Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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The finite field \( \mathrm{{GF}}\left( {2}^{4}\right) \) is isomorphic to the field \( {\mathbb{Z}}_{2}/\left\langle {1 + x + {x}^{4}}\right\rangle \) . Therefore, the elements of \( \operatorname{GF}\left( {2}^{4}\right) \) can be taken to be\n\n\[ \left\{ {{a}_{0} + {a}_{1}\alpha + {a}_{2}{\alpha }^{2} + {a}_{3}{\al... | \[ {\alpha }^{1} = \alpha \;{\alpha }^{6} = {\alpha }^{2} + {\alpha }^{3}\;{\alpha }^{11} = \alpha + {\alpha }^{2} + {\alpha }^{3} \]\n\n\[ {\alpha }^{2} = {\alpha }^{2}\;{\alpha }^{7} = 1 + \alpha + {\alpha }^{3}\;{\alpha }^{12} = 1 + \alpha + {\alpha }^{2} + {\alpha }^{3} \]\n\n\[ {\alpha }^{3} = {\alpha }^{3}\;{\alp... | Yes |
Consider the \( \left( {6,3}\right) \) -linear codes generated by the two matrices\n\n\[ \n{G}_{1} = \left( \begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \;\text{ and }\;{G}_{2} = \left( \begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 1 & ... | Messages in the first code are encoded as follows:\n\n\[ \n\left( {000}\right) \; \mapsto \;\left( {000000}\right) \;\left( {100}\right) \; \mapsto \;\left( {100100}\right) \n\]\n\n\[ \n\begin{matrix} \left( {001}\right) & \mapsto & \left( {001001}\right) & & & \left( {101}\right) & \mapsto & \left( {101101}\right) \en... | Yes |
If we let \( g\left( x\right) = 1 + {x}^{3} \), we can define a \( \left( {6,3}\right) \) -code \( C \) as follows. To encode a 3-tuple \( \left( {{a}_{0},{a}_{1},{a}_{2}}\right) \), we multiply the corresponding polynomial \( f\left( x\right) = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} \) by \( 1 + {x}^{3} \) . We are defin... | Observe that \( \phi \left( {{a}_{0},{a}_{1},{a}_{2}}\right) = \left( {000000}\right) \) exactly when\n\n\[ 0 + {0x} + 0{x}^{2} + 0{x}^{3} + 0{x}^{4} + 0{x}^{5} = \left( {1 + {x}^{3}}\right) \left( {{a}_{0} + {a}_{1}x + {a}_{2}{x}^{2}}\right) \]\n\n\[ = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + {a}_{0}{x}^{3} + {a}_{1}{x}^... | Yes |
Theorem 22.16. A linear code \( C \) in \( {\mathbb{Z}}_{2}^{n} \) is cyclic if and only if it is an ideal in \( {R}_{n} = \) \( \mathbb{Z}\left\lbrack x\right\rbrack /\left\langle {{x}^{n} - 1}\right\rangle \) | Proof. Let \( C \) be a linear cyclic code and suppose that \( f\left( t\right) \) is in \( C \) . Then \( {tf}\left( t\right) \) must also be in \( C \) . Consequently, \( {t}^{k}f\left( t\right) \) is in \( C \) for all \( k \in \mathbb{N} \) . Since \( C \) is a linear code, any linear combination of the codewords \... | Yes |
If we factor \( {x}^{7} - 1 \) into irreducible components, we have\n\n\[ {x}^{7} - 1 = \left( {1 + x}\right) \left( {1 + x + {x}^{3}}\right) \left( {1 + {x}^{2} + {x}^{3}}\right) . \] | We see that \( g\left( t\right) = \left( {1 + t + {t}^{3}}\right) \) generates an ideal \( C \) in \( {R}_{7} \) . This code is a \( \left( {7,4}\right) \) -block code. As in Example 22.15, it is easy to calculate a generator matrix by examining what \( g\left( t\right) \) does to the polynomials \( 1, t,{t}^{2} \), an... | No |
In Example 22.17, \[ {x}^{7} - 1 = g\left( x\right) h\left( x\right) = \left( {1 + x + {x}^{3}}\right) \left( {1 + x + {x}^{2} + {x}^{4}}\right) . \] | Therefore, a parity-check matrix for this code is \[ H = \left( \begin{array}{lllllll} 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 & 1 & 0 & 0 \end{array}\right) \] | Yes |
Lemma 22.20. Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be elements in a field \( F \) with \( n \geq 2 \) . Then\n\n\[ \n\det \left( \begin{matrix} 1 & 1 & \cdots & 1 \\ {\alpha }_{1} & {\alpha }_{2} & \cdots & {\alpha }_{n} \\ {\alpha }_{1}^{2} & {\alpha }_{2}^{2} & \cdots & {\alpha }_{n}^{2} \\ \vdots & \vdots & ... | Proof. We will induct on \( n \) . If \( n = 2 \), then the determinant is \( {\alpha }_{2} - {\alpha }_{1} \) . Let us assume the result for \( n - 1 \) and consider the polynomial \( p\left( x\right) \) defined by\n\n\[ \np\left( x\right) = \det \left( \begin{matrix} 1 & 1 & \cdots & 1 & 1 \\ {\alpha }_{1} & {\alpha ... | Yes |
Theorem 22.22. Let \( C = \langle g\left( t\right) \rangle \) be a cyclic code in \( {R}_{n} \) . The following statements are equivalent.\n\n1. The code \( C \) is a \( \mathrm{{BCH}} \) code whose minimum distance is at least \( d \) .\n\n2. A code polynomial \( f\left( t\right) \) is in \( C \) if and only if \( f\l... | Proof. \( \;\left( 1\right) \Rightarrow \left( 2\right) \) . If \( f\left( t\right) \) is in \( C \), then \( g\left( x\right) \mid f\left( x\right) \) in \( {\mathbb{Z}}_{2}\left\lbrack x\right\rbrack \) . Hence, for \( i = 1,\ldots ,{2r} \) , \( f\left( {\omega }^{i}\right) = 0 \) since \( g\left( {\omega }^{i}\right... | Yes |
Proposition 23.1. The set of all automorphisms of a field \( F \) is a group under composition of functions. | Proof. If \( \sigma \) and \( \tau \) are automorphisms of \( F \), then so are \( {\sigma \tau } \) and \( {\sigma }^{-1} \) . The identity is certainly an automorphism; hence, the set of all automorphisms of a field \( F \) is indeed a group. | Yes |
Proposition 23.2. Let \( E \) be a field extension of \( F \) . Then the set of all automorphisms of \( E \) that fix \( F \) elementwise is a group; that is, the set of all automorphisms \( \sigma : E \rightarrow E \) such that \( \sigma \left( \alpha \right) = \alpha \) for all \( \alpha \in F \) is a group. | Proof. We need only show that the set of automorphisms of \( E \) that fix \( F \) elementwise is a subgroup of the group of all automorphisms of \( E \) . Let \( \sigma \) and \( \tau \) be two automorphisms of \( E \) such that \( \sigma \left( \alpha \right) = \alpha \) and \( \tau \left( \alpha \right) = \alpha \) ... | Yes |
Complex conjugation, defined by \( \sigma : a + {bi} \mapsto a - {bi} \), is an automorphism of the complex numbers. | Since\n\n\[ \sigma \left( a\right) = \sigma \left( {a + {0i}}\right) = a - {0i} = a, \]\nthe automorphism defined by complex conjugation must be in \( G\left( {\mathbb{C}/\mathbb{R}}\right) \). | No |
Consider the fields \( \mathbb{Q} \subset \mathbb{Q}\left( \sqrt{5}\right) \subset \mathbb{Q}\left( {\sqrt{3},\sqrt{5}}\right) \) . Then for \( a, b \in \mathbb{Q}\left( \sqrt{5}\right) \) , \[ \sigma \left( {a + b\sqrt{3}}\right) = a - b\sqrt{3} \] is an automorphism of \( \mathbb{Q}\left( {\sqrt{3},\sqrt{5}}\right) \... | It will soon be clear that \( \{ \mathrm{{id}},\sigma ,\tau ,\mu \} \) is the Galois group of \( \mathbb{Q}\left( {\sqrt{3},\sqrt{5}}\right) \) over \( \mathbb{Q} \) . The following table shows that this group is isomorphic to \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \) . <table><thead><tr><th></th><th>id</th><th>\(... | Yes |
Proposition 23.5. Let \( E \) be a field extension of \( F \) and \( f\left( x\right) \) be a polynomial in \( F\left\lbrack x\right\rbrack \) . Then any automorphism in \( G\left( {E/F}\right) \) defines a permutation of the roots of \( f\left( x\right) \) that lie in \( E \) . | Proof. Let\n\n\[ f\left( x\right) = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + \cdots + {a}_{n}{x}^{n} \]\n\nand suppose that \( \alpha \in E \) is a zero of \( f\left( x\right) \) . Then for \( \sigma \in G\left( {E/F}\right) \) ,\n\n\[ 0 = \sigma \left( 0\right) \]\n\n\[ = \sigma \left( {f\left( \alpha \right) }\right) \]... | Yes |
Theorem 23.7. Let \( f\left( x\right) \) be a polynomial in \( F\left\lbrack x\right\rbrack \) and suppose that \( E \) is the splitting field for \( f\left( x\right) \) over \( F \) . If \( f\left( x\right) \) has no repeated roots, then \[ \left| {G\left( {E/F}\right) }\right| = \left\lbrack {E : F}\right\rbrack . \] | Proof. We will use mathematical induction on the degree of \( f\left( x\right) \) . If the degree of \( f\left( x\right) \) is 0 or 1, then \( E = F \) and there is nothing to show. Assume that the result holds for all polynomials of degree \( k \) with \( 0 \leq k < n \) . Suppose that the degree of \( f\left( x\right... | Yes |
Corollary 23.8. Let \( F \) be a finite field with a finite extension \( E \) such that \( \left\lbrack {E : F}\right\rbrack = k \) . Then \( G\left( {E/F}\right) \) is cyclic of order \( k \) . | Proof. Let \( p \) be the characteristic of \( E \) and \( F \) and assume that the orders of \( E \) and \( F \) are \( {p}^{m} \) and \( {p}^{n} \), respectively. Then \( {nk} = m \) . We can also assume that \( E \) is the splitting field of \( {x}^{{p}^{m}} - x \) over a subfield of order \( p \) . Therefore, \( E ... | Yes |
We can now confirm that the Galois group of \( \mathbb{Q}\left( {\sqrt{3},\sqrt{5}}\right) \) over \( \mathbb{Q} \) in Example 23.4 is indeed isomorphic to \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \). | Certainly the group \( H = \{ \mathrm{{id}},\sigma ,\tau ,\mu \} \) is a subgroup of \( G\left( {\mathbb{Q}\left( {\sqrt{3},\sqrt{5}}\right) /\mathbb{Q}}\right) \) ; however, \( H \) must be all of \( G\left( {\mathbb{Q}\left( {\sqrt{3},\sqrt{5}}\right) /\mathbb{Q}}\right) \), since\n\n\[ \left| H\right| = \left\lbrack... | Yes |
Let us compute the Galois group of \[ f\left( x\right) = {x}^{4} + {x}^{3} + {x}^{2} + x + 1 \] over \( \mathbb{Q} \). | We know that \( f\left( x\right) \) is irreducible by Exercise 17.4.20 in Chapter 17. Furthermore, since \( \left( {x - 1}\right) f\left( x\right) = {x}^{5} - 1 \), we can use DeMoivre’s Theorem to determine that the roots of \( f\left( x\right) \) are \( {\omega }^{i} \), where \( i = 1,\ldots ,4 \) and \[ \omega = \c... | Yes |
Proposition 23.11. Let \( f\left( x\right) \) be an irreducible polynomial over \( F \) . If the characteristic of \( F \) is 0, then \( f\left( x\right) \) is separable. If the characteristic of \( F \) is \( p \) and \( f\left( x\right) \neq g\left( {x}^{p}\right) \) for some \( g\left( x\right) \) in \( F\left\lbrac... | Proof. First assume that \( \operatorname{char}F = 0 \) . Since \( \deg {f}^{\prime }\left( x\right) < \deg f\left( x\right) \) and \( f\left( x\right) \) is irreducible, the only way \( \gcd \left( {f\left( x\right) ,{f}^{\prime }\left( x\right) }\right) \neq 1 \) is if \( {f}^{\prime }\left( x\right) \) is the zero p... | Yes |
Theorem 23.12 Primitive Element Theorem. Let \( E \) be a finite separable extension of a field \( F \) . Then there exists an \( \alpha \in E \) such that \( E = F\left( \alpha \right) \) . | Proof. We already know that there is no problem if \( F \) is a finite field. Suppose that \( E \) is a finite extension of an infinite field. We will prove the result for \( F\left( {\alpha ,\beta }\right) \) . The general case easily follows when we use mathematical induction. Let \( f\left( x\right) \) and \( g\left... | Yes |
Proposition 23.13. Let \( \left\{ {{\sigma }_{i} : i \in I}\right\} \) be a collection of automorphisms of a field \( F \) . Then\n\n\[ \n{F}_{\left\{ {\sigma }_{i}\right\} } = \left\{ {a \in F : {\sigma }_{i}\left( a\right) = a\text{ for all }{\sigma }_{i}}\right\} \n\]\n\nis a subfield of \( F \) . | Proof. Let \( {\sigma }_{i}\left( a\right) = a \) and \( {\sigma }_{i}\left( b\right) = b \) . Then\n\n\[ \n{\sigma }_{i}\left( {a \pm b}\right) = {\sigma }_{i}\left( a\right) \pm {\sigma }_{i}\left( b\right) = a \pm b \n\]\n\nand\n\n\[ \n{\sigma }_{i}\left( {ab}\right) = {\sigma }_{i}\left( a\right) {\sigma }_{i}\left... | Yes |
Proposition 23.16. Let \( E \) be a splitting field over \( F \) of a separable polynomial. Then \( {E}_{G\left( {E/F}\right) } = F \) . | Proof. Let \( G = G\left( {E/F}\right) \) . Clearly, \( F \subset {E}_{G} \subset E \) . Also, \( E \) must be a splitting field of \( {E}_{G} \) and \( G\left( {E/F}\right) = G\left( {E/{E}_{G}}\right) \) . By Theorem 23.7,\n\n\[ \left| G\right| = \left\lbrack {E : {E}_{G}}\right\rbrack = \left\lbrack {E : F}\right\rb... | Yes |
Theorem 23.18. Let \( E \) be a field extension of \( F \) . Then the following statements are equivalent.\n\n1. \( E \) is a finite, normal, separable extension of \( F \) .\n\n2. \( E \) is a splitting field over \( F \) of a separable polynomial.\n\n3. \( F = {E}_{G} \) for some finite group \( G \) of automorphisms... | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) . Let \( E \) be a finite, normal, separable extension of \( F \) . By the Primitive Element Theorem, we can find an \( \alpha \) in \( E \) such that \( E = F\left( \alpha \right) \) . Let \( f\left( x\right) \) be the minimal polynomial of \( \alpha \) over \( ... | Yes |
Corollary 23.19. Let \( K \) be a field extension of \( F \) such that \( F = {K}_{G} \) for some finite group of automorphisms \( G \) of \( K \) . Then \( G = G\left( {K/F}\right) \) . | Proof. Since \( F = {K}_{G}, G \) is a subgroup of \( G\left( {K/F}\right) \) . Hence, \[ \left\lbrack {K : F}\right\rbrack \leq \left| G\right| \leq \left| {G\left( {K/F}\right) }\right| = \left\lbrack {K : F}\right\rbrack . \] It follows that \( G = G\left( {K/F}\right) \), since they must have the same order. | Yes |
Theorem 23.22 Fundamental Theorem of Galois Theory. Let \( F \) be a finite field or a field of characteristic zero. If \( E \) is a finite normal extension of \( F \) with Galois group \( G\left( {E/F}\right) \), then the following statements are true.\n\n1. The map \( K \mapsto G\left( {E/K}\right) \) is a bijection ... | Proof. (1) Suppose that \( G\left( {E/K}\right) = G\left( {E/L}\right) = G \) . Both \( K \) and \( L \) are fixed fields of \( G \) ; hence, \( K = L \) and the map defined by \( K \mapsto G\left( {E/K}\right) \) is one-to-one. To show that the map is onto, let \( G \) be a subgroup of \( G\left( {E/F}\right) \) and \... | Yes |
In this example we will illustrate the Fundamental Theorem of Galois Theory by determining the lattice of subgroups of the Galois group of \( f\left( x\right) = {x}^{4} - 2 \) . We will compare this lattice to the lattice of field extensions of \( \mathbb{Q} \) that are contained in the splitting field of \( {x}^{4} - ... | The splitting field of \( f\left( x\right) \) is \( \mathbb{Q}\left( {\sqrt[4]{2}, i}\right) \) . To see this, notice that \( f\left( x\right) \) factors as \( \left( {{x}^{2} + \sqrt{2}}\right) \left( {{x}^{2} - \sqrt{2}}\right) \) ; hence, the roots of \( f\left( x\right) \) are \( \pm \sqrt[4]{2} \) and \( \pm \sqrt... | Yes |
Lemma 23.28. Let \( F \) be a field of characteristic zero and let\n\n\[ F = {F}_{0} \subset {F}_{1} \subset {F}_{2} \subset \cdots \subset {F}_{r} = E \]\n\n a radical extension of \( F \) . Then there exists a normal radical extension\n\n\[ F = {K}_{0} \subset {K}_{1} \subset {K}_{2} \subset \cdots \subset {K}_{r} = ... | Proof. Since \( E \) is a radical extension of \( F \), there exists a chain of subfields\n\n\[ F = {F}_{0} \subset {F}_{1} \subset {F}_{2} \subset \cdots \subset {F}_{r} = E \]\n\n such for \( i = 1,2,\ldots, r \), we have \( {F}_{i} = {F}_{i - 1}\left( {\alpha }_{i}\right) \) and \( {\alpha }_{i}^{{n}_{i}} \in {F}_{i... | Yes |
Theorem 23.29. Let \( f\left( x\right) \) be in \( F\left\lbrack x\right\rbrack \), where \( \operatorname{char}F = 0 \) . If \( f\left( x\right) \) is solvable by radicals, then the Galois group of \( f\left( x\right) \) over \( F \) is solvable. | Proof. Since \( f\left( x\right) \) is solvable by radicals there exists an extension \( E \) of \( F \) by radicals \( F = {F}_{0} \subset {F}_{1} \subset \cdots \subset {F}_{n} = E \) . By Lemma 23.28, we can assume that \( E \) is a splitting field \( f\left( x\right) \) and \( {F}_{i} \) is normal over \( {F}_{i - ... | Yes |
Lemma 23.30. If \( p \) is prime, then any subgroup of \( {S}_{p} \) that contains a transposition and a cycle of length \( p \) must be all of \( {S}_{p} \) . | Proof. Let \( G \) be a subgroup of \( {S}_{p} \) that contains a transposition \( \sigma \) and \( \tau \) a cycle of length \( p \) . We may assume that \( \sigma = \left( {12}\right) \) . The order of \( \tau \) is \( p \) and \( {\tau }^{n} \) must be a cycle of length \( p \) for \( 1 \leq n < p \) . Therefore, we... | No |
We will show that \( f\left( x\right) = {x}^{5} - 6{x}^{3} - {27x} - 3 \in \mathbb{Q}\left\lbrack x\right\rbrack \) is not solvable. | We claim that the Galois group of \( f\left( x\right) \) over \( \mathbb{Q} \) is \( {S}_{5} \) . By Eisenstein’s Criterion, \( f\left( x\right) \) is irreducible and, therefore, must be separable. The derivative of \( f\left( x\right) \) is \( {f}^{\prime }\left( x\right) = 5{x}^{4} - {18}{x}^{2} - \) 27; hence, setti... | Yes |
Theorem 23.33 Fundamental Theorem of Algebra. The field of complex numbers is algebraically closed; that is, every polynomial in \( \mathbb{C}\left\lbrack x\right\rbrack \) has a root in \( \mathbb{C} \) . | Proof. Suppose that \( E \) is a proper finite field extension of the complex numbers. Since any finite extension of a field of characteristic zero is a simple extension, there exists an \( \alpha \in E \) such that \( E = \mathbb{C}\left( \alpha \right) \) with \( \alpha \) the root of an irreducible polynomial \( f\l... | Yes |
Proposition 8.1.13. If \( B = \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{k}}\right\} \) is a basis of a vector space \( X \), then every point \( y \in X \) has a unique representation of the form\n\n\[ y = \mathop{\sum }\limits_{{j = 1}}^{k}{a}_{j}{x}_{j} \]\n\nfor some scalars \( {a}_{1},{a}_{2},\ldots ,{a}_{k} \) . | Proof. Every \( y \in X \) is a linear combination of elements of \( B \) since \( X \) is the span of \( B \) . For uniqueness suppose\n\n\[ y = \mathop{\sum }\limits_{{j = 1}}^{k}{a}_{j}{x}_{j} = \mathop{\sum }\limits_{{j = 1}}^{k}{b}_{j}{x}_{j} \]\n\nthen\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{k}\left( {{a}_{j} - {b... | Yes |
Proposition 8.1.14. Let \( X \) be a vector space and \( d \) a nonnegative integer.\n\n(i) If \( X \) is spanned by \( d \) vectors, then \( \dim X \leq d \) . | Proof. Let us start with (i). Suppose \( S = \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{d}}\right\} \) spans \( X \), and \( T = \left\{ {{y}_{1},{y}_{2},\ldots ,{y}_{m}}\right\} \) is a set of linearly independent vectors of \( X \) . We wish to show that \( m \leq d \) . Write\n\n\[{y}_{1} = \mathop{\sum }\limits_{{k = 1}... | Yes |
Proposition 8.1.16. If \( A \in L\left( {X, Y}\right) \) is invertible, then \( {A}^{-1} \) is linear. | Proof. Let \( a \in \mathbb{R} \) and \( y \in Y \) . As \( A \) is onto, then there is an \( x \) such that \( y = {Ax} \), and further as it is also one-to-one \( {A}^{-1}\left( {Az}\right) = z \) for all \( z \in X \) . So\n\n\[
{A}^{-1}\left( {ay}\right) = {A}^{-1}\left( {aAx}\right) = {A}^{-1}\left( {A\left( {ax}\... | Yes |
Proposition 8.1.17. If \( A \in L\left( {X, Y}\right) \) is linear, then it is completely determined by its values on a basis of \( X \) . Furthermore, if \( B \) is a basis of \( X \), then any function \( \widetilde{A} : B \rightarrow Y \) extends to a linear function on \( X \) . | Proof. Let \( \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} \) be a basis of \( X \), and let \( {y}_{j} \mathrel{\text{:=}} A{x}_{j} \) . Every \( x \in X \) has a unique representation\n\n\[ x = \mathop{\sum }\limits_{{j = 1}}^{n}{b}_{j}{x}_{j} \]\n\nfor some numbers \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) . By lin... | Yes |
Proposition 8.1.18. If \( X \) is a finite dimensional vector space and \( A \in L\left( X\right) \), then \( A \) is one-to-one if and only if it is onto. | Proof. Let \( \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} \) be a basis for \( X \) . Suppose \( A \) is one-to-one. Now suppose\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}A{x}_{j} = A\mathop{\sum }\limits_{{j = 1}}^{n}{c}_{j}{x}_{j} = 0. \]\n\nAs \( A \) is one-to-one, the only vector that is taken to 0 is ... | Yes |
Proposition 8.1.23. The intersection two convex sets is convex. In fact, if \( {\left\{ {C}_{\lambda }\right\} }_{\lambda \in I} \) is an arbitrary collection of convex sets, then\n\n\[ C \mathrel{\text{:=}} \mathop{\bigcap }\limits_{{\lambda \in I}}{C}_{\lambda } \]\n\nis convex. | Proof. If \( x, y \in C \), then \( x, y \in {C}_{\lambda } \) for all \( \lambda \in I \), and hence if \( t \in \left\lbrack {0,1}\right\rbrack \), then \( {tx} + \left( {1 - t}\right) y \in {C}_{\lambda } \) for all \( \lambda \in I \) . Therefore, \( {tx} + \left( {1 - t}\right) y \in C \) and \( C \) is convex. | Yes |
Proposition 8.1.24. Let \( T : V \rightarrow W \) be a linear mapping between two vector spaces and let \( C \subset V \) be a convex set. Then \( T\left( C\right) \) is convex. | Proof. Take any two points \( p, q \in T\left( C\right) \) . Pick \( x, y \in C \) such that \( {Tx} = p \) and \( {Ty} = q \) . As \( C \) is convex, then \( {tx} + \left( {1 - t}\right) y \in C \) for all \( t \in \left\lbrack {0,1}\right\rbrack \), so\n\n\[ \n{tp} + \left( {1 - t}\right) q = {tTx} + \left( {1 - t}\r... | Yes |
Theorem 8.2.2 (Cauchy-Schwarz inequality). Let \( x, y \in {\mathbb{R}}^{n} \), then\n\n\[ \left| {x \cdot y}\right| \leq \parallel x\parallel \parallel y\parallel = \sqrt{x \cdot x}\sqrt{y \cdot y} \]\n\nwith equality if and only if \( x = {\lambda y} \) or \( y = {\lambda x} \) for some \( \lambda \in \mathbb{R} \) . | Proof. If \( x = 0 \) or \( y = 0 \), then the theorem holds trivially. So assume \( x \neq 0 \) and \( y \neq 0 \) .\n\nIf \( x \) is a scalar multiple of \( y \), that is \( x = {\lambda y} \) for some \( \lambda \in \mathbb{R} \), then the theorem holds with equality:\n\n\[ \left| {x \cdot y}\right| = \left| {{\lamb... | Yes |
Proposition 8.2.4. Let \( X \) and \( Y \) be normed vector spaces. Suppose that \( X \) is finite dimensional. If \( A \in L\left( {X, Y}\right) \), then \( \parallel A\parallel < \infty \), and \( A \) is uniformly continuous (Lipschitz with constant \( \parallel A\parallel \) ). | Proof. As we said we only prove the proposition for euclidean space so suppose that \( X = {\mathbb{R}}^{n} \) and the norm is the standard euclidean norm. The general case is left as an exercise.\n\nLet \( \left\{ {{e}_{1},{e}_{2},\ldots ,{e}_{n}}\right\} \) be the standard basis of \( {\mathbb{R}}^{n} \) . Write \( x... | No |
Proposition 8.2.5. Let \( X, Y \), and \( Z \) be finite dimensional normed vector spaces*. (i) If \( A, B \in L\left( {X, Y}\right) \) and \( c \in \mathbb{R} \), then \[ \parallel A + B\parallel \leq \parallel A\parallel + \parallel B\parallel ,\;\parallel {cA}\parallel = \left| c\right| \parallel A\parallel . \] In ... | Proof. First, since all the spaces are finite dimensional, then all the operator norms are finite, and the statements make sense to begin with. For (i), \[ \parallel \left( {A + B}\right) x\parallel = \parallel {Ax} + {Bx}\parallel \leq \parallel {Ax}\parallel + \parallel {Bx}\parallel \leq \parallel A\parallel \parall... | Yes |
Proposition 8.2.6. Let \( X \) be a finite dimensional normed vector space. Let \( {GL}\left( X\right) \subset L\left( X\right) \) be the set of invertible linear operators.\n\n(i) If \( A \in {GL}\left( X\right), B \in L\left( X\right) \), and\n\n\[ \parallel A - B\parallel < \frac{1}{\begin{Vmatrix}{A}^{-1}\end{Vmatr... | Proof. Let us prove (i). We know something about \( {A}^{-1} \) and \( A - B \) . These are linear operators so let us apply them to a vector:\n\n\[ {A}^{-1}\left( {A - B}\right) x = x - {A}^{-1}{Bx}. \]\n\nTherefore,\n\n\[ \parallel x\parallel = \begin{Vmatrix}{{A}^{-1}\left( {A - B}\right) x + {A}^{-1}{Bx}}\end{Vmatr... | Yes |
Proposition 8.2.7. If \( f : S \rightarrow {\mathbb{R}}^{nm} \) is a continuous function for a metric space \( S \), then considering the components of \( f \) as the entries of a matrix, \( f \) is a continuous mapping from \( S \) to \( L\left( {{\mathbb{R}}^{n},{\mathbb{R}}^{m}}\right) \) . Conversely, if \( f : S \... | Let us prove the second part. Take \( f\left( x\right) {e}_{j} \), which is a continuous function of \( S \) to \( {\mathbb{R}}^{m} \) with standard euclidean norm: \( \begin{Vmatrix}{f\left( x\right) {e}_{j} - f\left( y\right) {e}_{j}}\end{Vmatrix} = \begin{Vmatrix}{\left( {f\left( x\right) - f\left( y\right) }\right)... | Yes |
Determinant is independent of the basis. In other words, if \( B \) is invertible, then\n\n\[ \det \left( A\right) = \det \left( {{B}^{-1}{AB}}\right) \] | The proof is to compute \( \det \left( {{B}^{-1}{AB}}\right) = \det \left( {B}^{-1}\right) \det \left( A\right) \det \left( B\right) = \frac{1}{\det \left( B\right) }\det \left( A\right) \det \left( B\right) = \det \left( A\right) . | Yes |
Proposition 8.3.2. Let \( U \subset {\mathbb{R}}^{n} \) be an open subset and \( f : U \rightarrow {\mathbb{R}}^{m} \) . Suppose \( x \in U \) and there exist \( A, B \in L\left( {{\mathbb{R}}^{n},{\mathbb{R}}^{m}}\right) \) such that\n\n\[ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\parallel f\left( {x + h}\right)... | Proof. Suppose \( h \in {\mathbb{R}}^{n}, h \neq 0 \) . Compute\n\n\[ \frac{\parallel \left( {A - B}\right) h\parallel }{\parallel h\parallel } = \frac{\parallel f\left( {x + h}\right) - f\left( x\right) - {Ah} - \left( {f\left( {x + h}\right) - f\left( x\right) - {Bh}}\right) \parallel }{\parallel h\parallel }\]\n\n\[... | Yes |
Proposition 8.3.5. Let \( U \subset {\mathbb{R}}^{n} \) be open and \( f : U \rightarrow {\mathbb{R}}^{m} \) be differentiable at \( p \in U \) . Then \( f \) is continuous at \( p \) . | Proof. Another way to write the differentiability of \( f \) at \( p \) is to first write\n\n\[ r\left( h\right) \mathrel{\text{:=}} f\left( {p + h}\right) - f\left( p\right) - {f}^{\prime }\left( p\right) h, \]\n\nand \( \frac{\parallel r\left( h\right) \parallel }{\parallel h\parallel } \) must go to zero as \( h \ri... | Yes |
Proposition 8.3.6. Suppose \( U \subset {\mathbb{R}}^{n} \) is open, \( f : U \rightarrow {\mathbb{R}}^{m} \) and \( g : U \rightarrow {\mathbb{R}}^{m} \) are differentiable at \( p \) , and \( \alpha \in \mathbb{R} \) . Then the functions \( f + g \) and \( {\alpha f} \) are differentiable at \( p \) and\n\n\[{\left( ... | Proof. Let \( h \in {\mathbb{R}}^{n}, h \neq 0 \) . Then\n\n\[ \frac{\begin{Vmatrix}f\left( p + h\right) + g\left( p + h\right) - \left( f\left( p\right) + g\left( p\right) \right) - \left( {f}^{\prime }\left( p\right) + {g}^{\prime }\left( p\right) \right) h\end{Vmatrix}}{\parallel h\parallel }\n\n\[ \leq \frac{\paral... | Yes |
Theorem 8.3.7 (Chain rule). Let \( U \subset {\mathbb{R}}^{n} \) be open and let \( f : U \rightarrow {\mathbb{R}}^{m} \) be differentiable at \( p \in U \) . Let \( V \subset {\mathbb{R}}^{m} \) be open, \( f\left( U\right) \subset V \) and let \( g : V \rightarrow {\mathbb{R}}^{\ell } \) be differentiable at \( f\lef... | Proof. Let \( A \mathrel{\text{:=}} {f}^{\prime }\left( p\right) \) and \( B \mathrel{\text{:=}} {g}^{\prime }\left( {f\left( p\right) }\right) \) . Take \( h \in {\mathbb{R}}^{n} \) and write \( q = f\left( p\right), k = f\left( {p + h}\right) - f\left( p\right) \) . Let\n\n\[ r\left( h\right) \mathrel{\text{:=}} f\le... | Yes |
Proposition 8.3.9. Let \( U \subset {\mathbb{R}}^{n} \) be open and let \( f : U \rightarrow {\mathbb{R}}^{m} \) be differentiable at \( p \in U \). Then all the partial derivatives at \( p \) exist and, in terms of the standard bases of \( {\mathbb{R}}^{n} \) and \( {\mathbb{R}}^{m},{f}^{\prime }\left( p\right) \) is ... | Proof. Fix a \( j \) and note that\n\n\[ \begin{Vmatrix}{\frac{f\left( {p + h{e}_{j}}\right) - f\left( p\right) }{h} - {f}^{\prime }\left( p\right) {e}_{j}}\end{Vmatrix} = \begin{Vmatrix}\frac{f\left( {p + h{e}_{j}}\right) - f\left( p\right) - {f}^{\prime }\left( p\right) h{e}_{j}}{h}\end{Vmatrix} \]\n\n\[ = \frac{\beg... | Yes |
Lemma 8.4.1. If \( \varphi : \left\lbrack {a, b}\right\rbrack \rightarrow {\mathbb{R}}^{n} \) is differentiable on \( \left( {a, b}\right) \) and continuous on \( \left\lbrack {a, b}\right\rbrack \), then there exists \( a{t}_{0} \in \left( {a, b}\right) \) such that\n\n\[ \parallel \varphi \left( b\right) - \varphi \l... | Proof. By mean value theorem on the scalar-valued function \( t \mapsto \left( {\varphi \left( b\right) - \varphi \left( a\right) }\right) \cdot \varphi \left( t\right) \), where the dot is the dot product, we obtain that there is a \( {t}_{0} \in \left( {a, b}\right) \) such that\n\n\[ \parallel \varphi \left( b\right... | Yes |
Proposition 8.4.2. Let \( U \subset {\mathbb{R}}^{n} \) be a convex open set, \( f : U \rightarrow {\mathbb{R}}^{m} \) a differentiable function, and an M be such that\n\n\[ \n\begin{Vmatrix}{{f}^{\prime }\left( x\right) }\end{Vmatrix} \leq M\;\text{ for all }x \in U.\n\]\n\nThen \( f \) is Lipschitz with constant \( M... | Proof. Fix \( x \) and \( y \) in \( U \) and note that \( \left( {1 - t}\right) x + {ty} \in U \) for all \( t \in \left\lbrack {0,1}\right\rbrack \) by convexity. Next\n\n\[ \n\frac{d}{dt}\left\lbrack {f\left( {\left( {1 - t}\right) x + {ty}}\right) }\right\rbrack = {f}^{\prime }\left( {\left( {1 - t}\right) x + {ty}... | Yes |
Corollary 8.4.4. If \( U \subset {\mathbb{R}}^{n} \) is open and connected, \( f : U \rightarrow {\mathbb{R}}^{m} \) is differentiable, and \( {f}^{\prime }\left( x\right) = 0 \) for all \( x \in U \), then \( f \) is constant. | Proof. For any \( x \in U \), there is a ball \( B\left( {x,\delta }\right) \subset U \) . The ball \( B\left( {x,\delta }\right) \) is convex. Since \( \begin{Vmatrix}{{f}^{\prime }\left( y\right) }\end{Vmatrix} \leq 0 \) for all \( y \in B\left( {x,\delta }\right) \), then by the proposition, \( \parallel f\left( x\r... | Yes |
Corollary 8.4.7. A polynomial \( p : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) in several variables\n\n\[ p\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) = \mathop{\sum }\limits_{{0 \leq {j}_{1} + {j}_{2} + \cdots + {j}_{n} \leq d}}{c}_{{j}_{1},{j}_{2},\ldots ,{j}_{n}}{x}_{1}^{{j}_{1}}{x}_{2}^{{j}_{2}}\cdots {x}_{n}^... | Proof. Consider the partial derivative of \( p \) in the \( {x}_{n} \) variable. Write \( p \) as\n\n\[ p\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{d}{p}_{j}\left( {{x}_{1},\ldots ,{x}_{n - 1}}\right) {x}_{n}^{j} \]\n\nwhere \( {p}_{j} \) are polynomials in one less variable. Then\n\n\[ \frac{\partial p}{\parti... | Yes |
Theorem 8.5.1 (Inverse function theorem). Let \( U \subset {\mathbb{R}}^{n} \) be an open set and let \( f : U \rightarrow {\mathbb{R}}^{n} \) be a continuously differentiable function. Suppose \( p \in U \) and \( {f}^{\prime }\left( p\right) \) is invertible (that is, \( {J}_{f}\left( p\right) \neq 0 \) ). Then there... | Proof. Write \( A = {f}^{\prime }\left( p\right) \). As \( {f}^{\prime } \) is continuous, there exists an open ball \( V \) around \( p \) such that\n\n\[ \begin{Vmatrix}{A - {f}^{\prime }\left( x\right) }\end{Vmatrix} < \frac{1}{2\begin{Vmatrix}{A}^{-1}\end{Vmatrix}}\;\text{ for all }x \in V. \]\n\nConsequently, the ... | Yes |
Corollary 8.5.2. Suppose \( U \subset {\mathbb{R}}^{n} \) is open and \( f : U \rightarrow {\mathbb{R}}^{n} \) is a continuously differentiable mapping such that \( {f}^{\prime }\left( x\right) \) is invertible for all \( x \in U \) . Then given any open set \( V \subset U, f\left( V\right) \) is open \( (f \) is said ... | Proof. Without loss of generality, suppose \( U = V \) . For each point \( y \in f\left( V\right) \), we pick \( x \in {f}^{-1}\left( y\right) \) (there could be more than one such point), then by the inverse function theorem there is a neighborhood of \( x \) in \( V \) that maps onto a neighborhood of \( y \) . Hence... | Yes |
Proposition 8.5.5. Let \( A = \left\lbrack {{A}_{x}{A}_{y}}\right\rbrack \in L\left( {{\mathbb{R}}^{n + m},{\mathbb{R}}^{m}}\right) \) and suppose \( {A}_{y} \) is invertible. If \( B = - {\left( {A}_{y}\right) }^{-1}{A}_{x} \) , then\n\n\[ 0 = A\left( {x,{Bx}}\right) = {A}_{x}x + {A}_{y}{Bx}. \]\n\nFurthermore, \( y =... | The proof is immediate: We solve and obtain \( y = {Bx} \) . Another way to solve is to \ | No |
Theorem 8.5.6 (Implicit function theorem). Let \( U \subset {\mathbb{R}}^{n + m} \) be an open set and let \( f : U \rightarrow {\mathbb{R}}^{m} \) be a \( {C}^{1}\left( U\right) \) mapping. Let \( \left( {p, q}\right) \in U \) be a point such that \( f\left( {p, q}\right) = 0 \) and such that \[ \frac{\partial \left( ... | Proof. Define \( F : U \rightarrow {\mathbb{R}}^{n + m} \) by \( F\left( {x, y}\right) \mathrel{\text{:=}} \left( {x, f\left( {x, y}\right) }\right) \) . It is clear that \( F \) is \( {C}^{1} \), and we want to show that the derivative at \( \left( {p, q}\right) \) is invertible. Let us compute the derivative. We know... | Yes |
Theorem 9.1.1 (Leibniz integral rule). Suppose \( f : \left\lbrack {a, b}\right\rbrack \times \left\lbrack {c, d}\right\rbrack \rightarrow \mathbb{R} \) is a continuous function, such that \( \frac{\partial f}{\partial y} \) exists for all \( \left( {x, y}\right) \in \left\lbrack {a, b}\right\rbrack \times \left\lbrack... | Proof. Fix \( y \in \left\lbrack {c, d}\right\rbrack \) and let \( \varepsilon > 0 \) be given. As \( \frac{\partial f}{\partial y} \) is continuous on \( \left\lbrack {a, b}\right\rbrack \times \left\lbrack {c, d}\right\rbrack \) it is uniformly continuous. In particular, there exists \( \delta > 0 \) such that whenev... | Yes |
Proposition 9.2.6. If \( \gamma : \left\lbrack {a, b}\right\rbrack \rightarrow {\mathbb{R}}^{n} \) is a piecewise smooth path, and \( \gamma \circ h : \left\lbrack {c, d}\right\rbrack \rightarrow {\mathbb{R}}^{n} \) is a piecewise smooth reparametrization, then \( \gamma \circ h \) is a piecewise smooth path. | Proof. Let us assume that \( h \) preserves orientation, that is, \( h \) is strictly increasing. If \( h : \left\lbrack {c, d}\right\rbrack \rightarrow \left\lbrack {a, b}\right\rbrack \) gives a piecewise smooth reparametrization, then for some partition \( {r}_{0} = c < {r}_{1} < {r}_{2} < \cdots < {r}_{\ell } = d \... | "No" |
Proposition 9.2.12. Let \( \gamma : \left\lbrack {a, b}\right\rbrack \rightarrow {\mathbb{R}}^{n} \) be a piecewise smooth path and \( \gamma \circ h : \left\lbrack {c, d}\right\rbrack \rightarrow {\mathbb{R}}^{n} \) a piecewise smooth reparametrization. Suppose \( \omega \) is a one-form defined on the set \( \gamma \... | Proof. Assume first that \( \gamma \) and \( h \) are both smooth. Write \( \omega = {\omega }_{1}d{x}_{1} + {\omega }_{2}d{x}_{2} + \cdots + {\omega }_{n}d{x}_{n} \) . Suppose that \( h \) is orientation preserving. Use the change of variables formula for the Riemann integral:\n\n\[ \n{\int }_{\gamma }\omega = {\int }... | No |
Proposition 9.2.15. Let \( \\gamma : \\left\\lbrack {a, b}\\right\\rbrack \\rightarrow {\\mathbb{R}}^{n} \) be a piecewise smooth path and \( \\gamma \\circ h : \\left\\lbrack {c, d}\\right\\rbrack \\rightarrow {\\mathbb{R}}^{n} \) a piecewise smooth reparametrization. Suppose \( f \) is a continuous function defined o... | Proof. Suppose first that \( h \) is orientation preserving and that \( \\gamma \) and \( h \) are both smooth. Then\n\n\[{\\int }_{\\gamma }{fds} = {\\int }_{a}^{b}f\\left( {\\gamma \\left( t\\right) }\\right) \\begin{Vmatrix}{{\\gamma }^{\\prime }\\left( t\\right) }\\end{Vmatrix}{dt}\]\n\n\[= {\\int }_{c}^{d}f\\left(... | No |
Proposition 9.3.4. Let \( U \subset {\mathbb{R}}^{n} \) be a path connected open set and \( \omega \) a one-form defined on \( U \) . Then \( \omega = {df} \) for some continuously differentiable \( f : U \rightarrow \mathbb{R} \) if and only if\n\n\[ \n{\int }_{\gamma }\omega = 0\;\text{ for every piecewise smooth clo... | Proof. Suppose \( \omega = {df} \) and let \( \gamma \) be a piecewise smooth closed path. Since \( \gamma \left( a\right) = \gamma \left( b\right) \) for a closed path, the previous proposition says\n\n\[ \n{\int }_{\gamma }\omega = f\left( {\gamma \left( b\right) }\right) - f\left( {\gamma \left( a\right) }\right) = ... | Yes |
Theorem 9.3.6 (Poincarè lemma). Let \( U \subset {\mathbb{R}}^{n} \) be a star-shaped domain and \( \omega \) a continuously differentiable one-form defined on \( U \) . That is, if\n\n\[ \omega = {\omega }_{1}d{x}_{1} + {\omega }_{2}d{x}_{2} + \cdots + {\omega }_{n}d{x}_{n} \]\n\nthen \( {\omega }_{1},{\omega }_{2},\l... | Proof. Suppose \( U \) is star-shaped with respect to \( p = \left( {{p}_{1},{p}_{2},\ldots ,{p}_{n}}\right) \in U \) .\n\nGiven \( x = \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \in U \), define the path \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow U \) as \( \gamma \left( t\right) \mathrel{\text{:=}} ... | Yes |
Proposition 10.1.2. Suppose \( R \subset {\mathbb{R}}^{n} \) is a closed rectangle and \( f : R \rightarrow \mathbb{R} \) is a bounded function. Let \( m, M \in \mathbb{R} \) be such that for all \( x \in R \) we have \( m \leq f\left( x\right) \leq M \) . For any partition \( P \) of \( R \) we have\n\n\[ \n{mV}\left(... | Proof. Let \( P \) be a partition. Then for all \( i \) we have \( m \leq {m}_{i} \) and \( {M}_{i} \leq M \) . Also \( {m}_{i} \leq {M}_{i} \) for all \( i \) . Finally \( \mathop{\sum }\limits_{{i = 1}}^{N}V\left( {R}_{i}\right) = V\left( R\right) \) . Therefore,\n\n\[ \n{mV}\left( R\right) = m\left( {\mathop{\sum }\... | Yes |
Proposition 10.1.5. Suppose \( R \subset {\mathbb{R}}^{n} \) is a closed rectangle, \( P \) is a partition of \( R \), and \( \widetilde{P} \) is a refinement of \( P \) . If \( f : R \rightarrow \mathbb{R} \) be a bounded function, then \[ L\left( {P, f}\right) \leq L\left( {\widetilde{P}, f}\right) \;\text{ and }\;U\... | Proof. We prove the first inequality, and the second follows similarly. Let \( {R}_{1},{R}_{2},\ldots ,{R}_{N} \) be the subrectangles of \( P \) and \( {\widetilde{R}}_{1},{\widetilde{R}}_{2},\ldots ,{\widetilde{R}}_{\widetilde{N}} \) be the subrectangles of \( \widetilde{R} \) . Let \( {I}_{k} \) be the set of all in... | Yes |
Proposition 10.1.6. Let \( R \subset {\mathbb{R}}^{n} \) be a closed rectangle and \( f : R \rightarrow \mathbb{R} \) a bounded function. Let \( m, M \in \mathbb{R} \) be such that for all \( x \in R \) we have \( m \leq f\left( x\right) \leq M \) . Then\n\n\[ \n{mV}\left( R\right) \leq {\int }_{\underline{R}}f \leq \o... | Proof. For any partition \( P \), via Proposition 10.1.2,\n\n\[ \n{mV}\left( R\right) \leq L\left( {P, f}\right) \leq U\left( {P, f}\right) \leq {MV}\left( R\right) .\n\]\n\nTaking supremum of \( L\left( {P, f}\right) \) and infimum of \( U\left( {P, f}\right) \) over all partitions \( P \), we obtain the first and the... | Yes |
Proposition 10.1.12. Let \( R \subset {\mathbb{R}}^{n} \) be a closed rectangle and \( f : R \rightarrow \mathbb{R} \) a bounded function. Then \( f \in \mathcal{R}\left( R\right) \) if and only if for every \( \varepsilon > 0 \), there exists a partition \( P \) of \( R \) such that\n\n\[ U\left( {P, f}\right) - L\lef... | Proof. First, if \( f \) is integrable, then clearly the supremum of \( L\left( {P, f}\right) \) and infimum of \( U\left( {P, f}\right) \) must be equal and hence the infimum of \( U\left( {P, f}\right) - L\left( {P, f}\right) \) is zero. Therefore for every \( \varepsilon > 0 \) there must be some partition \( P \) s... | Yes |
Proposition 10.1.13. Let \( S \subset {\mathbb{R}}^{n} \) be a closed rectangle. If \( f : S \rightarrow \mathbb{R} \) is integrable and \( R \subset S \) is a closed rectangle, then \( f \) is integrable on \( R \) . | Proof. Given \( \varepsilon > 0 \), we find a partition \( P \) of \( S \) such that \( U\left( {P, f}\right) - L\left( {P, f}\right) < \varepsilon \) . By making a refinement of \( P \) if necessary, we assume that the endpoints of \( R \) are in \( P \) . In other words, \( R \) is a union of subrectangles of \( P \)... | Yes |
Proposition 10.1.14. If a rectangle \( R \subset {\mathbb{R}}^{n} \) has longest side at most \( \alpha \), then for any \( x, y \in R \), \[ \parallel x - y\parallel \leq \sqrt{n}\alpha \] | Proof. \[ \parallel x - y\parallel = \sqrt{{\left( {x}_{1} - {y}_{1}\right) }^{2} + {\left( {x}_{2} - {y}_{2}\right) }^{2} + \cdots + {\left( {x}_{n} - {y}_{n}\right) }^{2}} \] \[ \leq \sqrt{{\left( {b}_{1} - {a}_{1}\right) }^{2} + {\left( {b}_{2} - {a}_{2}\right) }^{2} + \cdots + {\left( {b}_{n} - {a}_{n}\right) }^{2}... | Yes |
Theorem 10.1.15. Let \( R \subset {\mathbb{R}}^{n} \) be a closed rectangle and \( f : R \rightarrow \mathbb{R} \) a continuous function, then \( f \in \mathcal{R}\left( R\right) \) . | Proof. The proof is analogous to the one variable proof with some complications. The set \( R \) is a closed and bounded subset of \( {\mathbb{R}}^{n} \), and hence compact. So \( f \) is not just continuous, but in fact uniformly continuous by Theorem 7.5.11 from volume I. Let \( \varepsilon > 0 \) be given. Find a \(... | Yes |
Proposition 10.1.19. Suppose \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be a continuous function with compact support. If \( R \) and \( S \) are closed rectangles such that \( \operatorname{supp}\left( f\right) \subset R \) and \( \operatorname{supp}\left( f\right) \subset S \), then\n\n\[{\int }_{S}f = {\int }... | Proof. As \( f \) is continuous, it is automatically integrable on the rectangles \( R, S \), and \( R \cap S \) . Then Exercise 10.1.7 says \( {\int }_{S}f = {\int }_{S \cap R}f = {\int }_{R}f \) . | No |
Proposition 10.3.4. A countable union of measure zero sets is of measure zero. | Proof. Suppose\n\n\[ S = \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{S}_{j} \]\n\nwhere \( {S}_{j} \) are all measure zero sets. Let \( \varepsilon > 0 \) be given. For each \( j \) there exists a sequence of open rectangles \( {\left\{ {R}_{j, k}\right\} }_{k = 1}^{\infty } \) such that\n\n\[ {S}_{j} \subset \mathop{... | Yes |
Proposition 10.3.7. Suppose \( E \subset {\mathbb{R}}^{n} \) is a compact set of measure zero. Then for every \( \varepsilon > 0 \), there exist finitely many open rectangles \( {R}_{1},{R}_{2},\ldots ,{R}_{k} \) such that\n\n\[ E \subset {R}_{1} \cup {R}_{2} \cup \cdots \cup {R}_{k}\;\text{ and }\;\mathop{\sum }\limit... | Proof. Find a sequence of open rectangles \( \left\{ {R}_{j}\right\} \) such that\n\n\[ E \subset \mathop{\bigcup }\limits_{{j = 1}}^{\infty }{R}_{j}\;\text{ and }\;\mathop{\sum }\limits_{{j = 1}}^{\infty }V\left( {R}_{j}\right) < \varepsilon . \]\n\nBy compactness, there are finitely many of these rectangles that stil... | No |
Lemma 10.3.9. Suppose \( U \subset {\mathbb{R}}^{n} \) is an open set, \( B \subset U \) is an open (resp. closed) ball of radius at most \( r, f : B \rightarrow {\mathbb{R}}^{n} \) is continuously differentiable and suppose \( \begin{Vmatrix}{{f}^{\prime }\left( x\right) }\end{Vmatrix} \leq M \) for all \( x \in B \) ... | Proof. Without loss of generality assume \( B \) is a closed ball. The ball \( B \) is convex, and hence via Proposition 8.4.2, \( \parallel f\left( x\right) - f\left( y\right) \parallel \leq M\parallel x - y\parallel \) for all \( x, y \) in \( B \) . In particular, suppose \( B = C\left( {y, r}\right) \) , then \( f\... | Yes |
Proposition 10.3.10. Suppose \( U \subset {\mathbb{R}}^{n} \) is an open set and \( f : U \rightarrow {\mathbb{R}}^{n} \) is a continuously differentiable mapping. If \( E \subset U \) is a measure zero set, then \( f\left( E\right) \) is measure zero. | Proof. We leave the proof for a general measure zero set as an exercise, and we now prove the proposition for a compact measure zero set.\n\nSuppose \( E \) is compact. First let us replace \( U \) by a smaller open set to make \( \begin{Vmatrix}{{f}^{\prime }\left( x\right) }\end{Vmatrix} \) bounded. At each point \( ... | No |
Proposition 10.4.1. \( f : S \rightarrow \mathbb{R} \) is continuous at \( x \in S \) if and only if \( o\left( {f, x}\right) = 0 \) . | Proof. First suppose that \( f \) is continuous at \( x \in S \) . Then given any \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that for \( y \in {B}_{S}\left( {x,\delta }\right) \) we have \( \left| {f\left( x\right) - f\left( y\right) }\right| < \varepsilon \) . Therefore if \( {y}_{1},{y}_{2} \in {B}_{... | Yes |
Proposition 10.4.2. Let \( S \subset {\mathbb{R}}^{n} \) be closed, \( f : S \rightarrow \mathbb{R} \), and \( \varepsilon > 0 \) . The set \( \{ x \in S : o\left( {f, x}\right) \geq \varepsilon \} \) is closed. | Proof. Equivalently we want to show that \( G = \{ x \in S : o\left( {f, x}\right) < \varepsilon \} \) is open in the subspace topology. As \( \mathop{\inf }\limits_{{\delta > 0}}o\left( {f, x,\delta }\right) < \varepsilon \), find a \( \delta > 0 \) such that\n\n\[ o\left( {f, x,\delta }\right) < \varepsilon \]\n\nTak... | Yes |
Corollary 10.4.4. Let \( R \subset {\mathbb{R}}^{n} \) be a closed rectangle. Let \( \mathcal{R}\left( R\right) \) be the set of Riemann integrable functions on \( R \) . Then\n\n(i) \( \mathcal{R}\left( R\right) \) is a real algebra: If \( f, g \in \mathcal{R}\left( R\right) \) and \( a \in \mathbb{R} \), then af \( \... | The proof is contained in the exercises. | No |
Proposition 10.5.1. A bounded set \( S \subset {\mathbb{R}}^{n} \) is Jordan measurable if and only if the boundary \( \partial S \) is a measure zero set. | Proof. Suppose \( R \) is a closed rectangle such that \( S \) is contained in the interior of \( R \) . If \( x \in \partial S \), then for every \( \delta > 0 \), the sets \( S \cap B\left( {x,\delta }\right) \) (where \( {\chi }_{S} \) is 1) and the sets \( \left( {R \smallsetminus S}\right) \cap B\left( {x,\delta }... | Yes |
Proposition 10.5.2. Suppose \( S \) and \( T \) are bounded Jordan measurable sets. Then\n\n(i) The closure \( \bar{S} \) is Jordan measurable.\n\n(ii) The interior \( {S}^{ \circ } \) is Jordan measurable.\n\n(iii) \( S \cup T \) is Jordan measurable.\n\n(iv) \( S \cap T \) is Jordan measurable.\n\n(v) \( S \smallsetm... | The proof of the proposition is left as an exercise. | No |
Proposition 10.5.3. If \( S \subset {\mathbb{R}}^{n} \) is Jordan measurable, then \( V\left( S\right) = {m}^{ * }\left( S\right) \) . | Proof. Given \( \varepsilon > 0 \), let \( R \) be a closed rectangle that contains \( S \) . Let \( P \) be a partition of \( R \) such that\n\n\[ U\left( {P,{\chi }_{S}}\right) \leq {\int }_{R}{\chi }_{S} + \varepsilon = V\left( S\right) + \varepsilon \;\text{ and }\;L\left( {P,{\chi }_{S}}\right) \geq {\int }_{R}{\c... | Yes |
Proposition 10.5.5. If \( S \subset {\mathbb{R}}^{n} \) is a bounded Jordan measurable set and \( f : S \rightarrow \mathbb{R} \) is a bounded continuous function, then \( f \) is integrable on \( S \) . | Proof. Define the function \( \widetilde{f} \) as above for some closed rectangle \( R \) with \( S \subset R \) . If \( x \in R \smallsetminus \bar{S} \), then \( \widetilde{f} \) is identically zero in a neighborhood of \( x \) . Similarly if \( x \) is in the interior of \( S \), then \( \widetilde{f} = f \) on a ne... | Yes |
Proposition 10.5.6. Suppose \( S \subset {\mathbb{R}}^{n} \) is a closed bounded Jordan measurable set, and \( S \subset U \) for an open set \( U \subset {\mathbb{R}}^{n} \) . Suppose \( g : U \rightarrow {\mathbb{R}}^{n} \) is a one-to-one continuously differentiable mapping such that \( {J}_{g} \) is never zero on \... | Proof. Let \( T \mathrel{\text{:=}} g\left( S\right) \) . As \( S \subset {\mathbb{R}}^{n} \) is closed and bounded it is compact. By Lemma 7.5.5 from volume I, the set \( T \) is also compact and so closed and bounded. We claim \( \partial T \subset g\left( {\partial S}\right) \) . Suppose the claim is proved. As \( S... | Yes |
Theorem 10.6.4 (Green). Suppose \( U \subset {\mathbb{R}}^{2} \) is a bounded domain with piecewise smooth boundary with the boundary positively oriented. Suppose \( P \) and \( Q \) are continuously differentiable functions defined on some open set that contains the closure \( \bar{U} \) . Then\n\n\[ \n{\int }_{\parti... | Proof of Green’s theorem for \( U \) of type III. Let \( f, g, h, k \) be the functions defined above. By Exercise 10.5.3, \( U \) is Jordan measurable and as \( U \) is of type I, then\n\n\[ \n{\int }_{U}\left( {-\frac{\partial P}{\partial y}}\right) = {\int }_{a}^{b}{\int }_{g\left( x\right) }^{f\left( x\right) }\lef... | No |
Proposition 10.7.1. Suppose \( R \subset {\mathbb{R}}^{n} \) is a rectangle and \( A : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) is linear. Then \( A\left( R\right) \) is Jordan measurable and \( V\left( {A\left( R\right) }\right) = \left| {\det \left( A\right) }\right| V\left( R\right) \) . | Proof. It is enough to prove for elementary matrices. The proof is left as an exercise. | No |
Proposition 11.1.5. If a complex series \( \sum {z}_{n} \) converges absolutely, then it converges. | The series \( \sum \left| {z}_{n}\right| \) is a real series. All the convergence tests (ratio test, root test, etc.) that talk about absolute convergence work with the numbers \( \left| {z}_{n}\right| \), that is, they are really talking about convergence of series of nonnegative real numbers. You can directly apply t... | "No" |
Theorem 11.2.4 (Weierstrass \( M \) -test). Let \( X \) be any set. Suppose \( {f}_{n} : X \rightarrow \mathbb{C} \) are functions, \( {M}_{n} > 0 \) numbers such that\n\n\[ \left| {{f}_{n}\left( x\right) }\right| \leq {M}_{n}\;\text{ for all }x \in X,\;\text{ and }\;\mathop{\sum }\limits_{{n = 1}}^{\infty }{M}_{n}\;\t... | Proof. Suppose \( \sum {M}_{n} \) converges. Given \( \varepsilon > 0 \), we have that the partial sums of \( \sum {M}_{n} \) are Cauchy so for there is an \( N \) such that for all \( m, n \geq N \) with \( m \geq n \) we have\n\n\[ \mathop{\sum }\limits_{{k = n + 1}}^{m}{M}_{k} < \varepsilon \]\n\nNow let us look at ... | Yes |
Proposition 11.2.7. Let \( \left( {X,{d}_{X}}\right) \) and \( \left( {Y,{d}_{Y}}\right) \) be metric spaces, and let \( {f}_{n} : X \rightarrow Y \) be functions. Suppose \( Y \) is complete metric space. Suppose \( \left\{ {f}_{n}\right\} \) converges uniformly to \( f : X \rightarrow Y \) . Let \( \left\{ {x}_{k}\ri... | Proof. First we show that \( \left\{ {a}_{n}\right\} \) converges. As \( \left\{ {f}_{n}\right\} \) converges uniformly it is uniformly Cauchy. Let \( \varepsilon > 0 \) be given. There is an \( M \) such that for all \( m, n \geq M \) we have\n\n\[ \n{d}_{Y}\left( {{f}_{n}\left( {x}_{k}\right) ,{f}_{m}\left( {x}_{k}\r... | Yes |
Proposition 11.2.11. Suppose \( {f}_{n} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{C} \) are Riemann integrable and suppose that \( \left\{ {f}_{n}\right\} \) converges uniformly to \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{C} \) . Then \( f \) is Riemann integrable and\n\n\[{\int }_{a}^{b... | Since the integral of a complex-valued function is just the integral of the real and imaginary parts separately, the proof follows directly by the results of chapter 6 of volume I. We leave the details as an exercise. | No |
Proposition 11.3.1. Let \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{n}{\left( z - a\right) }^{n} \) be a power series. There exists a \( \rho \in \left\lbrack {0,\infty }\right\rbrack \) such that\n\n(i) the series diverges and \( \rho = 0 \) ,\n\n(ii) the series converges for all \( z \in \mathbb{C} \) and \( \rh... | Proof. We use the real version of this proposition, Proposition 2.6.10 in volume I. Let\n\n\[ R \mathrel{\text{:=}} \mathop{\limsup }\limits_{{n \rightarrow \infty }}\sqrt[n]{\left| {c}_{n}\right| }.\]\n\nIf \( R = 0 \), then \( \mathop{\sum }\limits_{{n = 0}}^{\infty }\left| {c}_{n}\right| {\left| z - a\right| }^{n} \... | Yes |
Proposition 11.3.4. If\n\n\[ f\left( z\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{n}{\left( z - a\right) }^{n} \]\n\nis convergent in \( B\left( {a,\rho }\right) \) for some \( \rho > 0 \), then \( f : B\left( {a,\rho }\right) \rightarrow \mathbb{C} \) is continuous. In particular, analyti... | Proof. For any \( {z}_{0} \in B\left( {a,\rho }\right) \) pick \( r < \rho \) such that \( {z}_{0} \in B\left( {a, r}\right) \) . On \( B\left( {a, r}\right) \) the partial sums (which are continuous) converge uniformly, and so the limit \( {\left. f\right| }_{B\left( {a, r}\right) } \) is continuous. Any sequence conv... | Yes |
Theorem 11.3.5 (Fubini for sums). Let \( {\left\{ {a}_{kj}\right\} }_{k = 1, j = 1}^{\infty } \) be a double sequence of complex numbers and suppose that for every \( k \) the series\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{\infty }\left| {a}_{kj}\right| \;\text{ converges } \]\n\nand furthermore that\n\n\[ \mathop{\sum ... | Proof. Let \( E \) be the set \( \{ 1/n : n \in \mathbb{N}\} \cup \{ 0\} \), and treat it as a metric space with the metric inherited from \( \mathbb{R} \) . Define the sequence of functions \( {f}_{k} : E \rightarrow \mathbb{C} \) by\n\n\[ {f}_{k}\left( {1/n}\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{j = 1}}^... | Yes |
Theorem 11.3.6 (Taylor's theorem for real-analytic functions). Let\n\n\\[ \nf\left( x\right) \mathrel{\\text{:=}} \\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{a}_{k}{x}^{k}\n\\]\n\nbe a power series converging in \\( \\left( {-\\rho ,\\rho }\\right) \\) for some \\( \\rho > 0 \\) . Given any \\( a \\in \\left( {-\\rho... | Proof. Given \\( a \\) and \\( x \\) as in the theorem, write\n\n\\[ \nf\left( x\right) = \\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{a}_{k}{\\left( \\left( x - a\\right) + a\\right) }^{k}\n\\]\n\n\\[ \n= \\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{a}_{k}\\mathop{\\sum }\\limits_{{m = 0}}^{k}\\left( \\begin{matrix}... | Yes |
Corollary 11.3.7. If \( \sum {c}_{k}{\left( z - a\right) }^{k} \) converges to \( f\left( z\right) \) in \( B\left( {a,\rho }\right) \) and \( b \in B\left( {a,\rho }\right) \), then there exists a power series \( \sum {d}_{k}{\left( z - b\right) }^{k} \) that converges to \( f\left( z\right) \) in \( B\left( {b,\rho -... | Proof. Without loss of generality assume that \( a = 0 \) . We can rotate to assume that \( b \) is real, but since that is harder to picture, let us do it explicitly. Let \( \alpha = \frac{\bar{b}}{\left| b\right| } \) . Notice that\n\n\[ \left| {1/\alpha }\right| = \left| \alpha \right| = 1 \]\n\nTherefore the series... | Yes |
Lemma 11.3.8. Suppose \( f\left( z\right) = \sum {a}_{k}{z}^{k} \) is a convergent power series and \( \left\{ {z}_{n}\right\} \) is a sequence of nonzero complex numbers converging to 0, such that \( f\left( {z}_{n}\right) = 0 \) for all \( n \) . Then \( {a}_{k} = 0 \) for every \( k \) . | Proof. By continuity we know \( f\left( 0\right) = 0 \) so \( {a}_{0} = 0 \) . Suppose there exists some nonzero \( {a}_{k} \) . Let \( m \) be the smallest \( m \) such that \( {a}_{m} \neq 0 \) . Then\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{k = m}}^{\infty }{a}_{k}{z}^{k} = {z}^{m}\mathop{\sum }\limits_{{k =... | Yes |
Theorem 11.3.9 (Identity theorem). Let \( U \subset \mathbb{C} \) be an open connected set. If \( f : U \rightarrow \mathbb{C} \) and \( g : U \rightarrow \mathbb{C} \) are analytic functions that are equal on a set \( E \subset U \), and \( E \) has a cluster point in \( U \) , then \( f\left( z\right) = g\left( z\rig... | Proof. Without loss of generality suppose \( E \) is the set of all points \( z \in U \) such that \( g\left( z\right) = f\left( z\right) \) . Note that \( E \) must be closed as \( f \) and \( g \) are continuous.\n\nSuppose \( E \) has a cluster point. Without loss of generality assume that 0 is this cluster point. N... | Yes |
Proposition 11.4.1. Let \( z, w \in \mathbb{C} \) be complex numbers. Then\n\n\[ \n{e}^{z + w} = {e}^{z}{e}^{w} \n\] | Proof. We know \( {e}^{x + y} = {e}^{x}{e}^{y} \) is true for real numbers \( x \) and \( y \) .\n\nNow for any fixed \( y \in \mathbb{R} \), we get by the identity theorem (Theorem 11.3.9) that \( {e}^{z + y} = {e}^{z}{e}^{y} \) for all \( z \in \mathbb{C} \) . Fixing an arbitrary \( z \in \mathbb{C} \), we get \( {e}... | Yes |
Lemma 11.5.1. Let \( p\left( z\right) \) be complex polynomial. If \( p\left( {z}_{0}\right) \neq 0 \), then there exist \( w \in \mathbb{C} \) such that \( \left| {p\left( w\right) }\right| < \left| {p\left( {z}_{0}\right) }\right| \). In fact, we can pick \( w \) to be arbitrarily close to \( {z}_{0} \). | Proof. Without loss of generality assume that \( {z}_{0} = 0 \) and \( p\left( 0\right) = 1 \). Write\n\n\[ p\left( z\right) = 1 + {a}_{k}{z}^{k} + {a}_{k + 1}{z}^{k + 1} + \cdots + {a}_{d}{z}^{d}, \]\n\nwhere \( {a}_{k} \neq 0 \). Pick \( t \) such that \( {a}_{k}{e}^{ikt} = - \left| {a}_{k}\right| \), which we can do... | Yes |
Lemma 11.5.4. Let \( p\left( z\right) \) be complex polynomial. Then for any \( M > 0 \) there exists an \( R > 0 \) such that \( \left| {p\left( z\right) }\right| \geq M \) whenever \( \left| z\right| \geq R \) . | Proof. Write \( p\left( z\right) = {a}_{0} + {a}_{1}z + \cdots + {a}_{d}{z}^{d} \) and suppose that \( {a}_{d} \neq 0 \) . Suppose \( \left| z\right| \geq R \) (so also \( \left. {{\left| z\right| }^{-1} \leq {R}^{-1}}\right) \) . We estimate:\n\n\[ \left| {p\left( z\right) }\right| \geq \left| {{a}_{d}{z}^{d}}\right| ... | Yes |
Theorem 11.5.5 (Fundamental theorem of algebra). Let \( p\left( z\right) \) be complex polynomial, then there exists a \( {z}_{0} \in \mathbb{C} \) such that \( p\left( {z}_{0}\right) = 0 \) . | Proof. Let \( \mu \mathrel{\text{:=}} \inf \{ \left| {p\left( z\right) }\right| : z \in \mathbb{C}\} \) . Find an \( R \) such that for all \( z \) with \( \left| z\right| \geq R \) we have \( \left| {p\left( z\right) }\right| \geq \mu + 1 \) . Therefore, any \( z \) with \( \left| {p\left( z\right) }\right| \) close t... | Yes |
Proposition 11.6.5. Let \( X \) be a countable set and \( {f}_{n} : X \rightarrow \mathbb{C} \) give a pointwise bounded sequence of functions. Then \( \left\{ {f}_{n}\right\} \) has a subsequence that converges pointwise. | Proof. Let \( {x}_{1},{x}_{2},{x}_{3},\ldots \) be an enumeration of the elements of \( X \) . The sequence \( {\left\{ {f}_{n}\left( {x}_{1}\right) \right\} }_{n = 1}^{\infty } \) is bounded and hence we have a subsequence of \( {\left\{ {f}_{n}\right\} }_{n = 1}^{\infty } \), which we denote by \( {\left\{ {f}_{1, k}... | Yes |
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