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Proposition 3. (Continuity of a composition of continuous mappings.) Let \( \left( {X,{\tau }_{X}}\right) ,\left( {Y,{\tau }_{Y}}\right) \) and \( \left( {Z,{\tau }_{Z}}\right) \) be topological spaces. If the mapping \( g : Y \rightarrow Z \) is continuous at a point \( b \in Y \) and the mapping \( f : X \rightarrow ... | This follows from the definition of continuity of a mapping and Proposition 1. | No |
Proposition 4. (Boundedness of a mapping in a neighborhood of a point of continuity.) If a mapping \( f : X \rightarrow Y \) of a topological space \( \left( {X,\tau }\right) \) into a metric space \( \left( {Y, d}\right) \) is continuous at a point \( a \in X \), then it is bounded in some neighborhood of that point. | This proposition follows from the ultimate boundedness (over a base) of a mapping that has a limit. | No |
Proposition 5. A mapping \( f : X \rightarrow Y \) of a metric space \( \left( {X,{d}_{X}}\right) \) into a metric space \( \left( {Y,{d}_{Y}}\right) \) is continuous at the point \( a \in X \) if and only if \( \omega \left( {f, a}\right) = 0 \) . | This proposition follows immediately from the definition of continuity of a mapping at a point. | No |
Theorem 2. The image of a compact set under a continuous mapping is compact. | Proof. Let \( f : K \rightarrow Y \) be a continuous mapping of the compact space \( \left( {K,{\tau }_{K}}\right) \) into a topological space \( \left( {Y,{\tau }_{Y}}\right) \), and let \( \left\{ {{G}_{Y}^{\alpha },\alpha \in A}\right\} \) be a covering of \( f\left( K\right) \) by sets that are open in \( Y \) . By... | Yes |
Theorem 3. (Uniform continuity.) A continuous mapping \( f : K \rightarrow Y \) of a compact metric space \( K \) into a metric space \( \left( {Y,{d}_{Y}}\right) \) is uniformly continuous. | In particular, if \( K \) is a closed interval in \( \mathbb{R} \) and \( Y = \mathbb{R} \), we again have the classical theorem of Cantor, the proof of which given in Subsect. 4.2.2. carries over with almost no changes to this general case. | No |
Theorem 4. The image of a connected topological space under a continuous mapping is connected. | Proof. Let \( f : X \rightarrow Y \) be a continuous mapping of a connected topological space \( \left( {X,{\tau }_{X}}\right) \) onto a topological space \( \left( {Y,{\tau }_{Y}}\right) \) . Let \( {E}_{Y} \) be an open-closed subset of \( Y \) . By Theorem 1, the pre-image \( {E}_{X} = {f}^{-1}\left( {E}_{Y}\right) ... | Yes |
As an important example of the application of the contraction mapping principle we shall prove, following Picard, an existence theorem for the solution of the differential equation \( {y}^{\prime }\left( x\right) = f\left( {x, y\left( x\right) }\right) \) satisfying an initial condition \( y\left( {x}_{0}\right) = {y}_... | Proof. Equation (9.23) and the condition (9.22) can be jointly written as a single relation \( y\left( x\right) = {y}_{0} + {\int }_{{x}_{0}}^{x}f\left( {t, y\left( t\right) }\right) {dt}. \) Denoting the right-hand side of this equality by \( A\left( y\right) \), we find that \( A \) : \( C\left( {V\left( {x}_{0}\righ... | Yes |
As an illustration of what was just said, we shall seek a solution of the familiar equation\n\n\\[ \n{y}^{\prime } = y \n\\]\n\nwith the initial condition (9.22) on the basis of the contraction mapping principle. | In this case\n\n\\[ \n{Ay} = {y}_{0} + {\\int }_{{x}_{0}}^{x}y\\left( t\\right) {dt} \n\\]\n\nand the principle is applicable at least for \\( \\left| {x - {x}_{0}}\\right| \\leq q < 1 \\) .\n\nStarting from the initial approximation \\( y\\left( x\\right) \\equiv 0 \\), we construct successively the sequence \\( 0,{y}... | Yes |
Newton’s method of finding a root of the equation \( f\left( x\right) = 0 \) . Suppose a real-valued function that is convex and has a positive derivative on a closed interval \( \left\lbrack {\alpha ,\beta }\right\rbrack \) assumes values of opposite signs at the endpoints of the interval. Then there is a unique point... | \[ {x}_{n + 1} = {x}_{n} - {\left\lbrack {f}^{\prime }\left( {x}_{n}\right) \right\rbrack }^{-1} \cdot f\left( {x}_{n}\right) \] (9.25) of points that, as one can verify, will tend monotonically to \( a \) in the present case. | Yes |
In analysis, besides the spaces \( {\mathbb{R}}^{n} \) and \( {\mathbb{C}}^{n} \) exhibited in Example 1, we encounter the space closest to them, which is the space \( \ell \) of sequences \( x = \left( {{x}^{1},\ldots ,{x}^{n},\ldots }\right) \) of real or complex numbers. The vector-space operations in \( \ell \), as... | The set of finite sequences (all of whose terms are zero from some point on) is a vector subspace \( \ell \) of the space \( \ell \), also infinite-dimensional. | No |
If for \( p \geq 1 \) we set\n\n\[ \parallel x{\parallel }_{p} \mathrel{\text{:=}} {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\left| {x}^{i}\right| }^{p}\right) }^{\frac{1}{p}} \]\n\nfor \( x = \left( {{x}^{1},\ldots ,{x}^{n}}\right) \in {\mathbb{R}}^{n} \), it follows from Minkowski’s inequality that we obtain a norm... | One can verify that\n\n\[ \parallel x{\parallel }_{{p}_{2}} \leq \parallel x{\parallel }_{{p}_{1}},\text{ if }1 \leq {p}_{1} \leq {p}_{2}, \]\n\nand that\n\n\[ \parallel x{\parallel }_{p} \rightarrow \max \left\{ {\left| {x}^{1}\right| ,\ldots ,\left| {x}^{n}\right| }\right\} \]\n\nas \( p \rightarrow + \infty \) . Thu... | Yes |
Example 6. The preceding example can be usefully generalized as follows. If \( X = {X}_{1} \times \cdots \times {X}_{n} \) is the direct product of normed vector spaces, one can introduce the norm of a vector \( x = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) in the direct product by setting\n\n\[ \parallel x{\parallel ... | Naturally, inequalities (10.8) remain valid in this case as well.\n\nFrom now on, when the direct product of normed spaces is considered, unless the contrary is explicitly stated, it is assumed that the norm is defined in accordance with formula (10.9) (including the case \( p = + \infty \) ). | No |
Let \( p \geq 1 \) . We denote by \( {\ell }_{p} \) the set of sequences \( x = \) \( \left( {{x}^{1},\ldots ,{x}^{n},\ldots }\right) \) of real or complex numbers such that the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{\left| {x}^{n}\right| }^{p} \) converges, and for \( x \in {\ell }_{p} \) we set\n\n\[ \pa... | Using Minkowski’s inequality, one can easily see that \( {\ell }_{p} \) is a normed vector space with respect to the standard vector-space operations and the norm (10.10). This is an infinite-dimensional space with respect to which \( {\mathbb{R}}_{p}^{n} \) is a vector subspace of finite dimension. | Yes |
In the vector space \( C\left\lbrack {a, b}\right\rbrack \) of numerical-valued functions that are continuous on the closed interval \( \left\lbrack {a, b}\right\rbrack \), one usually considers the following norm:\n\n\[ \parallel f\parallel \mathrel{\text{:=}} \mathop{\max }\limits_{{x \in \left\lbrack {a, b}\right\rb... | We leave the verification of the norm axioms to the reader. We remark that this norm generates a metric on \( C\left\lbrack {a, b}\right\rbrack \) that is already familiar to us (see Sect. 9.5), and we know that the metric space that thereby arises is complete. Thus the vector space \( C\left\lbrack {a, b}\right\rbrack... | No |
One can also introduce another norm in \( C\left\lbrack {a, b}\right\rbrack \)\n\n\[ \parallel f{\parallel }_{p} \mathrel{\text{:=}} {\left( {\int }_{a}^{b}{\left| f\right| }^{p}\left( x\right) dx\right) }^{\frac{1}{p}},\;p \geq 1, \] | which becomes (10.11) as \( p \rightarrow + \infty \) .\n\nIt is easy to see (for example, Sect. 9.5) that the space \( C\left\lbrack {a, b}\right\rbrack \) with the norm (10.12) is not complete for \( 1 \leq p < + \infty \) . | No |
In \( {\ell }_{2} \) the inner product of the vectors \( x \) and \( y \) can be defined as\n\n\[\n\langle x, y\rangle \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{\infty }{x}^{i}\overline{{y}^{i}}\n\] | The series in this expression converges absolutely since\n\n\[\n2\mathop{\sum }\limits_{{i = 1}}^{\infty }\left| {{x}^{i}\overline{{y}^{i}}}\right| \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }{\left| {x}^{i}\right| }^{2} + \mathop{\sum }\limits_{{i = 1}}^{\infty }{\left| {y}^{i}\right| }^{2}.\n\] | Yes |
The following important inequality, known as the Cauchy-Bunyakovskii inequality, holds for the inner product:\n\n\[ \n{\left| \langle x, y\rangle \right| }^{2} \leq \langle x, x\rangle \cdot \langle y, y\rangle \n\]\n\nwhere equality holds if and only if the vectors \( x \) and \( y \) are collinear. | Proof. Indeed, let \( a = \langle x, x\rangle, b = \langle x, y\rangle \), and \( c = \langle y, y\rangle \) . By hypothesis \( a \geq 0 \) and \( c \geq 0 \) . If \( c > 0 \), the inequalities\n\n\[ \n0 \leq \langle x + {\lambda y}, x + {\lambda y}\rangle = a + \bar{b}\lambda + b\bar{\lambda } + {c\lambda }\bar{\lambd... | Yes |
Example 4. We define the transformation \( A : C\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \rightarrow C\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \) by the formula\n\n\[ A\left( f\right) \mathrel{\text{:=}} {\int }_{a}^{x}f\left( t\right) {dt} \]\n\nwhere \( x \) is a point ranging ... | All of these transformations are obviously linear. | No |
Let \( X = {X}_{1} \times \cdots \times {X}_{m} \) be the vector space that is the direct product of the spaces \( {X}_{1},\ldots ,{X}_{m} \), and let \( A : X \rightarrow Y \) be a linear mapping of \( X \) into a vector space \( Y \) . Representing every vector \( x = \left( {{x}_{1},\ldots ,{x}_{m}}\right) \in X \) ... | Since the mapping \( A : X = {X}_{1} \times \cdots \times {X}_{m} \rightarrow Y \) is obviously linear for any linear mappings \( {A}_{i} : {X}_{i} \rightarrow Y \), we have shown that formula (10.22) gives the general form of any linear mapping \( A \in \mathcal{L}\left( {X = {X}_{1} \times \cdots \times {X}_{m};Y}\ri... | Yes |
Starting from the definition of the direct product \( Y = \) \( {Y}_{1} \times \cdots \times {Y}_{n} \) of the vector spaces \( {Y}_{1},\ldots ,{Y}_{n} \) and the definition of a linear mapping \( A : X \rightarrow Y \), one can easily see that any linear mapping | \[ A : X \rightarrow Y = {Y}_{1} \times \cdots \times {Y}_{n} \] has the form \( x \mapsto {Ax} = \left( {{A}_{1}x,\ldots ,{A}_{n}x}\right) = \left( {{y}_{1},\ldots ,{y}_{n}}\right) = y \in Y \), where \( {A}_{i} : X \rightarrow {Y}_{i} \) are linear mappings. | Yes |
Combining Examples 9 and 10, we conclude that any linear mapping\n\n\\[ \nA : {X}_{1} \\times \\cdots \\times {X}_{m} = X \\rightarrow Y = {Y}_{1} \\times \\cdots \\times {Y}_{n} \n\\]\n\nof the direct product \\( X = {X}_{1} \\times \\cdots \\times {X}_{m} \\) of vector spaces into another direct product \\( Y = {Y}_{... | In particular, if \\( {X}_{1} = {X}_{2} = \\cdots = {X}_{m} = \\mathbb{R} \\) and \\( {Y}_{1} = {Y}_{2} = \\cdots = {Y}_{n} = \\mathbb{R} \\), then \\( {A}_{ij} : {X}_{j} \\rightarrow {Y}_{i} \\) are the linear mappings \\( \\mathbb{R} \\ni x \\mapsto {a}_{ij}x \\in \\mathbb{R} \\), each of which is given by a single n... | Yes |
Proposition 1. For a multilinear transformation \( A : {X}_{1} \times \cdots \times {X}_{n} \rightarrow Y \) mapping a product of normed spaces \( {X}_{1},\ldots ,{X}_{n} \) into a normed space \( Y \) the following conditions are equivalent:\n\na) A has a finite norm,\n\nb) \( A \) is a bounded transformation,\n\nc) \... | Proof. We prove a closed chain of implications \( a) \Rightarrow b) \Rightarrow c) \Rightarrow d) \Rightarrow a) \) .\n\nIt is obvious from relation (10.27) that \( a) \Rightarrow b \) ).\n\nLet us verify that \( b) \Rightarrow c \) ), that is, that (10.29) implies that the operator \( A \) is continuous. Indeed, takin... | Yes |
Proposition 2. The norm of a multilinear transformation is a norm in the vector space \( \mathcal{L}\left( {{X}_{1},\ldots ,{X}_{n};Y}\right) \) of continuous multilinear transformations. | Proof. We observe first of all that by Proposition 1 the nonnegative number \( \parallel A\parallel < \infty \) is defined for every transformation \( A \in \mathcal{L}\left( {{X}_{1},\ldots ,{X}_{n};Y}\right) \). Inequality (10.27) shows that\n\n\[ \parallel A\parallel = 0 \Leftrightarrow A = 0. \]\n\nNext, by definit... | Yes |
Proposition 3. If \( Y \) is a complete normed space, then \( \mathcal{L}\left( {{X}_{1},\ldots ,{X}_{n};Y}\right) \) is also a complete normed space. | Proof. We shall carry out the proof for the space \( \mathcal{L}\left( {X;Y}\right) \) of continuous linear transformations. The general case, as will be clear from the reasoning below, differs only in requiring a more cumbersome notation.\n\nLet \( {A}_{1},{A}_{2},\ldots ,{A}_{n}\ldots \) be a Cauchy sequence in \( \m... | Yes |
Proposition 4. For each \( m \in \{ 1,\ldots, n\} \) there is a bijection between the spaces\n\n\[ \mathcal{L}\left( {{X}_{1},\ldots ,{X}_{m};\mathcal{L}\left( {{X}_{m + 1},\ldots ,{X}_{n};Y}\right) }\right) \text{and}\mathcal{L}\left( {{X}_{1},\ldots ,{X}_{n};Y}\right) \]\n\nthat preserves the vector-space structure a... | Proof. We shall exhibit this isomorphism.\n\nLet \( \mathfrak{B} \in \mathcal{L}\left( {{X}_{1},\ldots ,{X}_{m};\mathcal{L}\left( {{X}_{m + 1},\ldots ,{X}_{n};Y}\right) }\right) \), that is, \( \mathfrak{B}\left( {{x}_{1},\ldots ,{x}_{m}}\right) \in \) \( \mathcal{L}\left( {{X}_{m + 1},\ldots ,{X}_{n};Y}\right) \) We s... | No |
Proposition 1. If a mapping \( f : E \rightarrow Y \) is differentiable at an interior point \( x \) of a set \( E \subset X \) , its differential \( L\left( x\right) \) at that point is uniquely determined. | Proof. Thus we are verifying the uniqueness of the differential.\n\nLet \( {L}_{1}\left( x\right) \) and \( {L}_{2}\left( x\right) \) be linear mappings satisfying relation (10.31), that is\n\n\[ f\left( {x + h}\right) - f\left( x\right) - {L}_{1}\left( x\right) h = {\alpha }_{1}\left( {x;h}\right) ,\]\n\n(10.32)\n\n\[... | Yes |
Example 1. If \( f : U \rightarrow Y \) is a constant mapping of a neighborhood \( U = \) \( U\left( x\right) \subset X \) of the point \( x \), that is, \( f\left( U\right) = {y}_{0} \in Y \), then \( {f}^{\prime }\left( x\right) = 0 \in \mathcal{L}\left( {X;Y}\right) \) . | Proof. Indeed, in this case it is obvious that\n\n\[ f\left( {x + h}\right) - f\left( x\right) - {0h} = {y}_{0} - {y}_{0} - 0 = 0 = o\left( h\right) . \] | Yes |
If the mapping \( f : X \rightarrow Y \) is a continuous linear mapping of a normed vector space \( X \) into a normed vector space \( Y \), then \( {f}^{\prime }\left( x\right) = f \in \) \( \mathcal{L}\left( {X;Y}\right) \) at any point \( x \in A \) . | Indeed, \[ f\left( {x + h}\right) - f\left( x\right) - {fh} = {fx} + {fh} - {fx} - {fh} = 0. \] We remark that strictly speaking \( {f}^{\prime }\left( x\right) \in \mathcal{L}\left( {T{X}_{x};T{Y}_{f\left( x\right) }}\right) \) here and \( h \) is a vector of the tangent space \( T{X}_{x} \) . But parallel translation... | Yes |
From the chain rule for differentiating a composition of mappings and the result of Example 2 one can conclude that if \( f : U \rightarrow Y \) is a mapping of a neighborhood \( U = U\left( x\right) \subset X \) of the point \( x \in X \) and is differentiable at \( x \), while \( A \in \mathcal{L}\left( {Y;Z}\right) ... | \[ {\left( A \circ f\right) }^{\prime }\left( x\right) = A \circ {f}^{\prime }\left( x\right) . \] | Yes |
Example 4. Suppose once again that \( U = U\\left( x\\right) \) is a neighborhood of the point \( x \) in a normed space \( X \), and let\n\n\[ f : U \\rightarrow Y = {Y}_{1} \\times \\cdots \\times {Y}_{n} \]\n\nbe a mapping of \( U \) into the direct product of the normed spaces \( {Y}_{1},\\ldots ,{Y}_{n} \) .\n\nDe... | If we now take account of the fact that in formula (10.31) we have\n\n\[ f\\left( {x + h}\\right) - f\\left( x\\right) = \\left( {{f}_{1}\\left( {x + h}\\right) - {f}_{1}\\left( x\\right) ,\\ldots ,{f}_{n}\\left( {x + h}\\right) - {f}_{n}\\left( x\\right) }\\right) ,\]\n\n\[ L\\left( x\\right) h = \\left( {{L}_{1}\\lef... | Yes |
We shall prove that the mapping \[ A : {X}_{1} \times \cdots \times {X}_{n} = X \rightarrow Y \] is differentiable and find its differential. | Proof. Using the multilinearity of \( A \), we find that \[ A\left( {x + h}\right) - A\left( x\right) = A\left( {{x}_{1} + {h}_{1},\ldots ,{x}_{n} + {h}_{n}}\right) - A\left( {{x}_{1},\ldots ,{x}_{n}}\right) = \] \[ = A\left( {{x}_{1},\ldots ,{x}_{n}}\right) + A\left( {{h}_{1},{x}_{2},\ldots ,{x}_{n}}\right) + \cdots +... | Yes |
Let \( X \) be a complete normed vector space. The important mapping \[ \exp : \mathcal{L}\left( {X;X}\right) \rightarrow \mathcal{L}\left( {X;X}\right) \] is defined as follows: \[ \exp A \mathrel{\text{:=}} E + \frac{1}{1!}A + \frac{1}{2!}{A}^{2} + \cdots + \frac{1}{n!}{A}^{n} + \cdots , \] if \( A \in \mathcal{L}\le... | The series in (10.37) converges, since \( \mathcal{L}\left( {X;X}\right) \) is a complete space and \( \parallel \frac{1}{n!}{A}^{n}\parallel \leq \frac{\parallel A{\parallel }^{n}}{n!} \), while the numerical series \( \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{\parallel A{\parallel }^{n}}{n!} \) converges. It is ... | Yes |
We shall attempt to give a mathematical description of the instantaneous angular velocity of a rigid body with a fixed point \( o \) (a top). Consider an orthonormal frame \( \left\{ {{\mathbf{e}}_{1},{\mathbf{e}}_{2},{\mathbf{e}}_{3}}\right\} \) at the point \( o \) rigidly attached to the body. It is clear that the p... | Since \( O\left( t\right) \) is an orthogonal matrix, the relation\n\n\[ O\left( t\right) {O}^{ * }\left( t\right) = E \]\n\nholds at any time \( t \), where \( {O}^{ * }\left( t\right) \) is the transpose of \( O\left( t\right) \) and \( E \) is the identity matrix.\n\nWe remark that the product \( A \cdot B \) of mat... | Yes |
Consider the familiar situation when \( X = Y = \mathbb{R} \), and hence \( f : U \rightarrow \mathbb{R} \) is a real-valued function of a real argument. Since any linear mapping \( A \in \mathcal{L}\left( {\mathbb{R};\mathbb{R}}\right) \) reduces to multiplication by some number \( a \in \mathbb{R} \), that is, \( {Ah... | Next, since\n\n\[ \left( {{f}^{\prime }\left( {x + \delta }\right) - {f}^{\prime }\left( x\right) }\right) h = {f}^{\prime }\left( {x + \delta }\right) h - {f}^{\prime }\left( x\right) h = \]\n\n\[ = a\left( {x + \delta }\right) h - a\left( x\right) h = \left( {a\left( {x + \delta }\right) - a\left( x\right) }\right) h... | Yes |
If the mapping \( f \) is differentiable at \( x \in U \), its differential \( \mathrm{d}f\left( x\right) \) at that point is an element of the space \( \mathcal{L}\left( {{X}_{1} \times \cdots \times {X}_{m} = X;Y}\right) \). | The action of \( \mathrm{d}f\left( x\right) \) on a vector \( h = \left( {{h}_{1},\ldots ,{h}_{m}}\right) \), by formula (10.45), can be represented as\n\n\[ \mathrm{d}f\left( x\right) h = {\partial }_{1}f\left( x\right) {h}_{1} + \cdots + {\partial }_{m}f\left( x\right) {h}_{m}, \]\n\nwhere \( {\partial }_{i}f\left( x... | Yes |
Proposition 1. If \( K \) is a convex compact set in a normed space \( X \) and \( f \in {C}^{\left( 1\right) }\left( {K, Y}\right) \), where \( Y \) is also a normed space, then the mapping \( f : K \rightarrow Y \) satisfies a Lipschitz condition on \( K \), that is, there exists a constant \( M > 0 \) such that the ... | Proof. By hypothesis \( {f}^{\prime } : K \rightarrow \mathcal{L}\left( {X;Y}\right) \) is a continuous mapping of the compact set \( K \) into the metric space \( \mathcal{L}\left( {X;Y}\right) \) . Since the norm is a continuous function on a normed space with its natural metric, the mapping \( x \mapsto \begin{Vmatr... | Yes |
Proposition 2. Under the hypotheses of Proposition 1 there exists a nonnegative function \( \omega \left( \delta \right) \) tending to 0 as \( \delta \rightarrow + 0 \) such that\n\n\[ \left| {f\left( {x + h}\right) - f\left( x\right) - {f}^{\prime }\left( x\right) h}\right| \leq \omega \left( \delta \right) \left| h\r... | Proof. By the corollary to the finite-increment theorem we can write\n\n\[ \left| {f\left( {x + h}\right) - f\left( x\right) - {f}^{\prime }\left( x\right) h}\right| \leq \mathop{\sup }\limits_{{0 < \theta < 1}}\begin{Vmatrix}{{f}^{\prime }\left( {x + {\theta h}}\right) - {f}^{\prime }\left( x\right) }\end{Vmatrix}\lef... | Yes |
Theorem 2. Let \( U \) be a neighborhood of the point \( x \) in a normed space \( X = {X}_{1} \times \cdots \times {X}_{m} \), which is the direct product of the normed spaces \( {X}_{1} \times \cdots \times {X}_{m} \), and let \( f : U \rightarrow Y \) be a mapping of \( U \) into a normed space \( Y \) . If the mapp... | Proof. To simplify the writing we carry out the proof for the case \( m = 2 \) . We verify immediately that the mapping\n\n\[ {Lh} = {\partial }_{1}f\left( x\right) {h}_{1} + {\partial }_{2}f\left( x\right) {h}_{2}, \]\n\nwhich is linear in \( h = \left( {{h}_{1},{h}_{2}}\right) \), is the total differential of \( f \)... | Yes |
Theorem 1. If a mapping \( f : U \rightarrow Y \) from a neighborhood \( U = U\\left( x\\right) \) of a point \( x \) in a normed space \( X \) into a normed space \( Y \) has derivatives up to order \( n - 1 \) inclusive in \( U \) and has an \( n \) -th order derivative \( {f}^{\\left( n\\right) }\\left( x\\right) \)... | Proof. We prove Taylor's formula by induction.\n\nFor \( n = 1 \) it is true by definition of \( {f}^{\\prime }\\left( x\\right) \) .\n\nAssume formula (10.70) is true for some \( \\left( {n - 1}\\right) \\in \\mathbb{N} \) .\n\nThen by the mean-value theorem, formula (10.69) of Sect. 10.5, and the induction hypothesis... | Yes |
Theorem 2. Let \( f : U \rightarrow \mathbb{R} \) be a real-valued function defined on an open set \( U \) in a normed space \( X \) and having continuous derivatives up to order \( k - 1 \geq 1 \) inclusive in a neighborhood of a point \( x \in U \) and a derivative \( {f}^{\left( k\right) }\left( x\right) \) of order... | Proof. For the proof we consider the Taylor expansion (10.70) of \( f \) in a neighborhood of \( x \) . The assumptions enable us to write\n\n\[ f\left( {x + h}\right) - f\left( x\right) = \frac{1}{k!}{f}^{\left( k\right) }\left( x\right) {h}^{k} + \alpha \left( h\right) {\left| h\right| }^{k}, \]\n\nwhere \( \alpha \l... | Yes |
We shall show that the functional (10.72) is a differentiable mapping and find its differential. | We remark that the function (10.72) can be regarded as the composition of the mappings\n\n\[ \n{F}_{1} : {C}^{\left( 1\right) }\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \rightarrow C\left( {\left\lbrack {a, b}\right\rbrack ,\mathbb{R}}\right) \n\]\n\n\[ \n{F}_{1}\left( f\right) \left( x\right) = L\le... | No |
Among all the curves in a plane joining two fixed points, find the curve that has minimal length. | We shall assume that a fixed Cartesian coordinate system has been chosen in the plane, in which the two points are, for example, \( \\left( {0,0}\\right) \) and \( \\left( {1,0}\\right) \) . We confine ourselves to just the curves that are the graphs of functions \( f \\in \) \( {C}^{\\left( 1\\right) }\\left( {\\left\... | Yes |
Lemma 1. The measure of an interval in \( {\mathbb{R}}^{n} \) has the following properties.\n\na) It is homogeneous, that is, if \( \lambda {I}_{a, b} \mathrel{\text{:=}} {I}_{{\lambda a},{\lambda b}} \), where \( \lambda \geq 0 \), then\n\n\[ \left| {\lambda {I}_{a, b}}\right| = {\lambda }^{n}\left| {I}_{a, b}\right| ... | All these assertions follow easily from Definitions 1 and 2. | No |
Proposition 1. \( f \in \mathcal{R}\left( I\right) \Rightarrow f \) is bounded on \( I \) . | Proof. Let \( P \) be an arbitrary partition of the interval \( I \) . If the function \( f \) is unbounded on \( I \), then it must be unbounded on some interval \( {I}_{{i}_{0}} \) of the partition \( P \) . If \( \left( {P,{\xi }^{\prime }}\right) \) and \( \left( {P,{\xi }^{\prime \prime }}\right) \) are partitions... | Yes |
Let \( f : I \rightarrow \mathbb{R} \) be a continuous real-valued function defined on an \( \left( {n - 1}\right) \) -dimensional interval \( I \subset {\mathbb{R}}^{n - 1} \). We shall show that its graph in \( {\mathbb{R}}^{n} \) is a set of \( n \) -dimensional measure zero. | Proof. Since the function \( f \) is uniformly continuous on \( I \), for \( \varepsilon > 0 \) we find \( \delta > 0 \) such that \( \left| {f\left( {x}_{1}\right) - f\left( {x}_{2}\right) }\right| < \varepsilon \) for any two points \( {x}_{1},{x}_{2} \in I \) such that \( \left| {{x}_{1} - {x}_{2}}\right| < \delta \... | Yes |
Lemma 3. a) The class of sets of measure zero remains the same whether the intervals covering the set \( E \) in Definition 9, that is, \( E \subset \mathop{\bigcup }\limits_{i}{I}_{i} \), are interpreted as an ordinary system of intervals \( \left\{ {I}_{i}\right\} \), or in a stricter sense, requiring that each point... | Proof. a) If \( \left\{ {I}_{i}\right\} \) is a covering of \( E \) (that is, \( E \subset \mathop{\bigcup }\limits_{i}{I}_{i} \) and \( \mathop{\sum }\limits_{i}\left| {I}_{i}\right| < \varepsilon \) ), then, replacing each \( {I}_{i} \) by a dilation of it from its center, which we denote \( {\widetilde{I}}_{i} \), w... | Yes |
Lemma 4. If the relation \( \omega \left( {f;x}\right) \leq {\omega }_{0} \) holds at each point of a compact set \( K \) for the function \( f : K \rightarrow \mathbb{R} \), then for every \( \varepsilon > 0 \) there exists \( \delta > 0 \) such that \( \omega \left( {f;{U}_{K}^{\delta }\left( x\right) }\right) < {\om... | When \( {\omega }_{0} = 0 \), this assertion becomes Cantor’s theorem on uniform continuity of a function that is continuous on a compact set. The proof of Lemma 4 is a verbatim repetition of the proof of Cantor's theorem (Subsect. 6.2.2) and therefore we do not take the time to give it here. | No |
Lemma 5. The following relations hold between the Darboux sums of a function \( f : I \rightarrow \mathbb{R} \) :\n\n\[ \n\text{a)}s\left( {f, P}\right) = \mathop{\inf }\limits_{\xi }\sigma \left( {f, P,\xi }\right) \leq \sigma \left( {f, P,\xi }\right) \leq \mathop{\sup }\limits_{\xi }\sigma \left( {f, P,\xi }\right) ... | Proof. Relations \( a \) ) and \( b \) ) follow immediately from Definitions 6 and 10, taking account, of course, of the definition of the greatest lower bound and least upper bound of a set of numbers. | No |
Theorem 2. (Darboux). For any bounded function \( f : I \rightarrow \mathbb{R} \) ,\n\n\[ \left( {\exists \mathop{\lim }\limits_{{\lambda \left( P\right) \rightarrow 0}}s\left( {f, P}\right) }\right) \land \left( {\mathop{\lim }\limits_{{\lambda \left( P\right) \rightarrow 0}}s\left( {f, P}\right) = \underline{\mathcal... | Proof. If we compare these assertions with Definition 11, it becomes clear that in essence all we have to prove is that the limits exist. We shall verify this for the lower Darboux sums.\n\nFix \( \varepsilon > 0 \) and a partition \( {P}_{\varepsilon } \) of the interval \( I \) for which \( s\left( {f;{P}_{\varepsilo... | Yes |
Theorem 3. (The Darboux criterion). A real-valued function \( f : I \rightarrow \mathbb{R} \) defined on an interval \( I \subset {\mathbb{R}}^{n} \) is integrable over that interval if and only if it is bounded on \( I \) and its upper and lower Darboux integrals are equal. Thus,\n\n\[ f \in \mathcal{R}\left( I\right)... | Proof. Necessity. If \( f \in \mathcal{R}\left( I\right) \), then by Proposition 1 the function \( f \) is bounded on \( I \) . It follows from Definition 7 of the integral, Definition 11 of the quantities \( \underline{\mathcal{J}} \) and \( \overline{\mathcal{J}} \), and part \( a \) ) of Lemma 5 that in this case \(... | Yes |
Suppose the functions \( {\varphi }_{i} : I \rightarrow \mathbb{R}, i = 1,2 \), defined on an \( \left( {n - 1}\right) \) - dimensional interval \( I \subset {\mathbb{R}}^{n} \) are such that \( {\varphi }_{1}\left( x\right) < {\varphi }_{2}\left( x\right) \) at every point \( x \in I \) . If these functions are contin... | We recall that the boundary \( \partial E \) of a set \( E \subset {\mathbb{R}}^{n} \) consists of the points \( x \) such that every neighborhood of \( x \) contains both points of \( E \) and points of the complement of \( E \) in \( {\mathbb{R}}^{n} \) . Hence we have the following lemma. | No |
Lemma 3. If \( {I}_{1} \) and \( {I}_{2} \) are two intervals, both containing the set \( E \), then the integrals\n\n\[{\int }_{{I}_{1}}f{\chi }_{E}\left( x\right) \mathrm{d}x\text{ and }{\int }_{{I}_{2}}f{\chi }_{E}\left( x\right) \mathrm{d}x\]\n\neither both exist or both fail to exist, and in the first case their v... | Proof. Consider the interval \( I = {I}_{1} \cap {I}_{2} \) . By hypothesis \( I \supset E \) . The points of discontinuity of \( f{\chi }_{E} \) are either points of discontinuity of \( f \) on \( E \), or the result of discontinuities of \( {\chi }_{E} \), in which case they lie on \( \partial E \) . In any case, all... | Yes |
Theorem 1. A function \( f : E \rightarrow \mathbb{R} \) is integrable over an admissible set if and only if it is bounded and continuous at almost all points of \( E \) . | Proof. Compared with \( f \), the function \( f{\chi }_{E} \) may have additional points of discontinuity only on the boundary \( \partial E \) of \( E \), which by hypothesis is a set of measure zero. | No |
Proposition 1. a) The set \( \\mathcal{R}\\left( E\\right) \) of functions that are Riemann-integrable over a bounded set \( E \\subset {\\mathbb{R}}^{n} \) is a vector space with respect to the standard operations of addition of functions and multiplication by constants. | Proof. Noting that the union of two sets of measure zero is also a set of measure zero, we see that assertion \( a \) ) follows immediately from the definition of the integral and the Lebesgue criterion for existence of the integral of a function over an interval. | No |
Proposition 2. Let \( {E}_{1} \) and \( {E}_{2} \) be admissible sets in \( {\mathbb{R}}^{n} \) and \( f \) a function defined on \( {E}_{1} \cup {E}_{2} \) . a) The following relations hold: \[ \left( {\exists {\int }_{{E}_{1} \cup {E}_{2}}f\left( x\right) \mathrm{d}x}\right) \Leftrightarrow \left( {\exists {\int }_{{... | Proof. Assertion \( a \) ) follows from Lebesgue’s criterion for existence of the Riemann integral over an admissible set (Theorem 1 of Sect. 11.2). Here it is only necessary to recall that the union and intersection of admissible sets are also admissible sets (Lemma 2 of Sect. 11.2). | No |
Proposition 3. If \( f \in \mathcal{R}\left( E\right) \), then \( \left| f\right| \in \mathcal{R}\left( E\right) \), and the inequality\n\n\[ \left| {{\int }_{E}f\left( x\right) \mathrm{d}x}\right| \leq {\int }_{E}\left| f\right| \left( x\right) \mathrm{d}x \]\n\nholds. | Proof. The relation \( \left| f\right| \in \mathcal{R}\left( E\right) \) follows from the definition of the integral over a set and the Lebesgue criterion for integrability of a function over an interval.\n\nThe inequality now follows from the corresponding inequality for Riemann sums and passage to the limit. | No |
Proposition 4. The following implication holds for a function \( f : E \rightarrow \mathbb{R} \) :\n\n\[ \n\left( {f \in \mathcal{R}\left( E\right) }\right) \land \left( {\forall x \in E\left( {f\left( x\right) \geq 0}\right) }\right) \Rightarrow {\int }_{E}f\left( x\right) \mathrm{d}x \geq 0.\n\] | Proof. Indeed, if \( f\left( x\right) \geq 0 \) on \( E \), then \( f{\chi }_{E}\left( x\right) \geq 0 \) in \( {\mathbb{R}}^{n} \) . Then, by definition,\n\n\[ \n{\int }_{E}f\left( x\right) \mathrm{d}x = {\int }_{I \supset E}f{\chi }_{E}\left( x\right) \mathrm{d}x.\n\]\n\nThis last integral exists by hypothesis. But i... | Yes |
Corollary 2. If \( f \in \mathcal{R}\left( E\right) \) and the inequalities \( m \leq f\left( x\right) \leq M \) hold at every point of the admissible set \( E \), then | \[ {m\mu }\left( E\right) \leq {\int }_{E}f\left( x\right) \mathrm{d}x \leq {M\mu }\left( E\right) . \] | Yes |
Corollary 3. If \( f \in \mathcal{R}\left( E\right), m = \mathop{\inf }\limits_{{x \in E}}f\left( x\right) \), and \( M = \mathop{\sup }\limits_{{x \in E}}f\left( x\right) \), then there is a number \( \theta \in \left\lbrack {m, M}\right\rbrack \) such that | \[ {\int }_{E}f\left( x\right) \mathrm{d}x = {\theta \mu }\left( E\right) \] | Yes |
Corollary 5. If in addition to the hypotheses of Corollary 2 the function \( g \in \mathcal{R}\left( E\right) \) is nonnegative on \( E \), then\n\n\[ m{\int }_{E}g\left( x\right) \mathrm{d}x \leq {\int }_{E}{fg}\left( x\right) \mathrm{d}x \leq M{\int }_{E}g\left( x\right) \mathrm{d}x. \] | Proof. Corollary 5 follows from the inequalities \( {mg}\left( x\right) \leq f\left( x\right) g\left( x\right) \leq {Mg}\left( x\right) \) taking account of the linearity of the integral and Corollary 1. It can also be proved directly by passing from integrals over \( E \) to the corresponding integrals over an interva... | No |
Corollary 1. If \( f \in \mathcal{R}\left( {X \times Y}\right) \), then for almost all \( x \in X \) (in the sense of Lebesgue) the integral \( {\int }_{Y}f\left( {x, y}\right) \mathrm{d}y \) exists, and for almost all \( y \in Y \) the integral \( {\int }_{X}f\left( {x, y}\right) \mathrm{d}x \) exists. | Proof. By the theorem just proved,\n\n\[{\int }_{X}\left( {{\int }_{Y}f\left( {x, y}\right) \mathrm{d}y - {\int }_{\bar{Y}}f\left( {x, y}\right) \mathrm{d}y}\right) \mathrm{d}x = 0.\]\n\nBut the difference of the upper and lower integrals in parentheses is nonnegative. We can therefore conclude by the lemma of Sect. 11... | Yes |
Corollary 2. If the interval \( I \subset {\mathbb{R}}^{n} \) is the direct product of the closed intervals \( {I}_{i} = \left\lbrack {{a}^{i},{b}^{i}}\right\rbrack, i = 1,\ldots, n \), then | \[ {\int }_{I}f\left( x\right) \mathrm{d}x = {\int }_{{a}^{n}}^{{b}^{n}}\mathrm{\;d}{x}^{n}{\int }_{{a}^{n - 1}}^{{b}^{n - 1}}\mathrm{\;d}{x}^{n - 1}\cdots {\int }_{{a}^{1}}^{{b}^{1}}f\left( {{x}^{1},{x}^{2},\ldots ,{x}^{n}}\right) \mathrm{d}{x}^{1}. \] Proof. This formula obviously results from repeated application of... | No |
Let \( f\left( {x, y, z}\right) = z\sin \left( {x + y}\right) \) . We shall find the integral of the restriction of this function to the interval \( I \subset {\mathbb{R}}^{3} \) defined by the relations \( 0 \leq x \leq \pi ,\left| y\right| \leq \pi /2,0 \leq z \leq 1. \) | By Corollary 2\n\n\[ \n{\iiint }_{I}f\left( {x, y, z}\right) \mathrm{d}x\mathrm{\;d}y\mathrm{\;d}z = {\int }_{0}^{1}\mathrm{\;d}z{\int }_{-\pi /2}^{\pi /2}\mathrm{\;d}y{\int }_{0}^{\pi }z\sin \left( {x + y}\right) \mathrm{d}x = \n\]\n\n\[ \n= {\int }_{0}^{1}\mathrm{\;d}z{\int }_{-\pi /2}^{\pi /2}\left( {-{\left. z\cos ... | Yes |
Corollary 3. Let \( D \) be a bounded set in \( {\mathbb{R}}^{n - 1} \) and \( E = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{n} \mid (x \in }\right. \) \( D) \land \left( {{\varphi }_{1}\left( x\right) \leq y \leq {\varphi }_{2}\left( x\right) }\right) \} \) . If \( f \in \mathcal{R}\left( E\right) \), then\n\n\[... | Proof. Let \( {E}_{x} = \left\{ {y \in \mathbb{R} \mid {\varphi }_{1}\left( x\right) \leq y \leq {\varphi }_{2}\left( x\right) }\right\} \) if \( x \in D \) and \( {E}_{x} = \varnothing \) if \( x \notin D \) . We remark that \( {\chi }_{E}\left( {x, y}\right) = {\chi }_{D}\left( x\right) \cdot {\chi }_{{E}_{x}}\left( ... | Yes |
For the disk \( E = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} \mid {x}^{2} + {y}^{2} \leq {r}^{2}}\right\} \) we obtain by this formula | \[\n\mu \left( E\right) = {\int }_{-r}^{r}\left( {\sqrt{{r}^{2} - {y}^{2}} - \left( {-\sqrt{{r}^{2} - {y}^{2}}}\right) }\right) \mathrm{d}y = 2{\int }_{-r}^{r}\sqrt{{r}^{2} - {y}^{2}}\mathrm{\;d}y =\n\]\n\[= 4{\int }_{0}^{r}\sqrt{{r}^{2} - {y}^{2}}\mathrm{\;d}y = 4{\int }_{0}^{\pi /2}r\cos \varphi \mathrm{d}\left( {r\s... | Yes |
Corollary 5. Let \( E \) be a measurable set contained in the interval \( I \subset {\mathbb{R}}^{n} \) . Represent \( I \) as the direct product \( I = {I}_{x} \times {I}_{y} \) of the \( \left( {n - 1}\right) \) -dimensional interval \( {I}_{x} \) and the closed interval \( {I}_{y} \) . Then for almost all values \( ... | Proof. Corollary 5 follows immediately from the theorem and Corollary 1, if we set \( f = {\chi }_{E} \) in both of them and take account of the relation \( {\chi }_{E}\left( {x, y}\right) = \) \( {\chi }_{{E}_{y}}\left( x\right) \) . | Yes |
Corollary 6. (Cavalieri’s principle.) \( {}^{6} \) Let \( A \) and \( B \) be two solids in \( {\mathbb{R}}^{3} \) having volume (that is, Jordan-measurable). Let \( {A}_{c} = \{ \left( {x, y, z}\right) \in A \mid z = c\} \) and \( {B}_{c} = \{ \left( {x, y, z}\right) \in B \mid z = c\} \) be the sections of the solids... | It is clear that Cavalieri’s principle can be stated for spaces \( {\mathbb{R}}^{n} \) of any dimension. | No |
Using formula (11.3), let us compute the volume \( {V}_{n} \) of the ball \( B = \\left\\{ {x \\in {\\mathbb{R}}^{n}\\left| \\right| x \\mid \\leq r}\\right\\} \) of radius \( r \) in the Euclidean space \( {\\mathbb{R}}^{n} \) . | It is obvious that \( {V}_{1} = 2 \) . In Example 2 we found that \( {V}_{2} = \\pi {r}^{2} \) . We shall show that \( {V}_{n} = {c}_{n}{r}^{n} \), where \( {c}_{n} \) is a constant (which we shall compute below). Let us choose some diameter \( \\left\\lbrack {-r, r}\\right\\rbrack \) of the ball and for each point \( ... | Yes |
Theorem 1. If \( \varphi : {D}_{t} \rightarrow {D}_{x} \) is a diffeomorphism of a bounded open set \( {D}_{t} \subset {\mathbb{R}}^{n} \) onto a set \( {D}_{x} = \varphi \left( {D}_{t}\right) \subset {\mathbb{R}}^{n} \) of the same type, \( f \in \mathcal{R}\left( {D}_{x}\right) \) , and \( \operatorname{supp}f \) is ... | \[ {\int }_{{D}_{x} = \varphi \left( {D}_{t}\right) }f\left( x\right) \mathrm{d}x = {\int }_{{D}_{t}}f \circ \varphi \left( t\right) \left| {\det {\varphi }^{\prime }\left( t\right) }\right| \mathrm{d}t. \] | Yes |
Lemma 2. a) If \( \varphi : {I}_{t} \rightarrow {I}_{x} \) is a diffeomorphism of a closed interval \( {I}_{t} \subset {\mathbb{R}}^{1} \) onto a closed interval \( {I}_{x} \subset {\mathbb{R}}^{1} \) and \( f \in \mathcal{R}\left( {I}_{x}\right) \), then \( f \circ \varphi \cdot \left| {\varphi }^{\prime }\right| \in ... | Proof. Although we essentially already know assertion \( a \) ) of this lemma, we shall use the Lebesgue criterion for the existence of an integral, which is now at our disposal, to give a short proof here that is independent of the proof given in Part 1.\n\nSince \( f \in \mathcal{R}\left( {I}_{x}\right) \) and \( \va... | Yes |
Lemma 4. If \( {D}_{\tau }\overset{\psi }{ \rightarrow }{D}_{t}\overset{\varphi }{ \rightarrow }{D}_{x} \) are two diffeomorphisms for each of which formula (11.10) for change of variable in the integral holds, then it holds also for the composition \( \varphi \circ \psi : {D}_{\tau } \rightarrow {D}_{x} \) of these ma... | Proof. It suffices to recall that \( {\left( \varphi \circ \psi \right) }^{\prime } = {\varphi }^{\prime } \circ {\psi }^{\prime } \) and that \( \det {\left( \varphi \circ \psi \right) }^{\prime }\left( \tau \right) = \) \( \det {\varphi }^{\prime }\left( t\right) \det {\psi }^{\prime }\left( \tau \right) \), where \(... | Yes |
Proposition 1. Let \( \varphi : {D}_{t} \rightarrow {D}_{x} \) be a diffeomorphism of a bounded open set \( {D}_{t} \subset {\mathbb{R}}^{n} \) onto a set \( {D}_{x} \subset {\mathbb{R}}^{n} \) of the same type; let \( {E}_{t} \) and \( {E}_{x} \) be subsets of \( {D}_{t} \) and \( {D}_{x} \) respectively and such that... | Proof. Indeed,\n\n\[{\int }_{{E}_{x}}f\left( x\right) \mathrm{d}x = {\int }_{{D}_{x}}\left( {f{\chi }_{{E}_{x}}}\right) \left( x\right) \mathrm{d}x = {\int }_{{D}_{t}}\left( {\left( {\left( {f{\chi }_{{E}_{x}}}\right) \circ \varphi }\right) \left| {\det {\varphi }^{\prime }}\right| }\right) \left( t\right) \mathrm{d}t ... | Yes |
Proposition 2. The value of the integral of a function \( f \) over a set \( E \subset {\mathbb{R}}^{n} \) is independent of the choice of Cartesian coordinate system in \( {\mathbb{R}}^{n} \). | Proof. In fact the transition from one Cartesian coordinate system in \( {\mathbb{R}}^{n} \) to another Cartesian system has a Jacobian constantly equal to 1 in absolute value. By Proposition 1 this implies the equality\n\n\[ \n{\int }_{{E}_{x}}f\left( x\right) \mathrm{d}x = {\int }_{{E}_{t}}\left( {f \circ \varphi }\r... | Yes |
Theorem 2. Let \( \varphi : {D}_{t} \rightarrow {D}_{x} \) be a mapping of a (Jordan) measurable set \( {D}_{t} \subset {\mathbb{R}}_{t}^{n} \) onto a set \( {D}_{x} \subset {\mathbb{R}}_{x}^{n} \) of the same type. Suppose that there are subsets \( {S}_{t} \) and \( {S}_{x} \) of \( {D}_{t} \) and \( {D}_{x} \) respec... | Proof. By Lebesgue’s criterion the function \( f \) can have discontinuities in \( {D}_{x} \) and hence also in \( {D}_{x} \smallsetminus {S}_{x} \) only on a set of measure zero. By Lemma 1, the image of this set of discontinuities under the mapping \( {\varphi }^{-1} : {D}_{x} \smallsetminus {S}_{x} \rightarrow {D}_{... | Yes |
Proposition 1. If a function \( f : E \rightarrow \mathbb{R} \) is nonnegative and the limit in Definition 2 exists for even one exhaustion \( \left\{ {E}_{n}\right\} \) of the set \( E \), then the improper integral of \( f \) over \( E \) converges. | Proof. Let \( \left\{ {E}_{k}^{\prime }\right\} \) be a second exhaustion of \( E \) into elements on which \( f \) is integrable. The sets \( {E}_{n}^{k} \mathrel{\text{:=}} {E}_{k}^{\prime } \cap {E}_{n}, n = 1,2,\ldots \) form an exhaustion of the set \( {E}_{k}^{\prime } \), and so it follows from part \( b \) ) of... | Yes |
Let us find the improper integral \( {\iint }_{{\mathbb{R}}^{2}}{e}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mathrm{d}x\mathrm{\;d}y \) . | We shall exhaust the plane \( {\mathbb{R}}^{2} \) by the sequence of disks \( {E}_{n} = \{ \left( {x, y}\right) \in \) \( \left. {{\mathbb{R}}^{2} \mid {x}^{2} + {y}^{2} < {n}^{2}}\right\} \) . After passing to polar coordinates we find easily that\n\n\[ \n{\iint }_{{E}_{n}}{e}^{-\left( {{x}^{2} + {y}^{2}}\right) }\mat... | Yes |
Proposition 2. Let \( f \) and \( g \) be functions defined on the set \( E \) and integrable over exactly the same measurable subsets of it, and suppose \( \left| {f\left( x\right) }\right| \leq g\left( x\right) \) on E. If the improper integral \( {\int }_{E}g\left( x\right) \mathrm{d}x \) converges, then the integra... | Proof. Let \( \left\{ {E}_{n}\right\} \) be an exhaustion of \( E \) on whose elements both \( g \) and \( f \) are integrable. It follows from the Lebesgue criterion that the function \( \left| f\right| \) is integrable on the sets \( {E}_{n}, n \in \mathbb{N} \), and so we can write \[ {\int }_{{E}_{n + k}}\left| f\r... | Yes |
In the deleted \( n \) -dimensional ball of radius \( 1, B \subset {\mathbb{R}}^{n} \) with its center at 0 removed, consider the function \( 1/{r}^{\alpha } \), where \( r = d\left( {0, x}\right) \) is the distance from the point \( x \in B \smallsetminus 0 \) to the point 0 . Let us determine the values of \( \alpha ... | To do this we construct an exhaustion of the domain by the annular regions \( B\left( \varepsilon \right) = \{ x \in B \mid \varepsilon < d\left( {0, x}\right) < 1\} . \)\n\nPassing to polar coordinates with center at 0 , by Fubini's theorem, we obtain\n\n\[ \n{\int }_{B\left( \varepsilon \right) }\frac{\mathrm{d}x}{{r... | Yes |
Let \( I = \left\{ {x \in {\mathbb{R}}^{n} \mid 0 \leq {x}^{i} \leq 1, i = 1,\ldots, n}\right\} \) be the \( n \) -dimensional cube and \( {I}_{k} \) the \( k \) -dimensional face of it defined by the conditions \( {x}^{k + 1} = \) \( \cdots = {x}^{n} = 0 \) . On the set \( I \smallsetminus {I}_{k} \) we consider the f... | We remark that if \( x = \left( {{x}^{1},\ldots ,{x}^{k},{x}^{k + 1},\ldots ,{x}^{n}}\right) \) then\n\n\[ d\left( x\right) = \sqrt{{\left( {x}^{k + 1}\right) }^{2} + \cdots + {\left( {x}^{n}\right) }^{2}}.\]\n\nLet \( I\left( \varepsilon \right) \) be the cube \( I \) from which the \( \varepsilon \) -neighborhood of ... | Yes |
Let the function \( f : {\mathbb{R}}_{ + } \rightarrow \mathbb{R} \) be defined on the set \( {\mathbb{R}}_{ + } \) of nonnegative numbers by the following conditions: \( f\left( x\right) = \frac{{\left( -1\right) }^{n - 1}}{n} \), if \( n - 1 \leq x < n \) , \( n \in \mathbb{N} \) . | Since the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n} \) converges, the integral \( \mathop{\int }\limits_{0}^{A}f\left( x\right) \;\mathrm{d}x \) has a limit as \( A \rightarrow \infty \) equal to the sum of this series. However, this series does not converge absolutely, an... | Yes |
Theorem 1. Let \( \varphi : {D}_{t} \rightarrow {D}_{x} \) be a diffeomorphism of the open set \( {D}_{t} \subset {\mathbb{R}}_{t}^{n} \) onto the set \( {D}_{x} \subset {\mathbb{R}}_{x}^{n} \) of the same type, and let \( f : {D}_{x} \rightarrow \mathbb{R} \) be integrable on all measurable compact subsets of \( {D}_{... | Proof. The open set \( {D}_{t} \subset {\mathbb{R}}_{t}^{n} \) can be exhausted by a sequence of compact sets \( {E}_{t}^{k}, k \in \mathbb{N} \), contained in \( \mathbb{N} \), each of which is the union of a finite number of intervals in \( {\mathbb{R}}_{t}^{n} \) (in this connection, see the beginning of the proof o... | Yes |
Theorem 2. Let \( \varphi : {D}_{t} \rightarrow {D}_{x} \) be a mapping of the open sets \( {D}_{t} \) and \( {D}_{x} \) . Assume that there are subsets \( {S}_{t} \) and \( {S}_{x} \) of measure zero contained in \( {D}_{t} \) and \( {D}_{x} \) respectively such that \( {D}_{t} \smallsetminus {S}_{t} \) and \( {D}_{x}... | Proof. The assertion is a direct corollary of Theorem 1 and Theorem 2 of Sect. 11.5, provided we take account of the fact that when finding an improper integral over an open set one may restrict consideration to exhaustions that consist of measurable compact sets (see Remark 3). | Yes |
Let us compute the integral \( {\iint }_{{x}^{2} + {y}^{2} < 1}\frac{\mathrm{d}x\mathrm{\;d}y}{{\left( 1 - {x}^{2} - {y}^{2}\right) }^{\alpha }} \), which is an improper integral when \( \alpha > 0 \), since the integrand is unbounded in that case in a neighborhood of the disk \( {x}^{2} + {y}^{2} = 1 \) . | Passing to polar coordinates, we obtain from Theorem 2\n\n\[ \n{\iint }_{{x}^{2} + {y}^{2} < 1}\frac{\mathrm{d}x\mathrm{\;d}y}{{\left( 1 - {x}^{2} - {y}^{2}\right) }^{\alpha }} = {\iint }_{\begin{matrix} {0 < \varphi < {2\pi }} \\ {0 < r < 1} \end{matrix}}\frac{r\mathrm{\;d}r\mathrm{\;d}\varphi }{{\left( 1 - {r}^{2}\ri... | Yes |
We recall that if \( {F}^{i} \in {C}^{\left( m\right) }\left( {{\mathbb{R}}^{n},\mathbb{R}}\right), i = 1,\ldots, n - k \), is a set of smooth functions such that the system of equations\n\n\[ \left\{ \begin{matrix} {F}^{1}\left( {{x}^{1},\ldots ,{x}^{k},{x}^{k + 1},\ldots ,{x}^{n}}\right) = 0, \\ \ldots \ldots \ldots ... | Proof. We shall verify that if \( S \neq \varnothing \), then \( S \) does indeed satisfy Definition 4. This follows from the implicit function theorem, which says that in some neighborhood of each point \( {x}_{0} \in S \) the system (12.2) is equivalent, up to a relabeling of the variables, to a system\n\n\[ \left\{ ... | Yes |
In particular, the sphere defined in \( {\mathbb{R}}^{n} \) by the equation\n\n\[ \n{\left( {x}^{1}\right) }^{2} + \cdots + {\left( {x}^{n}\right) }^{2} = {r}^{2}\;\left( {r > 0}\right) \n\]\n\nis an \( \left( {n - 1}\right) \) -dimensional smooth surface in \( {\mathbb{R}}^{n} \) since the set \( S \) of solutions of ... | When \( n = 2 \), we obtain the circle in \( {\mathbb{R}}^{2} \) given by\n\n\[ \n{\left( {x}^{1}\right) }^{2} + {\left( {x}^{2}\right) }^{2} = {r}^{2} \n\]\n\nwhich can easily be parametrized locally by the polar angle \( \theta \) using the polar\n\ncoordinates\n\[ \n\left\{ \begin{array}{l} {x}^{1} = r\cos \theta \\... | No |
The cylinder\n\n\[ \n{\left( {x}^{1}\right) }^{2} + \cdots + {\left( {x}^{k}\right) }^{2} = {r}^{2}\;\left( {r > 0}\right) , \n\]\n\nfor \( k < n \) is an \( \left( {n - 1}\right) \) -dimensional surface in \( {\mathbb{R}}^{n} \) that is the direct product of the \( \left( {k - 1}\right) \) -dimensional sphere in the p... | A local parametrization of this surface can obviously be obtained if we take the first \( k - 1 \) of the \( n - 1 \) parameters \( \left( {{t}^{1},\ldots ,{t}^{n - 1}}\right) \) to be the polar coordinates \( {\theta }_{1},\ldots ,{\theta }_{k - 1} \) of a point of the \( \left( {k - 1}\right) \) -dimensional sphere i... | No |
If we take a curve (a one-dimensional surface) in the plane \( x = 0 \) of \( {\mathbb{R}}^{3} \) endowed with Cartesian coordinates \( \left( {x, y, z}\right) \), and the curve does not intersect the \( z \) -axis, we can rotate the curve about the \( z \) -axis and obtain a 2-dimensional surface. The local coordinate... | In particular, if the original curve is a circle of radius \( a \) with center at \( \left( {b,0,0}\right) \), for \( a < b \) we obtain the two-dimensional torus (Fig. 12.1). Its parametric equation can be represented in the form\n\n\[ \left\{ \begin{array}{l} x = \left( {b + a\cos \psi }\right) \cos \varphi , \\ y = ... | Yes |
Proposition 1. The mutual transitions from one curvilinear coordinate system to another on a smooth surface \( S \subset {\mathbb{R}}^{n} \) are diffeomorphisms of the same degree of smoothness as the charts of the surface. | Proof. In fact, by the proposition in Sect. 12.1, we can regard any chart \( {I}^{k} \rightarrow U \subset S \) locally as the restriction to \( {I}^{k} \cap O\left( t\right) \) of a diffeomorphism \( \mathcal{F} : O\left( t\right) \rightarrow O\left( x\right) \) from some \( n \) -dimensional neighborhood \( O\left( t... | Yes |
Example 2. The Klein bottle is also a nonorientable surface, since it contains a Möbius band. | This last fact can be seen immediately from the construction of the Klein bottle shown in Fig. 12.5. | No |
The two-dimensional torus studied in Example 4 of Sect. 12.1 is also an orientable surface. | Indeed, using the parametric equations of the torus exhibited in Example 4 of Sect. 12.1, one can easily exhibit an orienting atlas for it. | No |
Proposition 1. The boundary of a \( k \) -dimensional surface of class \( {C}^{\left( m\right) } \) is itself a surface of the same smoothness class, and is a surface without boundary having dimension one less than the dimension of the original surface with boundary. | Proof. Indeed, if \( A\left( S\right) = \left\{ \left( {{H}^{k},{\varphi }_{i},{U}_{i}}\right) \right\} \cup \left\{ \left( {{\mathbb{R}}^{k},{\varphi }_{j},{U}_{j}}\right) \right\} \) is an atlas for the surface \( S \) with boundary, then \( A\left( {\partial S}\right) = \left\{ \left( {{\mathbb{R}}^{k - 1},{\left. {... | Yes |
A closed \( n \) -dimensional ball \( {\bar{B}}^{n} \) in \( {\mathbb{R}}^{n} \) is an \( n \) -dimensional surface with boundary. Its boundary \( \partial {\bar{B}}^{n} \) is the \( \left( {n - 1}\right) \) -dimensional sphere (see Figs. 12.8 and 12.9, a). | The ball \( {\bar{B}}^{n} \), which is often called in analogy with the two-dimensional case an \( n \) -dimensional disk, can be homeomorphically mapped to half of an \( n \) -dimensional sphere whose boundary is the equatorial \( \left( {n - 1}\right) \) -dimensional sphere (Fig. 12.9, b). | No |
The closed cube \( {\bar{I}}^{n} \) in \( {\mathbb{R}}^{n} \) can be homeomorphically mapped to the closed ball \( \partial {\bar{B}}^{n} \) along rays emanating from its center. | Consequently \( {\bar{I}}^{n} \) , like \( {\bar{B}}^{n} \) is an \( n \) -dimensional surface with boundary, which in this case is formed by the faces of the cube (Fig. 12.10). We note that on the edges, which are the intersections of the faces, it is obvious that no mapping of the cube onto the ball can be regular (t... | No |
If the Möbius band is obtained by gluing together two opposite sides of a closed rectangle, as described in Example 5 of Sect. 12.1, the result is obviously a surface with boundary in \( {\mathbb{R}}^{3} \), and the boundary is homeomorphic to a circle (to be sure, the circle is knotted in \( {\mathbb{R}}^{3} \)). | Under the other possible gluing of these sides the result is a cylindrical surface whose boundary consists of two circles. This surface is homeomorphic to the usual planar annulus (see Fig. 12.3 and Example 5 of Sect. 12.1). | No |
Example 5. The surface of a three-dimensional cube, as one can easily verify, is an orientable piecewise-smooth surface. | In general, all the piecewise-smooth surfaces exhibited in Example 4 are orientable. | No |
The mapping \( \rbrack 0,{2\pi }\lbrack \ni t \mapsto \left( {R\cos t, R\sin t}\right) \in {\mathbb{R}}^{2} \) is a chart for the arc \( \widetilde{S} \) of the circle \( {x}^{2} + {y}^{2} = {R}^{2} \) obtained by removing the single point \( E = \left( {R,0}\right) \) from that circle. | Since \( E \) is a set of measure zero on \( S \), we can write\n\n\[ \n{V}_{1}\left( S\right) = {V}_{1}\left( \widetilde{S}\right) = {\int }_{0}^{2\pi }\sqrt{{R}^{2}{\sin }^{2}t + {R}^{2}{\cos }^{2}t}\mathrm{\;d}t = {2\pi R}. \n\] | No |
In Example 4 of Sect. 12.1 we exhibited the following parametric representation of the two-dimensional torus \( S \) in \( {\mathbb{R}}^{3} \):\n\n\[ \mathbf{r}\left( {\varphi ,\psi }\right) = \left( {\left( {b + a\cos \psi }\right) \cos \varphi ,\left( {b + a\cos \psi }\right) \sin \varphi, a\sin \psi }\right) .\n\]\n... | Let us carry out the necessary computations:\n\n\[ {\dot{\mathbf{r}}}_{\varphi } = \left( {-\left( {b + a\cos \psi }\right) \sin \varphi ,\left( {b + a\cos \psi }\right) \cos \varphi ,0}\right) ,\n\]\n\n\[ {\dot{\mathbf{r}}}_{\psi } = \left( {-a\sin \psi }\right) \cos \varphi , - a\sin \psi \sin \varphi, a\cos \psi )\;... | Yes |
Example 3. Let \( {\pi }^{i} \in \mathcal{L}\left( {{\mathbb{R}}^{n},\mathbb{R}}\right), i = 1,\ldots, n \), be the projections. More precisely, the linear function \( {\pi }^{i} : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is such that on each vector \( \xi = \left( {{\xi }^{1},\ldots ,{\xi }^{n}}\right) \in {\mathbb{... | \[ {\pi }^{{i}_{1}} \land \cdots \land {\pi }^{{i}_{k}}\left( {{\mathbf{\xi }}_{1},\ldots ,{\mathbf{\xi }}_{k}}\right) = \left| \begin{array}{lll} {\xi }_{1}^{{i}_{1}} & \cdots & {\xi }_{1}^{{i}_{k}} \\ \cdots \cdots \cdots \cdots & & \\ {\xi }_{k}^{{i}_{1}} & \cdots & {\xi }_{k}^{{i}_{k}} \end{array}\right| . \] | Yes |
The Cartesian coordinates of the vector product \( \left\lbrack {{\mathbf{\xi }}_{1},{\mathbf{\xi }}_{2}}\right\rbrack \) of the vectors \( {\mathbf{\xi }}_{1} = \left( {{\xi }_{1}^{1},{\xi }_{1}^{2},{\xi }_{1}^{3}}\right) \) and \( {\mathbf{\xi }}_{2} = \left( {{\xi }_{2}^{1},{\xi }_{2}^{2},{\xi }_{2}^{3}}\right) \) i... | \[ \left\lbrack {{\mathbf{\xi }}_{1},{\mathbf{\xi }}_{2}}\right\rbrack = \left( {\left| \begin{array}{ll} {\xi }_{1}^{2} & {\xi }_{1}^{3} \\ {\xi }_{2}^{2} & {\xi }_{2}^{3} \end{array}\right| ,\left| \begin{array}{ll} {\xi }_{1}^{3} & {\xi }_{1}^{1} \\ {\xi }_{2}^{3} & {\xi }_{2}^{1} \end{array}\right| ,\left| \begin{a... | Yes |
Let \( f : D \rightarrow \mathbb{R} \) be a function that is defined in a domain \( D \subset {\mathbb{R}}^{n} \) and differentiable at \( {x}_{0} \in D \) . As is known, the differential \( \mathrm{d}f\left( {x}_{0}\right) \) of the function at a point is a linear function defined on displacement vectors \( \mathbf{\x... | \[ \mathrm{d}f\left( {x}_{0}\right) \left( \mathbf{\xi }\right) = \frac{\partial f}{\partial {x}^{1}}\left( {x}_{0}\right) {\xi }^{1} + \cdots + \frac{\partial f}{\partial {x}^{n}}\left( {x}_{0}\right) {\xi }^{n} = {D}_{\mathbf{\xi }}f\left( {x}_{0}\right) . \] | Yes |
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