1}\r... | Yes |
Proposition 15. Suppose \( \overline{\mathrm{K}} \) has positive characteristic. Let \( \mathrm{V} \) be the group of units of \( {\widehat{\mathbf{L}}}_{nr} \) . For every integer \( m \geq 1 \), the homomorphism \( x \mapsto {x}^{\mathrm{F} - 1} \) maps \( {\mathbf{V}}^{\left( m\right) } \) onto itself. If \( x \) is... | The group \( {\mathrm{V}}^{\left( m\right) } \) is complete Hausdorff for the filtration by the \( {\mathrm{V}}^{\left( m + k\right) } \) , and the quotients \( {\mathrm{V}}^{\left( m + k\right) }/{\mathrm{V}}^{\left( m + k + 1\right) } \) can be identified with the additive group \( {\overline{\mathrm{K}}}_{nr} \) . A... | Yes |
Proposition 1\n\n(i) \( \left( {\chi + {\chi }^{\prime }, b}\right) = \left( {\chi, b}\right) + \left( {{\chi }^{\prime }, b}\right) \) .\n\n(ii) \( \left( {\chi, b{b}^{\prime }}\right) = \left( {\chi, b}\right) + \left( {\chi ,{b}^{\prime }}\right) \) . | Obvious. | No |
Proposition 2. Let \( b \in {\mathrm{K}}^{ * } \) . The element of \( {\mathrm{H}}^{2}\left( {{\mathrm{\;L}}_{\chi }/\mathrm{K}}\right) \) which corresponds to \( b \) is none other than \( \left( {\chi, b}\right) \) . | This follows from what has been said in Chap. VIII, §4. | No |
Proposition 3. Let \( {s}_{b} = \left( {b, * /\mathrm{K}}\right) \) be the element of \( {\mathrm{G}}^{a} \) defined by the reciprocity map. Then\n\n\[ \n{\left( \chi, b\right) }_{v} = \chi \left( {s}_{b}\right) \n\] | This follows from prop. 2 of Chap. XI, §3. | No |
Proposition 5. \( \left( {a, b}\right) = i\left( {{\varphi }_{a} \cdot {\varphi }_{b}}\right) \) . | To facilitate the computation, we will write all the G-modules additively-in particular, \( {\mathrm{K}}_{s}^{ * } \) . Moreover, if \( \varphi \) denotes a function on \( \mathrm{G} \) with values in \( \mathbf{Z}/n\mathbf{Z} \), we will denote by \( \bar{\varphi } \) a lifting of this function to \( \mathbf{Z} \) . L... | Yes |
Proposition 6. If \( \alpha \in {\mathrm{K}}_{s}^{ * } \) is an \( n \) th root of \( a \), and if \( {s}_{b} = \left( {b,{}^{ * }/\mathrm{K}}\right) \), then\n\n\[{\left( a, b\right) }_{v} = {s}_{b}\left( \alpha \right) /\alpha \] | Indeed\n\n\[{\left( a, b\right) }_{v} = {w}^{n \cdot \operatorname{inv}\left( {a, b}\right) } = {w}^{n \cdot \operatorname{inv}\left( {{\chi }_{a}, b}\right) }\]\n\n\[= {w}^{n{\chi }_{a}\left( {s}_{b}\right) ,}\;\text{by prop. 3,}\]\n\n\[= {w}^{{\varphi }_{a}\left( {s}_{b}\right) } = {s}_{b}\left( \alpha \right) /\alph... | Yes |
Lemma 1. K contains the nth roots of unity if and only if \( \overline{\mathrm{K}} \) does; in that case the canonical homomorphism \( {\mathrm{U}}_{\mathrm{K}} \rightarrow {\overline{\mathrm{K}}}^{ * } \) induces an isomorphism of the group \( {\mu }_{n} \) of \( n \) th roots of unity of \( \mathrm{K} \) onto the gro... | If \( \overline{\mathrm{K}} \) has characteristic zero, then \( \mathrm{K} \) is isomorphic to \( \overline{\mathrm{K}}\left( \left( \mathrm{T}\right) \right) \), and the lemma is obvious. So suppose \( \overline{\mathrm{K}} \) has characteristic \( p \neq 0 \) . If \( \overline{\mathrm{K}} \) contains the group \( {\b... | Yes |
Lemma 2. Let \( k \) be a quasi-finite field containing the group \( {\mu }_{n} \) of \( n \) th roots of unity (n being prime to the characteristic of \( k \) ). Given \( x \in {k}^{ * } \), let \( y \in {k}_{s}^{ * } \) be a solution to the equation \( {y}^{n} = x \), and let \( z = \mathrm{F}y/y \) . Then \( z \) be... | This follows-via Kummer theory-from the fact that \( k \) has one and only one cyclic extension of degree \( n \) . | No |
Proposition 8. Let \( a \) and \( b \) be two elements of \( {\mathrm{K}}^{ * } \), and let \( \alpha \) and \( \beta \) be their valuations. Set\n\n\[ c = {\left( -1\right) }^{\alpha \beta }\frac{{a}^{\beta }}{{b}^{\alpha }} \]\n\nThen \( c \) is a root of unity in \( \mathrm{K} \), and if \( \bar{c} \) denotes its im... | PROOF OF PROP. 8. It is clear that \( c \) is a unit, and that \( {\mathrm{P}}_{n}\left( \overline{\mathrm{c}}\right) \) depends bi-linearly (in the multiplicative sense) on \( a \) and \( b \) ; as this is also the case for the symbol \( {\left( a, b\right) }_{v} \), we are reduced to proving the formula when \( a \) ... | Yes |
Proposition 9. Let \( \mathrm{K} \) be a field complete under a discrete valuation \( v \) ; suppose that \( \mathrm{K} \) has characteristic zero and its residue field \( \overline{\mathrm{K}} \) has characteristic \( p \neq 0 \) . Let \( e = v\left( p\right) \) be the absolute ramification index of \( \mathrm{K} \) (... | Let \( \pi \) be a uniformizer of \( \mathrm{K} \), and let \( y \in {\mathrm{U}}^{\left( m + e\right) } \) . We can write \( y \) in the form\n\n\[ y = 1 + a{\pi }^{m + e},\;\text{ with }v\left( a\right) \geq 0. \]\n\nWe have \( p = b{\pi }^{e} \), with \( v\left( b\right) = 0 \) . We seek a root \( x \in {\mathrm{U}}... | Yes |
Proposition 10. With the hypotheses and notation of prop. 9, assume further that \( \mathrm{K} \) is a finite extension of degree \( n \) of the field \( {\mathbf{Q}}_{p} \) . If \( m > e/\left( {p - 1}\right) \) , then the group \( {\mathbf{U}}^{\left( m\right) } \) is a free \( {\mathbf{Z}}_{p} \) -module of rank \( ... | Let \( q = {p}^{f} \) be the cardinality of \( \overline{\mathrm{K}} \) ; we have \( {ef} = n \) . The group \( {\mathrm{U}}^{\left( m\right) }/{\mathrm{U}}^{\left( m + e\right) } \) is an abelian group of type \( \left( {p,\ldots, p}\right) \) and of order \( {q}^{e} = {p}^{n} \) . Let \( {x}_{1},\ldots ,{x}_{n} \) be... | Yes |
Proposition 12. Let \( \mathrm{K} = k\left( \left( t\right) \right) \), where \( k \) is a perfect field of characteristic \( p \neq 0 \) . If \( a \in \mathrm{K} \) and \( b \in {\mathrm{K}}^{ * } \), let \( c = \operatorname{Res}\left( {{adb}/b}\right) \in k \) . Then \[ \lbrack a, b) = \lbrack c, t)\;\text{ in }{\ma... | As both sides are bilinear, it suffices to consider the case where \( b \) is a uniformizer, i.e., the case \( b = t \) (changing the variable if needed). If \( a = \sum {a}_{n}{t}^{n} \) , we decompose \( a \) into \( {a}_{0},\mathop{\sum }\limits_{{n < 0}}{a}_{n}{t}^{n},\mathop{\sum }\limits_{{n > 0}}{a}_{n}{t}^{n} \... | Yes |
Proposition 13. If \( \alpha \in {\mathrm{K}}_{s} \) is a root of the equation \( \wp \left( \alpha \right) = a \), and if \( {s}_{b} = \left( {b, * /\mathrm{K}}\right) \) , then\n\n\[ \lbrack a, b{)}_{v} = {s}_{b}\left( \alpha \right) - \alpha \] | This follows from prop. 3. | No |
Lemma 4. Let \( k \) be a quasi-finite field of characteristic \( p \neq 0 \) . Given \( x \in k \) , let \( y \in {k}_{s} \) be a solution to the equation \( \wp \left( y\right) = x \), and let \( z = \mathrm{F}y - y \) . Then \( z \in {\mathbf{F}}_{p} = \mathbf{Z}/p\mathbf{Z} \), and \( z \) does not depend on the ch... | That is a consequence of Artin-Schreier theory and of the fact that \( k \) has one and only one cyclic extension of degree \( p \) . EXAMPLE. If \( k \) is a finite field with \( q = {p}^{f} \) elements, the map \( \mathrm{S} \) is given by the formula \[ \mathrm{S}\left( x\right) = {x}^{{p}^{f - 1}} + {x}^{{p}^{f - 2... | No |
Proposition 15. Let \( \mathrm{K} = k\left( \left( t\right) \right) \), where \( k \) is a quasi-finite field of characteristic p. If \( a \in \mathrm{K} \) and \( b \in {\mathrm{K}}^{ * } \), then \[ \lbrack a, b{)}_{v} = \mathrm{S}\left( {\operatorname{Res}\left( {a\frac{db}{b}}\right) }\right) \] | In view of prop. 12, it suffices to consider the case \( a \) constant (i.e., an element of \( k \) ) and \( b = t \) . We must show that \( \lbrack a, t{)}_{v} = \mathrm{S}\left( a\right) \) . Let \( {k}^{\prime } \) be a finite extension of \( k \) containing a root \( \alpha \) of the equation \( \wp \left( \alpha \... | Yes |
Proposition 16. Let \( \mathrm{K} = k\left( \left( t\right) \right) \), where \( k \) is a quasi-finite field of characteristic p. If an element \( b \in {\mathrm{K}}^{ * } \) is a norm in every cyclic extension of \( \mathrm{K} \) of degree \( p \) , then \( b \in {\mathrm{K}}^{*p} \) . | The hypothesis amounts to saying that \( \lbrack a, b{)}_{v} = 0 \) for all \( a \in \mathrm{K} \) . If \( b \) were not a \( p \) th power, the differential form \( {db}/b \) would not be identically zero; for every constant \( c \), one could then choose \( a \) so that \( {adb}/b = {cdt}/t \) ; by prop. 15 we would ... | Yes |
Theorem 1. Every closed subgroup of finite index in \( {\mathrm{K}}^{ * } \) is a norm group (in the sense of Chap. XI, §4). | (In other words, if \( \mathrm{H} \) is such a subgroup, there exists a finite abelian extension \( \mathrm{L}/\mathrm{K} \) such that \( {\mathrm{N}}_{\mathrm{L}/\mathrm{K}}\left( {\mathrm{L}}^{ * }\right) = \mathrm{H} \) ; moreover, if such an extension exists, we know that it is unique.) | Yes |
Corollary 1. In order that a subgroup of finite index of \( {\mathrm{K}}^{ * } \) be a norm group, it is necessary and sufficient that it contain \( {\mathrm{U}}_{\mathrm{K}}^{\left( n\right) } \) for \( n \) sufficiently large. | Indeed, to say that a subgroup \( \mathrm{H} \) of \( {\mathrm{K}}^{ * } \) contains \( {\mathrm{U}}_{\mathrm{K}}^{\left( n\right) } \) is equivalent to saying that \( \mathrm{H} \) is open; if \( \mathrm{H} \) has finite index in \( {\mathrm{K}}^{ * } \), the latter is equivalent to \( \mathrm{H} \) being closed. | Yes |
Theorem 2. Let \( p \) be prime, and let \( \mathrm{E} \) be the field obtained by adjoining all the roots of unity to \( {\mathbf{Q}}_{p} \) . Then \( \mathbf{E} \) is the maximal abelian extension of \( {\mathbf{Q}}_{p} \) . | Put \( \mathrm{K} = {\mathbf{Q}}_{p} \) and \( {\mathrm{K}}^{a} = \) the maximal abelian extension of \( \mathrm{K} \) . Obviously \( \mathrm{E} \subset {\mathrm{K}}^{a} \) . We have seen in Chap. IV, \( §4 \) that \( \mathrm{E} \) is the composite of the linearly disjoint extensions \( {\mathrm{K}}_{nr} \) and \( {\ma... | Yes |
Let \( \mathrm{N} \) be the kernel of \( \mathrm{P} \) in \( {k}_{s} \) . If \( x \in k \), let \( y \in {k}_{s} \) be a solution to the equation \( \mathbf{P}\left( y\right) = x \), and let \( z = \mathbf{F}y - y \) . Then \( z \in \mathbf{N} \), and the image \( \bar{z} \) of \( z \) in \( \mathrm{N}/\left( {\mathrm{... | Changing \( y \) amounts to replacing it with \( y + t \), with \( t \in \mathbf{N} \), and \( z \) is then replaced with \( z + \left( {\mathrm{F}t - t}\right) \) ; this shows that \( \bar{z} \) does not depend on the choice of \( y \) . It is immediate that \( {\delta }_{\mathrm{P}}\left( x\right) = 0 \) if and only ... | Yes |
Corollary 1. If \( \mathrm{N} \) is contained in \( k \), the map \( {\delta }_{\mathrm{P}} \) is an isomorphism of \( k/\mathrm{P}\left( k\right) \) onto \( \mathrm{N} \) . | Indeed, we then have \( \mathrm{F}z = z \) for all \( z \in \mathbf{N} \) . | No |
Corollary 2. The order of \( k/\mathrm{P}\left( k\right) \) is finite; it is a divisor of the separable degree \( {d}_{s}\left( \mathbf{P}\right) \) of \( \mathbf{P} \), and is equal to the latter if and only if \( \mathbf{N} \) is contained in \( k \) . | Indeed, we know that the order of \( \mathrm{N} \) is equal to \( {d}_{\mathrm{s}}\left( \mathrm{P}\right) \) -cf. Chap. V,§5. | No |
Corollary 3. The kernel and cokernel of \( \mathrm{P} : k \rightarrow k \) are finite groups of the same order. | Let \( {\mathrm{N}}_{k} \) be the kernel of \( \mathrm{P} : k \rightarrow k \), i.e., the intersection of \( \mathrm{N} \) with \( k \) . There is an exact sequence\n\n\[ 0 \rightarrow {\mathrm{N}}_{k} \rightarrow \mathrm{N}\xrightarrow[]{\mathrm{F} - 1}\mathrm{\;N} \rightarrow \mathrm{N}/\left( {\mathrm{F} - 1}\right)... | No |
Proposition 2. Let \( \mathrm{P} = {\mathrm{X}}^{n} \), with \( n \geq 1 \), and let \( \mathrm{N} \) be the kernel of \( \mathrm{P} : {k}_{s}^{ * } \rightarrow {k}_{s}^{ * } \) . If \( x \in k \), let \( y \in {k}_{s}^{ * } \) be a solution to the equation \( \mathbf{P}\left( y\right) = x \), and let \( z = {y}^{\math... | The proof is identical to that of prop. 1. | No |
Corollary 3. The kernel and cokernel of \( \mathbf{P} : {k}^{ * } \rightarrow {k}^{ * } \) are finite groups of the same order. | The proofs are the same as for the corresponding corollaries in the additive case. | No |
Proposition 3. The group \( {\mathrm{U}}_{\mathrm{K}}^{n}/{\mathrm{U}}_{\mathrm{K}}^{n + 1}{\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( n\right) } \) is isomorphic to \( {\mathrm{G}}_{\psi \left( n\right) }/{\mathrm{G}}_{\psi \left( n\right) + 1} \) . | \[ \text{ Put }{h}_{n} = \left( {{\mathrm{G}}_{\psi \left( n\right) } : {\mathrm{G}}_{\psi \left( n\right) + 1}}\right) . \] | No |
Corollary 1. \( \left( {{\mathrm{U}}_{\mathrm{K}}^{n} : {\mathrm{U}}_{\mathrm{K}}^{n + 1}{\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( n\right) }}\right) = {h}_{n} \) | Clear. | No |
Corollary 2. The group \( {\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( n\right) } \) is a subgroup of finite index of \( {\mathrm{U}}_{\mathrm{K}}^{n} \) . If \( {v}_{n} \) denotes this index, then \( {v}_{n} = 1 \) for sufficiently large \( n \) ; moreover, \( {v}_{n} \) divides \( {v}_{n + 1}{h}_{n} \), with equality ta... | We know that \( {\mathrm{U}}_{\mathrm{K}}^{n} = {\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( n\right) } \) if \( n \) is sufficiently large (cf. Chap. V. §6, cor. 3 to prop. 9). On the other hand, there is an exact sequence\n\n\[ \n{\mathrm{U}}_{\mathrm{K}}^{n + 1}/{\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( {n + 1}\right) ... | Yes |
Corollary 3. The integer \( {v}_{0} = \left( {{\mathrm{U}}_{\mathrm{K}} : {\mathrm{{NU}}}_{\mathrm{L}}}\right) = \left( {{\mathrm{K}}^{ * } : {\mathrm{{NL}}}^{ * }}\right) \) divides the product of the \( {h}_{n} \) . | Indeed, \( {v}_{0} \) divides \( {v}_{1}{h}_{0} \), which divides \( {v}_{2}{h}_{1}{h}_{0},\ldots \), which divides \( {v}_{n}{h}_{n - 1}\cdots {h}_{0} \) . Taking \( n \) large enough so that \( {v}_{n} = 1 \), we get the result we seek. | Yes |
Theorem 1. Suppose \( \mathrm{G} \) is abelian. Then:\na) \( {\mathrm{G}}_{m} = {\mathrm{G}}_{m + 1} \) if \( \varphi \left( m\right) \) is not an integer.\nb) \( {v}_{n} = {v}_{n + 1}{h}_{n} \) for all \( n \).\nc) The canonical map of \( {\mathrm{U}}_{\mathrm{K}}^{n}/{\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( n\right)... | By prop. 9 of Chap. XIII, \( {v}_{0} = \left\lbrack {\mathrm{L} : \mathrm{K}}\right\rbrack \), or, what comes to the same,\n\n\[ {v}_{0} = \mathop{\prod }\limits_{{m = 0}}^{\infty }\left( {{\mathrm{G}}_{m} : {\mathrm{G}}_{m + 1}}\right) \]\n\nOn the other hand, cor. 3 to prop. 3 shows that \( {v}_{0} \) divides the pro... | Yes |
Corollary 1. The groups \( {\mathrm{U}}_{\mathrm{K}}^{n}/{\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( n\right) } \) form a decreasing filtration of \( {\mathrm{K}}^{ * }/{\mathrm{{NL}}}^{ * } \) . We have \( {\mathrm{U}}_{\mathrm{K}}^{n}/{\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( n\right) } = 0 \) if and only if \( {\mathr... | The first assertion is just a reformulation of c). As for the second, note that \( {v}_{n} = 1 \) is equivalent to \( {h}_{n} = {h}_{n + 1} = \cdots = 1 \), i.e., \( {\mathrm{G}}^{n} = {\mathrm{G}}^{n + 1} = \cdots = \{ 1\} \) . | Yes |
Let \( c \) be the largest integer for which \( {\mathrm{G}}_{c} \neq \{ 1\} \), and let \( f = \) \( \varphi \left( c\right) + 1 \) . Then \( {\mathrm{U}}_{\mathrm{K}}^{f} \subset {\mathrm{{NL}}}^{ * } \), and \( f \) is the least integer enjoying this property. | We know that \( {\mathrm{U}}_{\mathrm{K}}^{f} \) is contained in \( {\mathrm{{NL}}}^{ * } \) (cf. Chap. V,§6, cor. 3 to prop. 9). On the other hand, corollary 1 shows that \( {\mathrm{U}}_{\mathrm{K}}^{f - 1} \) is not contained in \( {\mathrm{{NL}}}^{ * } \), because \( {\mathrm{G}}^{f - 1} = {\mathrm{G}}_{c} \) is no... | Yes |
Corollary 3. The reciprocity map \( \omega : {\mathrm{K}}^{ * }/{\mathrm{{NL}}}^{ * } \rightarrow \mathrm{G} \) transforms the filtration by the \( {\mathrm{U}}_{\mathrm{K}}^{n}/{\mathrm{{NU}}}_{\mathrm{L}}^{\psi \left( n\right) } \) into the filtration by the \( {\mathrm{G}}^{n} \) . | It suffices to show that for every subgroup \( \mathrm{H} \) of \( \mathrm{G} \), the relations\n\n\[ \n\omega \left( {\mathrm{U}}_{\mathrm{K}}^{n}\right) \subset \mathrm{H}\;\mathrm{{and}}\;{\mathrm{G}}^{n} \subset \mathrm{H} \n\]\n\nare equivalent. Passing to the quotient by \( \mathrm{H} \), this comes down to sayin... | Yes |
Theorem 2. Let \( \mathrm{L}/\mathrm{K} \) be an abelian extension, with Galois group \( \mathrm{G} \). Then the image of \( {\mathrm{U}}_{\mathrm{K}}^{n} \) under the reciprocity map \( \omega : {\mathrm{K}}^{ * } \rightarrow \mathrm{G} \) is dense in \( {\mathrm{G}}^{n} \). | In view of the definition of \( {\mathrm{G}}^{n} \), that comes down to saying that \( \omega \left( {\mathrm{U}}_{\mathrm{K}}^{n}\right) = {\mathrm{G}}^{n} \) when \( \mathrm{L}/\mathrm{K} \) is finite. Let \( \mathrm{T} \) then be the inertia group of \( \mathrm{G} \), and \( {\mathrm{K}}^{\prime }/\mathrm{K} \) the ... | Yes |
Proposition 5. Let \( \mathrm{L}/\mathrm{K} \) be a cyclic totally ramified extension of prime degree \( p \) equal to the characteristic of \( \overline{\mathrm{K}} \) ; let \( \mathrm{G} \) be its Galois group, and let \( s \) be a generator of \( \mathrm{G} \) ; let \( t \) be the largest integer for which \( {\math... | Let \( {\pi }^{\prime } \) be an element of \( \mathrm{K} \) such that \( {v}_{\mathrm{K}}\left( {\pi }^{\prime }\right) = t \) . Then \( \operatorname{Tr}\left( \mathrm{M}\right) = b{\pi }^{\prime } \) and \( \mathrm{N}\left( \mathrm{M}\right) = a{\pi }^{\prime } \), with \( a, b \in {\mathrm{U}}_{\mathrm{K}} \) -cf. ... | Yes |
Proposition 6. Suppose that \( \mathrm{K} \) (resp. \( \overline{\mathrm{K}} \) ) has characteristic zero (resp. p), and that \( \mathrm{K} \) contains the group \( {\mu }_{p} \) of all \( p \) th roots of unity. Let \( w \) be a generator of \( {\mu }_{p} \) . Let \( e = {v}_{\mathrm{K}}\left( p\right) \) be the absol... | We know from Chap. IV, prop. 17 that \( {v}_{\mathrm{K}}\left( {w - 1}\right) = e/\left( {p - 1}\right) \), which shows that \( t \) is indeed an integer. On the other hand, the bilinearity of the symbol\n\n\( {\left( a, b\right) }_{v} \) allows us to restrict to the case where \( a \) is a uniformizer of \( \mathrm{K}... | Yes |
Theorem 1.1 (Polarization identity). Let \( H \) be a complex vector space, \( s \) a sesquilinear form on \( H \), and \( q \) the quadratic form generated by \( s \) . Then for all \( f, g \in H \) we have\n\n\[ s\left( {f, g}\right) = \frac{1}{4}\{ q\left( {f + g}\right) - q\left( {f - g}\right) + {iq}\left( {f - {i... | The proof of this identity may be given by calculating the right side of (1.4) according to the rules (1.1) and (1.2). | No |
Theorem 1.2 (Parallelogram law). Let \( s \) be a sesquilinear form on a vector space \( H \), and let \( q \) be the corresponding quadratic form on \( H \). Then for all \( f, g \in H \) we have\n\n\[ q\left( {f + g}\right) + q\left( {f - g}\right) = 2\left\lbrack {q\left( f\right) + q\left( g\right) }\right\rbrack .... | Proof. For every \( f, g \in H \) we have\n\n\[ q\left( {f + g}\right) + q\left( {f - g}\right) = s\left( {f, f}\right) + s\left( {f, g}\right) + s\left( {g, f}\right) + s\left( {g, g}\right) \]\n\n\[ + s\left( {f, f}\right) - s\left( {f, g}\right) - s\left( {g, f}\right) + s\left( {g, g}\right) \]\n\n\[ = {2q}\left( f... | Yes |
Theorem 1.3. Let \( H \) be a vector space over \( \mathbb{K}, s \) a sesquilinear form on \( H \), and \( q \) the quadratic form generated by \( s \). (a) If \( \mathbb{K} = \mathbb{C} \), then the following statements are equivalent: (i) \( s \) is symmetric, (ii) \( q \) is real, (iii) for all \( f, g \in H \) we h... | Proof. (a) (ii) follows from (i): \( q{\left( f\right) }^{ * } = s{\left( f, f\right) }^{ * } = s\left( {f, f}\right) = q\left( f\right) \), i.e., \( q\left( f\right) \) is real. (iii) follows from (ii): Because \( q\left( h\right) \in \mathbb{R} \) for all \( h \in H \), it follows from (1.4) that \[ \operatorname{Re}... | Yes |
Theorem 1.4. If \( s \) is a non-negative sesquilinear form on \( H \), and \( q \) denotes the quadratic form generated by \( s \), then for every \( f, g \in H \) we have the Schwarz inequality\n\n\[ \left| {s\left( {f, g}\right) }\right| \leq {\left\lbrack q\left( f\right) q\left( g\right) \right\rbrack }^{1/2}. \] | Proof. Let \( f, g \in H \) . For all \( t \in \mathbb{R} \) we have\n\n\[ 0 \leq q\left( {f + {tg}}\right) = q\left( f\right) + {2t}\operatorname{Re}\left( {f, g}\right) + {t}^{2}q\left( g\right) . \]\n\nThis second degree polynomial in \( t \) has either no root or a double root. Since this holds for a polynomial \( ... | Yes |
Theorem 1.5. If \( s \) is a semi-scalar product on \( H \), then \( p\left( f\right) = {\left\lbrack s\left( f, f\right) \right\rbrack }^{1/2} \) defines a seminorm on \( H \) . | Proof. Property [(1.14) (i)] follows immediately from [(1.13) (iv)]; [(1.14) (iv)] follows from \( \left\lbrack {\left( {1.13}\right) \left( \mathrm{v}\right) }\right\rbrack \) . It is sufficient to prove the remaining properties for the first case. Because of \( \left\lbrack {\left( {1.13}\right) \left( \mathrm{{ii}}\... | Yes |
We show that not every Cauchy sequence is convergent. For this let the sequence \( \left( {f}_{n}\right) \) in \( C\left\lbrack {0,1}\right\rbrack \) be defined in the following way: \( {f}_{1}\left( x\right) = 1 \) for all \( x \in \left\lbrack {0,1}\right\rbrack \), and\n\n\[ \n{f}_{n}\left( x\right) = \left\{ \begin... | To prove that the sequence \( \left( {f}_{n}\right) \) is not convergent let us assume that there exists an \( f \in C\left\lbrack {0,1}\right\rbrack \) such that \( {f}_{n} \rightarrow f \), i.e., \( \begin{Vmatrix}{{f}_{n} - f}\end{Vmatrix} \rightarrow 0 \) . Then for \( 2 < n < m \) we have\n\n\[ \n{\int }_{0}^{1/2}... | Yes |
Example 3. \( {\mathbb{C}}^{m} \) and \( {\mathbb{R}}^{m} \) are Banach spaces with the norms \( \parallel \cdot {\parallel }_{1},\parallel \cdot {\parallel }_{\infty } \) and \( \parallel \) . \( \parallel \) from Section 1.2, Examples 1 and 3. This follows easily from the fact that a sequence \( \left( {f}_{n}\right)... | This follows easily from the fact that a sequence \( \left( {f}_{n}\right) \) is a Cauchy sequence (convergent sequence) in \( {\mathbb{C}}^{m} \) or \( {\mathbb{R}}^{m} \) if and only if it converges componentwise. (The proof can also be obtained as a special case of Example 4.) | No |
Theorem 2.2. \( \bar{A} \) is closed. \( A \) is closed if and only if \( A = \bar{A} \) . The set \( \bar{A} \) is the smallest closed subset of \( H \) that contains \( A \) . | Proof. First we show that \( \bar{A} \) is closed, i.e., \( C\bar{A} \) is open. Let \( f \in C\bar{A} \) . Since \( \bar{A} = \bar{A} \), then we have \( f \in \mathbf{C}\bar{A} \), i.e., \( f \) is not a contact point of \( \bar{A} \) . Therefore there is an \( \epsilon > 0 \) such that \( K\left( {f,\epsilon }\right... | Yes |
Theorem 2.3. The closure of a subspace of \( H \) is a subspace. | Proof. Let \( T \) be a subspace of \( H \), let \( f, g \in \bar{T} \) and let \( a, b \in \mathbb{K} \) . Then there are sequences \( \left( {f}_{n}\right) \) and \( \left( {g}_{n}\right) \) in \( T \) such that \( {f}_{n} \rightarrow f,{g}_{n} \rightarrow g \) . It follows that\n\n\[ \n{af} + {bg} = a\lim {f}_{n} + ... | Yes |
Theorem 2.4. A subspace \( T \) of a Banach space \( \left( {H,\langle \rangle }\right) \) (respectively a Hilbert space \( \left( {H,\langle .,.\rangle }\right) ) \) is closed if and only if \( \left( {T,\parallel .\parallel }\right) \) is a Banach space (respectively \( \left( {T,\langle .,.\rangle }\right) \) is a H... | Proof. If \( T \) is closed, and \( \left( {f}_{n}\right) \) is a Cauchy sequence in \( T \), then there exists an \( f \in H \) such that \( {f}_{n} \rightarrow f \) ; therefore \( f \in \bar{T} = T \), i.e., \( T \) is complete. If \( T \) is complete and \( f \in \bar{T} \), then there exists a sequence \( \left( {f... | Yes |
Theorem 2.5. Let \( \left( {H,\parallel \cdot \parallel }\right) \) be a normed space.\n\n(a) If \( A \) is a separable subset of \( H \), then \( \bar{A} \) is separable, also. | (a) Let \( B \) be at most countable and dense in \( A \) . Since \( A \) is dense in \( \bar{A} \), the set \( B \) is also dense in \( \bar{A} \) . | Yes |
Example 13. Let \( \rho \) be a measure on \( {\mathbb{R}}^{m} \) (cf. Appendix A) and let \( M \) be a \( \rho \) -measurable subset of \( {\mathbb{R}}^{m} \) . The Hilbert space \( {L}_{2}\left( {M,\rho }\right) \) (cf. Section 2.1, Example 6) is separable. | This can be proved for \( M = {\mathbb{R}}^{m} \) as in Example 11 (in the course of the proof of \( S \subset {\bar{S}}_{0} \) one has to notice that the boundaries of intervals in general have measures different from zero). For a general \( M \) we obtain the assertion by considering \( {L}_{2}\left( {M,\rho }\right)... | Yes |
Theorem 3.1. Let \( H \) be a Hilbert space and let \( A \) be a non-empty closed convex subset of \( H \) . Then for each \( f \in H \) there exists a unique \( g \in A \) such that\n\n\[ \parallel f - g\parallel = d\left( {f, A}\right) = \inf \{ \parallel f - h\parallel : h \in A\} . \] | Proof. There always exists a sequence \( \left( {g}_{n}\right) \) of elements of \( A \) such that \( \begin{Vmatrix}{{g}_{n} - f}\end{Vmatrix} \rightarrow d = d\left( {f, A}\right) \) . If we replace \( f \) by \( {g}_{n} - f \) and \( g \) by \( {g}_{m} - f \) in the parallelogram identity (1.16), then on account of ... | Yes |
Theorem 3.2 (Projection theorem). Let \( H \) be a Hilbert space, and let \( T \) be a closed subspace of \( H \) . Then we have \( {T}^{ \bot \bot } = T \) . Each \( f \in H \) can be uniquely decomposed in the form \( f = g + h \) with \( g \in T \) and \( h \in {T}^{ \bot } \) . This \( g \) is called the (orthogona... | Proof. As \( T \) is convex and closed, by Theorem 3.1 there exists a \( g \in T \) such that \( \parallel f - g\parallel = d\left( {f, T}\right) \) . Let us set \( h = f - g \) .\n\n\( h \in {T}^{ \bot } \) : We have to prove that for all \( w \in T \) we have \( \langle w, h\rangle = 0 \) . For \( w = 0 \) this is cl... | Yes |
In \( {L}_{2}\left( {-a, a}\right) \), two subspaces are defined by \( {T}_{ \pm } = \left\{ {f \in {L}_{2}\left( {-a, a}\right) : f\left( x\right) = \pm f\left( {-x}\right) \text{ almost everywhere in }\left( {-a, a}\right) }\right\} \). Show that \( {L}_{2}\left( {-a, a}\right) = {T}_{ + } \oplus {T}_{ - } \). | For \( g \in {T}_{ + } \) and \( h \in {T}_{ - } \) we have\n\n\[ \langle g, h\rangle = {\int }_{-a}^{a}g{\left( x\right) }^{ * }h\left( x\right) \mathrm{d}x = {\int }_{-a}^{0}g{\left( x\right) }^{ * }h\left( x\right) \mathrm{d}x - {\int }_{-a}^{0}g{\left( x\right) }^{ * }h\left( x\right) \mathrm{d}x = 0. \]\n\nLet us ... | Yes |
Theorem 3.4. Let \( H \) be a Hilbert space. If \( T \) is a closed subspace and \( S \) is a finite dimensional subspace, then \( T + S \) is closed. | Proof. The problem can be reduced by induction, to the case where \( S \) is one dimensional; \( S = L\left( f\right) \) . If we write \( f = {f}_{1} + {f}_{2} \) with \( {f}_{1} \in T \) and \( {f}_{2} \in {T}^{ \bot } \) , then we have \( T + S = T \oplus L\left( {f}_{2}\right) \) . Therefore \( T + S \) is closed by... | No |
In \( {L}_{2}\left( {0,1}\right) \) the set \( M = \left\{ {{e}_{n} : n \in \mathbb{Z}}\right\} \) with \( {e}_{n}\left( x\right) = \exp \left( {2i\pi nx}\right) \) is an ONS, as one can verify by a simple calculation. We show that \( M \) is an ONB, i.e., that \( M \) is total. | For this let \( \widehat{C}\left\lbrack {0,1}\right\rbrack = \{ f \in C\left\lbrack {0,1}\right\rbrack : f\left( 0\right) = f\left( 1\right) \} \) . For each \( f \in \widehat{C}\left\lbrack {0,1}\right\rbrack \) by Fejér’s theorem there exists a sequence \( \left( {f}_{n}\right) \) of trigonometric polynomials \( \lef... | Yes |
Let \( {F}_{0} = L\left\{ {{e}_{\lambda } : \lambda \in \mathbb{R}}\right\} \) with \( {e}_{\lambda } : \mathbb{R} \rightarrow \mathbb{C},{e}_{\lambda }\left( x\right) = \exp \left( {i\lambda x}\right) \). On \( {F}_{0} \) by \[ \langle f, g\rangle = \mathop{\lim }\limits_{{T \rightarrow \infty }}\frac{1}{2T}{\int }_{-... | For these we have \[ \langle f, g\rangle = \mathop{\lim }\limits_{{T \rightarrow \infty }}\frac{1}{2T}{\int }_{-T}^{T}{e}^{i\left( {\mu - \lambda }\right) x}\mathrm{\;d}x = \mathop{\lim }\limits_{{T \rightarrow \infty }}\frac{1}{2T}\frac{-i}{\mu - \lambda }\left\lbrack {{e}^{i\left( {\mu - \lambda }\right) T} - {e}^{-i... | Yes |
Theorem 3.5. Let \( H \) be a pre-Hilbert space. For each finite or countably infinite set \( F = \left\{ {f}_{n}\right\} \) from \( H \) there exists a finite or countably infinite orthonormal system \( M = \left\{ {e}_{n}\right\} \) such that \( L\left( F\right) = L\left( M\right) \) . If \( F \) is linearly independ... | Proof. It is obviously enough to prove only the last part of the assertion. Every normed element from \( L\left( {f}_{1}\right) \) has the form \( {b}_{1}{\begin{Vmatrix}{f}_{1}\end{Vmatrix}}^{-1}{f}_{1} \) with \( \left| {b}_{1}\right| = 1 \) . The additional condition \( {a}_{1} = {b}_{1}{\begin{Vmatrix}{f}_{1}\end{V... | Yes |
If \( \left\{ {{e}_{1},\ldots ,{e}_{n}}\right\} \) is a (finite) ONS in \( H \), then for each \( f \in H \) there exists a \( g \in L\left( {{e}_{1},\ldots ,{e}_{n}}\right) \) such that \( \parallel g - f\parallel = d\left( {f, L\left( {{e}_{1},\ldots ,{e}_{n}}\right) }\right) \) ; we have\n\n\[ g = \mathop{\sum }\lim... | (a) For all \( {c}_{1},\ldots ,{c}_{n} \in \mathbb{K} \) we have\n\n\[ {\begin{Vmatrix}f - \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}{e}_{j}\end{Vmatrix}}^{2} = \parallel f{\parallel }^{2} - \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}\left\langle {f,{e}_{j}}\right\rangle - \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}^{ * ... | Yes |
Theorem 3.7 (Expansion theorem). Let \( H \) be a pre-Hilbert space and let \( M = \\left\\{ {{e}_{\\alpha } : \\alpha \\in A}\\right\\} \) be an ONS in \( H \). (a) If \( \\left( {\\alpha }_{n}\\right) \) is a sequence of pairwise different elements from \( A,\\left( {c}_{n}\\right) a \) sequence from \( \\mathbb{K} \... | Proof. (a) The sequence \( {\\left( \\mathop{\\sum }\\limits_{{n = 1}}^{m}{c}_{n}{e}_{{\\alpha }_{n}}\\right) }_{m \\in \\mathbb{N}} \) is a Cauchy sequence if and only if we have \[ \\mathop{\\sum }\\limits_{{n = m + 1}}^{k}{\\left| {c}_{n}\\right| }^{2} = {\\begin{Vmatrix}\\mathop{\\sum }\\limits_{{n = m + 1}}^{k}{c}... | Yes |
Theorem 3.8. Let \( {M}_{1} \) and \( {M}_{2} \) be measurable subsets of \( {\mathbb{R}}^{p} \) and \( {\mathbb{R}}^{q} \), respectively and let \( \left\{ {{e}_{n} : n \in \mathbb{N}}\right\} \) and \( \left\{ {{f}_{m} : m \in \mathbb{N}}\right\} \) be orthonormal bases of \( {L}_{2}\left( {M}_{1}\right) \) and \( {L... | Proof. It is obvious that the functions \( {g}_{nm} \) are in \( {L}_{2}\left( {{M}_{1} \times {M}_{2}}\right) \) and they form an orthonormal system. It remains to prove that \( \left\{ {{g}_{nm} : \left( {n, m}\right) \in \mathbb{N} \times }\right. \) \( \mathbb{N}\} \) is total. Let \( h \) be an element of \( {L}_{... | Yes |
Theorem 3.9. Let \( H \) be a separable pre-Hilbert space.\n\n(a) \( H \) possesses an \( {ONB} \) . | (a) follows from (b) if we choose \( {M}_{1} = \varnothing \) . | No |
(b) If \( {M}_{0} \) is an ONS, then there exists an ONB \( M \) in \( H \) such that \( M \supset {M}_{0} \) . | (b) Let \( \mathfrak{M} \) be the set of all those ONS which contain \( {M}_{0} \) . \( \mathfrak{M} \) is partially ordered by the inclusion \ | No |
Theorem 3.11. If \( H \) is a Hilbert space and \( S \) and \( T \) are closed subspaces of \( H \) such that \( S \cap {T}^{ \bot } = \{ 0\} \), then we have \( \dim S \leq \dim T(\dim = \) Hilbert space dimension). | Proof. Let us distinguish between two different cases.\n\n(a) \( \dim T = k < \infty \) : Assume that \( \dim S > \dim T \) holds. If \( \left\{ {{e}_{1},\ldots ,{e}_{k}}\right\} \) is an ONB of \( T \) and \( \left\{ {{f}_{1},\ldots ,{f}_{k + 1}}\right\} \) is an ONS in \( S \), then the system of homogeneous equation... | Yes |
Theorem 3.12. Let \( {H}_{1} \) and \( {H}_{2} \) be Hilbert spaces.\n\n(a) If \( {M}_{1} \) and \( {M}_{2} \) are total subsets of \( {H}_{1} \) and \( {H}_{2} \), respectively, then the set \( \left\{ {f \otimes g : f \in {M}_{1}, g \in {M}_{2}}\right\} \) is total in \( {H}_{1}\widehat{ \otimes }{H}_{2} \) . | Proof.\n\n(a) Let \( \mathop{\sum }\limits_{{j = 1}}^{n}{f}_{j} \otimes {g}_{j} \in {H}_{1} \otimes {H}_{2},\epsilon > 0 \) . For each \( j \in \{ 1,2,\ldots, n\} \) there exist elements \( {f}_{j}^{\prime } \in L\left( {M}_{1}\right) \) and \( {g}_{j}^{\prime } \in L\left( {M}_{2}\right) \) such that \( \begin{Vmatrix... | Yes |
Theorem 4.1. Let \( H \) be a vector space over \( \mathbb{K} \), and let \( {T}_{1},\ldots ,{T}_{n}, T \) be linear functionals such that \( D\left( {T}_{1}\right) = \ldots = D\left( {T}_{n}\right) = D\left( T\right) = H \) . If we have \( N\left( T\right) \supset { \cap }_{i = 1}^{n}N\left( {T}_{i}\right) \), then th... | Proof. We prove this by induction on \( n \) . Let \( n \) be equal to 1 . If \( {T}_{1} = 0 \) , then \( N\left( T\right) \supset N\left( {T}_{1}\right) = H \), therefore \( T = 0 \) . If \( {T}_{1} \neq 0 \), then there exists an \( {f}_{0} \in H \) such that \( {T}_{1}{f}_{0} = 1 \) . For every \( f \in H \) we then... | Yes |
Theorem 4.2. Let \( T \) be an operator from \( {H}_{1} \) into \( {H}_{2} \) . Then the following assertions are equivalent:\n\n(a) \( T \) is continuous,\n\n(b) \( T \) is continuous at 0,\n\n(c) \( T \) is bounded. | Proof. (b) obviously follows from (a).\n\n(c) follows from (b): Let us assume that \( T \) is not bounded. Then for every \( n \in \mathbb{N} \) there exists an \( {f}_{n} \in D\left( T\right) \) such that \( \begin{Vmatrix}{T{f}_{n}}\end{Vmatrix} > n\begin{Vmatrix}{f}_{n}\end{Vmatrix} \) . From this it follows that, i... | Yes |
The functional \( T \) of Example 2 is bounded if \( \psi \in {L}_{2}\left( \mathbb{R}\right) \) | \[ \left| {Tf}\right| \leq \parallel \psi \parallel \parallel f\parallel \] From Theorem 4.8 (theorem of Riesz) it will follow that \( T \) is continuous if and only if \( \psi \in {L}_{2}\left( \mathbb{R}\right) \) | Yes |
Theorem 4.3. Let \( {H}_{1} \) and \( {H}_{2} \) be normed spaces. Let \( T \) be an operator from \( {H}_{1} \) into \( {\mathrm{H}}_{2} \) . (a) We have\n\n\[ \sup \{ \parallel {Tf}\parallel : f \in D\left( T\right) ,\parallel f\parallel \leq 1\} = \sup \{ \parallel {Tf}\parallel : f \in D\left( T\right) ,\parallel f... | Proof.\n\n(a) If we define \( \parallel T\parallel \) to be equal to \( \infty \) if \( T \) is unbounded, then we only have to prove that the three values, which we denote by \( {c}_{1},{c}_{2},{c}_{3} \) , respectively, are all equal to \( \parallel T\parallel \) . As \( \parallel T\parallel \) is a bound for \( T \)... | Yes |
Theorem 4.4. Let \( T \) be a densely defined operator on a complex pre-Hilbert space, or a symmetric operator defined on an arbitrary pre-Hilbert space. \( T \) is bounded if and only if \[ C = \sup \{ \left| {\langle f,{Tf}\rangle }\right| : f \in D\left( T\right) ,\parallel f\parallel < 1\} < \infty . \] If \( T \) ... | Proof. Let us set \( \parallel T\parallel = \infty \) for an unbounded \( T \) . Then we only have to prove inequality (a) and equality (b). (a) For all \( f, g \in D\left( T\right) \) such that \( \parallel f\parallel \leq 1 \) and \( \parallel g\parallel \leq 1 \), from (1.4) with \( s\left( {f, g}\right) = \langle f... | Yes |
Theorem 4.5. Let \( T \) be a bounded operator from a normed space \( {H}_{1} \) into a Banach space \( {\mathrm{H}}_{2} \). Then there exists a unique bounded extension \( S \) of \( T \) such that \( D\left( S\right) = \overline{D\left( T\right) } \). We have \( \parallel S\parallel = \parallel T\parallel \) . | Proof. Uniqueness: Assume \( S \) is a continuous extension of \( T \) such that \( D\left( S\right) = \overline{D\left( T\right) } \). If \( f \in D\left( S\right) \), then there exists a sequence \( \left( {f}_{n}\right) \) from \( D\left( T\right) \) such that \( {f}_{n} \rightarrow f \). As \( S \) is continuous, w... | Yes |
Theorem 4.6. Let \( \parallel \) . \( \parallel \) be defined as in (4.4). Then \( \left( {B\left( {{H}_{1},{H}_{2}}\right) ,\parallel \cdot \parallel }\right) \) is a normed space. If \( {\mathrm{H}}_{2} \) is a Banach space, then \( \left( {B\left( {{\mathrm{H}}_{1},{\mathrm{H}}_{2}}\right) ,\parallel \text{. ||}}\ri... | Proof. It is clear that \( \parallel \cdot \parallel \) is a semi-norm. If \( \parallel T\parallel = 0 \), then \( \parallel {Tf}\parallel = 0 \) for all \( f \in {H}_{1} \) such that \( \parallel f\parallel < 1 \) ; therefore \( {Tf} = 0 \) for all \( f \in {H}_{1} \), and thus \( T = 0 \) , the zero element in \( B\l... | Yes |
Theorem 4.8 (F. Riesz). Let \( H \) be a Hilbert space. Every \( g \in H \) induces a continuous linear functional on \( H \) by \( {T}_{g}\left( f\right) = \langle g, f\rangle \) . We have \( \begin{Vmatrix}{T}_{g}\end{Vmatrix} = \parallel g\parallel \) . This mapping of \( H \) onto \( B\left( {H,\mathbb{K}}\right) \... | Proof. The first part has already been shown. The antilinearity follows from\n\n\[ \n{T}_{{ag} + {bh}}\left( f\right) = \langle {ag} + {bh}, f\rangle = {a}^{ * }\langle g, f\rangle + {b}^{ * }\langle h, f\rangle = {a}^{ * }{T}_{g}\left( f\right) + {b}^{ * }{T}_{h}\left( f\right) .\n\]\n\nAs \( \begin{Vmatrix}{T}_{g}\en... | Yes |
Theorem 4.9. Let \( {H}_{1} \) and \( {H}_{2} \) be isomorphic normed spaces. \( {H}_{1} \) is a Banach space (Hilbert space) if and only if \( {\mathrm{H}}_{2} \) is a Banach space (Hilbert space). | Proof. Let \( {H}_{1} \) be a Banach space and let \( U \) be an isomorphism of \( {H}_{1} \) onto \( {H}_{2} \) . If \( \left( {f}_{n}\right) \) is a Cauchy sequence in \( {H}_{2} \), then \( \left( {{U}^{-1}{f}_{n}}\right) \) is a Cauchy sequence in \( {H}_{1} \) ; hence there exists a \( g \in {H}_{1} \) such that \... | Yes |
Theorem 4.10. Let \( H \) be a Hilbert space, and let \( A \) be a set, the cardinality of which equals the (Hilbert space) dimension of \( H \) . Then \( H \) is isomorphic to \( {l}_{2}\left( A\right) \) . In particular, all infinite dimensional separable Hilbert spaces are isomorphic to \( {l}_{2} \) . Hilbert space... | Proof. Let \( \left\{ {{e}_{\alpha } : \alpha \in A}\right\} \) be an ONB of \( H \) . For every \( f = \sum {f}_{\alpha }{e}_{\alpha } \in H \) let \( {Uf} \) be the function \( A \rightarrow \mathbb{C} \) with \( \left( {Uf}\right) \left( \alpha \right) = {f}_{\alpha } \) . It is easy to see that \( U \) is an isomor... | Yes |
Theorem 4.12. Let \( H \) be a pre-Hilbert space, and let \( {T}_{1},\ldots ,{T}_{n} \) be linear functionals such that \( D\left( {T}_{j}\right) = H \) and \( L\left( {{T}_{1},\ldots ,{T}_{n}}\right) \cap B\left( {H, K}\right) = \{ 0\} \) . Then\n\n\[ M = \mathop{\bigcap }\limits_{{j = 1}}^{n}N\left( {T}_{j}\right) = ... | Proof. Without loss of generality we may assume that \( H \) is a dense subspace of a Hilbert space \( \left( {{H}_{0},\langle .,.\rangle }\right) \) . (If \( \widehat{H} \) is a completion of \( H \) and \( U \) is an isomorphism of \( H \) onto a dense subspace of \( A \), then we replace \( H \) by \( U\left( H\righ... | Yes |
Theorem 4.13. Let \( T \) be a densely defined operator from \( {H}_{1} \) into \( {H}_{2} \). (a) If \( {T}^{ * } \) is also densely defined, then \( {T}^{* * } \) is an extension of \( T \). (b) We have \( N\left( {T}^{ * }\right) = R{\left( T\right) }^{ \bot } \). | Proof. (a) As \( T \) and \( {T}^{ * } \) are formal adjoints of each other, \( T \) is a restriction of the adjoint operator \( {T}^{* * } \) of \( {T}^{ * } \). (b) We have \( g \in N\left( {T}^{ * }\right) \) if and only if \( g \in D\left( {T}^{ * }\right) \) and \( {T}^{ * }g = 0 \) hold. Since \( D\left( T\right)... | Yes |
Theorem 4.14. Let \( T \) be a densely defined operator from \( {H}_{1} \) into \( {H}_{2} \) .\n\n(a) \( T \) is bounded if and only if \( {T}^{ * } \in B\left( {{H}_{2},{H}_{1}}\right) \) . | Proof.\n\n(a) and (b): Let \( T \) be bounded. Then for all \( g \in {H}_{2} \) and \( f \in D\left( T\right) \) we have\n\n\[ \left| {{L}_{g}f}\right| = \left| {\langle g,{Tf}\rangle }\right| \leq \parallel g\parallel \parallel T\parallel \parallel f\parallel \]\n\ni.e., \( {L}_{g} \) is continuous for all \( g \in {H... | Yes |
Let \( T \) be a continuous linear functional on a Hilbert space \( H \) , i.e., a continuous operator from \( H \) into \( \mathbb{K} \) . We want to compute \( {T}^{ * } \) . There exists a uniquely determined \( g \in H \) such that\n\n\[ \n{Tf} = \langle g, f\rangle \text{ for all }f \in H.\n\] | Hence for all \( z \in \mathbb{K} \) and all \( f \in H \) we have\n\n\[ \n{z}^{ * }{Tf} = \langle {zg}, f\rangle\n\]\n\ni.e., \( {T}^{ * }z = {zg} \) for all \( z \in \mathbb{K} \) . | Yes |
Theorem 4.15. A subset \( G \) of \( {H}_{1} \times {H}_{2} \) is the graph of an operator from \( {H}_{1} \) into \( {\mathrm{H}}_{2} \) if and only if \( G \) is a subspace possessing the following property: \( \left( {0, g}\right) \in G \) implies \( g = 0 \) . Each subspace of a graph is a graph. | Proof. If \( T \) is an operator from \( {H}_{1} \) into \( {H}_{2} \), then \( G\left( T\right) \) is obviously a subspace, as for \( \left( {{f}_{i},{g}_{i}}\right) \in G\left( T\right) ,{a}_{i} \in \mathbb{K}\left( {i = 1,2}\right) \) we have\n\n\[ \n{a}_{1}\left( {{f}_{1},{g}_{1}}\right) + {a}_{2}\left( {{f}_{2},{g... | Yes |
Theorem 4.16. Let \( T \) be a densely defined operator from \( {H}_{1} \) into \( {H}_{2} \). Then we have\n\n\[ G\left( {T}^{ * }\right) = U\left( {G{\left( T\right) }^{ \bot }}\right) = {\left( UG\left( T\right) \right) }^{ \bot } \]\n\n(here the symbol \( \bot \) has to be understood in the sense of \( {H}_{1} \opl... | Proof. By the definition of \( {T}^{ * } \) we have\n\n\[ G\left( {T}^{ * }\right) = \left\{ {\left( {g, h}\right) \in {H}_{2} \times {H}_{1} : \langle g,{Tf}{\rangle }_{2} = \langle h, f{\rangle }_{1}\text{ for all }f \in D\left( T\right) }\right\} \]\n\n\[ = \left\{ {\left( {g, h}\right) \in {H}_{2} \times {H}_{1} : ... | Yes |
Theorem 4.17. Let \( T \) be a densely defined injective operator from \( {H}_{1} \) into \( {H}_{2} \) . (a) We have \( G\left( {T}^{-1}\right) = {VG}\left( T\right) \) . (b) If \( R\left( T\right) \) is dense, then \( {T}^{ * } \) is also injective, and we have \( {T}^{* - 1} = {T}^{-1 * } \) . | Proof. Part (a) is obvious. (b) By Theorem 4.13(b) we have \( N\left( {T}^{ * }\right) = R{\left( T\right) }^{ \bot } = \{ 0\} \), i.e., \( {T}^{ * } \) is injective. As \( G\left( {T}^{-1}\right) = {VG}\left( T\right) \), it follows (cf. Exercise 4.14) that \[ G\left( {T}^{-1 * }\right) = {U}^{-1}\left( {G{\left( {T}^... | No |
Theorem 4.18. An operator \( T \) on a complex Hilbert space \( H \) is Hermitian if and only if the quadratic form \( q\left( f\right) = \langle f,{Tf}\rangle \) defined on \( D\left( T\right) \) is real. | Proof. By definition, \( T \) is Hermitian if and only if the sesquilinear form \( s\left( {f, g}\right) = \langle f,{Tg}\rangle \) is Hermitian on \( D\left( T\right) \) . The assertion follows from this by Theorem 1.3(a). | Yes |
Theorem 4.19. Let \( {T}_{1} \) and \( {T}_{2} \) be densely defined operators from \( {H}_{1} \) into \( {H}_{2} \) and from \( {\mathrm{H}}_{2} \) into \( {\mathrm{H}}_{3} \), respectively.\n\n(a) If \( {T}_{2}{T}_{1} \) is densely defined, then we have \( {T}_{1}^{ * }{T}_{2}^{ * } \subset {\left( {T}_{2}{T}_{1}\rig... | Proof.\n\n(a) We have to show that the operators \( {T}_{1}^{ * }{T}_{2}^{ * } \) and \( {T}_{2}{T}_{1} \) are formal adjoints of each other. Let \( f \in D\left( {{T}_{1}^{ * }{T}_{2}^{ * }}\right) \) and \( g \in D\left( {{T}_{2}{T}_{1}}\right) \) . Then \( f \in \) \( D\left( {T}_{2}^{ * }\right) ,{T}_{2}^{ * }f \in... | Yes |
Theorem 4.20. Let \( S \) and \( T \) be operators from \( {H}_{1} \) into \( {H}_{2} \) . (a) If \( T \) is densely defined, then we have \( {\left( aT\right) }^{ * } = {a}^{ * }{T}^{ * } \) for all \( a \in \mathbb{K} \) such that \( a \neq 0 \) . | Proof. (a) is evident (it follows from Theorem 4.19). | No |
Theorem 4.21. Let \( T \) be self-adjoint and injective. Then \( {T}^{-1} \) is self-adjoint, too. | Proof. \( R\left( T\right) \) is dense, since we have \( \{ 0\} = N\left( T\right) = N\left( {T}^{ * }\right) = R{\left( T\right) }^{ \bot } \) . Thus the assertion follows from Theorem 4.17(b). | No |
Theorem 4.23. Let \( {H}_{1} \) and \( {H}_{2} \) be normed spaces.\n\n(a) If \( \left( {T}_{n}\right) \) is a strongly convergent sequence in \( B\left( {{H}_{1},{H}_{2}}\right) \) and \( T = s - \) \( \lim {T}_{n} \), then \( \parallel T\parallel \leq \liminf \begin{Vmatrix}{T}_{n}\end{Vmatrix} \) . | Proof.\n\n(a) Let \( C = \liminf \begin{Vmatrix}{T}_{n}\end{Vmatrix} \) . Then there exists a subsequence \( \left( {T}_{{n}_{k}}\right) \) of \( \left( {T}_{n}\right) \) such that \( \begin{Vmatrix}{T}_{{n}_{k}}\end{Vmatrix} \rightarrow C \) as \( k \rightarrow \infty \) . Hence for all \( f \in {H}_{1} \) we have\n\n... | Yes |
Theorem 4.24. Let \( H \) be a pre-Hilbert space.\n\n(a) If \( \\left( {f}_{n}\\right) \) is a weakly convergent sequence in \( H \), and \( f = w - \\lim {f}_{n} \), then we have \( \\parallel f\\parallel \\leq \\lim \\inf \\begin{Vmatrix}{f}_{n}\\end{Vmatrix} \) . | The proof immediately follows from Theorem 4.23, if we notice that the weak convergence of \( \\left( {f}_{n}\\right) \) is equivalent to the strong convergence of the sequence \( \\left( {T}_{{f}_{n}}\\right) \) of the linear functionals induced by \( {f}_{n} \) . | Yes |
Theorem 4.25. Let \( H \) be a Hilbert space. Every bounded sequence \( \left( {f}_{n}\right) \) in \( H \) contains a weakly convergent subsequence \( \left( {f}_{{n}_{k}}\right) \) . | Proof. Let \( M = L\left\{ {{f}_{n} : n \in \mathbb{N}}\right\} \) . Then \( M \oplus {M}^{ \bot } \) is dense in \( H \) . For every \( k \in \mathbb{N} \) the sequence \( {\left( \left\langle {f}_{n},{f}_{k}\right\rangle \right) }_{n \in \mathbb{N}} \) is bounded. Consequently, by induction we can find, for all \( j ... | Yes |
Every orthonormal sequence \( \left( {f}_{n}\right) \) weakly converges to zero. | This follows from the Bessel inequality \( \parallel f{\parallel }^{2} \geq \sum {\left| \left\langle {f}_{n}, f\right\rangle \right| }^{2} \) . In particular, the sequence of unit vectors \( \left( {{e}_{j} = \left( {\delta }_{jn}\right) }\right) \) in \( {l}_{2} \) tends to zero weakly. This example also shows that w... | Yes |
Theorem 4.26. Let \( {H}_{1} \) and \( {H}_{2} \) be pre-Hilbert spaces.\n\n(a) If \( \left( {T}_{n}\right) \) is a weakly convergent sequence in \( B\left( {{H}_{1},{H}_{2}}\right) \) and \( T = w - \lim {T}_{n} \), then we have \( \parallel T\parallel \leq \liminf \begin{Vmatrix}{T}_{n}\end{Vmatrix} \) . | Proof.\n\n(a) Let \( C = \lim \inf \begin{Vmatrix}{T}_{n}\end{Vmatrix} \) and let \( \left( {T}_{{n}_{k}}\right) \) be a subsequence such that \( \begin{Vmatrix}{T}_{{n}_{k}}\end{Vmatrix} \rightarrow \) \( C \) . Then for all \( f \in {H}_{1} \) and \( g \in {\widehat{H}}_{2} \) we have\n\n\[ \left| {\langle {Tf}, g\ra... | Yes |
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