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Example 3. Find the minimum distance from a point to a line in \( {\mathbb{R}}^{3} \) . Let the line be given as the intersection of two planes whose equations are \( \langle a, x\rangle = k \) and \( \langle b, x\rangle = \ell \) . (Here, \( x, a \), and \( b \) belong to \( {\mathbb{R}}^{3} \) .) Let the point be \( ... | This \( H \) is a function of \( \left( {{x}_{1},{x}_{2},{x}_{3},\lambda ,\mu }\right) \) . The five equations to solve are\n\n\[ 2\left( {{x}_{1} - {c}_{1}}\right) + \lambda {a}_{1} + \mu {b}_{1} = 2\left( {{x}_{2} - {c}_{2}}\right) + \lambda {a}_{2} + \mu {b}_{2} = 2\left( {{x}_{3} - {c}_{3}}\right) + \lambda {a}_{3}... | Yes |
Theorem 2. Lagrange Multiplier. Let \( f \) and \( g \) be continuously differentiable real-valued functions on an open set \( \Omega \) in a Banach space. Let \( M = \{ x \in \Omega : g\left( x\right) = 0\} \) . If \( {x}_{0} \) is a local minimum point of \( f \mid M \) and if \( {g}^{\prime }\left( {x}_{0}\right) \n... | Proof. Let \( X \) be the Banach space in question. Select a neighborhood \( U \) of \( {x}_{0} \) such that\n\n\[ x \in U \cap M \Rightarrow f\left( {x}_{0}\right) \leq f\left( x\right) \]\n\nWe can assume \( U \subset \Omega \) . Define \( F : U \rightarrow {\mathbb{R}}^{2} \) by \( F\left( x\right) = \left( {f\left(... | Yes |
Theorem 3. Lagrange Multipliers. Let \( f,{g}_{1},\ldots ,{g}_{n} \) be continuously differentiable real-valued functions defined on an open set \( \Omega \) in a Banach space \( X \) . Let \( M = \left\{ {x \in \Omega : {g}_{1}\left( x\right) = \cdots = {g}_{n}\left( x\right) = 0}\right\} \) . If \( {x}_{0} \) is a lo... | Proof. Select a neighborhood \( U \) of \( {x}_{0} \) such that \( U \subset \Omega \) and such that \( f\left( {x}_{0}\right) \leq \) \( f\left( x\right) \) for all \( x \in U \cap M \) . Define \( F : U \rightarrow {\mathbb{R}}^{n + 1} \) by the equation\n\n\[ F\left( x\right) = \left( {f\left( x\right) ,{g}_{1}\left... | Yes |
Example 4. Let \( A \) be a compact Hermitian operator on a Hilbert space \( X \) . Then \( \parallel A\parallel = \max \{ \left| \lambda \right| : \lambda \in \Lambda \left( A\right) \} \), where \( \Lambda \left( A\right) \) is the set of eigenvalues of \( A \) . | This is proved by Lemma 2, page 92, together with Problem 22, page 101. Then by Lemma 2 in Section 2.3, page 85, we have \( \parallel A\parallel = \sup \{ \left| {\langle {Ax}, x\rangle }\right| : \parallel x\parallel = 1\} \) . Hence we can find an eigenvalue of \( A \) by determining an extremum of \( \langle {Ax}, x... | No |
Theorem 4. If \( A \) is a Hermitian operator on a Hilbert space, then each local constrained minimum or maximum point of \( \langle {Ax}, x\rangle \) on the unit sphere is an eigenvector of \( A \). The value of \( \langle {Ax}, x\rangle \) is the corresponding eigenvalue. | Proof. Use \( F\left( x\right) = \langle {Ax}, x\rangle \) and \( G\left( x\right) = \parallel x{\parallel }^{2} - 1 \). Then\n\n\[ \n{F}^{\prime }\left( x\right) h = 2\langle {Ax}, h\rangle \;{G}^{\prime }\left( x\right) h = 2\langle x, h\rangle \n\]\n\nOur theorem about Lagrange multipliers gives a necessary conditio... | Yes |
Find a function \( y \) in \( {C}^{1}\left\lbrack {a, b}\right\rbrack \), satisfying \( y\left( a\right) = \alpha \) and \( y\left( b\right) = \beta \) , such that the surface of revolution obtained by rotating the graph of \( y \) about the \( x \) -axis has minimum area. | \[ {\int }_{a}^{b}{2\pi y}\left( x\right) {ds} = {2\pi }{\int }_{a}^{b}y\left( x\right) \sqrt{1 + {y}^{\prime }{\left( x\right) }^{2}}{dx} \] | No |
Theorem 1. The Euler Equation. Let \( F \) be a mapping from \( {\mathbb{R}}^{3} \) to \( {\mathbb{R}}^{1} \), possessing piecewise continuous partial derivatives of the second order. In order that a function \( y \) in \( {C}^{1}\left\lbrack {a, b}\right\rbrack \) minimize \( {\int }_{a}^{b}F\left( {x, y\left( x\right... | Proof. Let \( u \in {C}^{1}\left\lbrack {a, b}\right\rbrack \) and \( u\left( a\right) = u\left( b\right) = 0 \) . Assume that \( y \) is a solution of the problem. For all real \( \theta, y + {\theta u} \) is a competing function. Hence\n\n\[ {\left. \frac{d}{d\theta }{\int }_{a}^{b}F\left( x, y\left( x\right) + \thet... | Yes |
Theorem 2. In Theorem 1, if \( {F}_{1} = 0 \), then the Euler equation\nimplies that\n\n(5)\n\n\[ \n{y}^{\prime }\left( x\right) {F}_{3}\left( {x, y\left( x\right) ,{y}^{\prime }\left( x\right) }\right) - F\left( {x, y\left( x\right) ,{y}^{\prime }\left( x\right) }\right) = \text{ constant } \n\] | Proof.\n\n\[ \n\frac{d}{dx}\left\lbrack {{y}^{\prime }{F}_{3} - F}}\right\rbrack = {y}^{\prime \prime }{F}_{3} + {y}^{\prime }\frac{d}{dx}{F}_{3} - {F}_{1} - {F}_{2}{y}^{\prime } - {F}_{3}{y}^{\prime \prime } = {y}^{\prime }\left\lbrack {\frac{d}{dx}{F}_{3} - {F}_{2}}\right\rbrack = 0. \n\] | Yes |
Example 4. This is the Brachistochrone Problem, except that the terminal point is allowed to be anywhere on a given vertical line. Following the previous discussion, we are led to minimize the expression\n\n\[ \n{\int }_{0}^{b}\sqrt{\frac{1 + {y}^{\prime }{\left( x\right) }^{2}}{y\left( x\right) }}{dx} \n\]\n\nsubject ... | Returning now to Example 4, we conclude that \( {F}_{3}\left( {b, y\left( b\right) ,{y}^{\prime }\left( b\right) }\right) = 0 \) . This entails \( {y}^{\prime }\left( b\right) /\sqrt{y\left( b\right) \left\lbrack {1 + {y}^{\prime }{\left( b\right) }^{2}}\right\rbrack } = 0 \), or \( {y}^{\prime }\left( b\right) = 0 \) ... | Yes |
Theorem 3. Any function \( y \) in \( {C}^{2}\left\lbrack {a, b}\right\rbrack \) that minimizes\n\n\[{\int }_{a}^{b}F\left( {x, y\left( x\right) ,{y}^{\prime }\left( x\right) }\right) {dx}\]\n\nsubject to the constraint \( y\left( a\right) = \alpha \) must satisfy the two conditions\n\n(7)\n\n\[\\frac{d}{dx}{F}_{3}\lef... | Proof. This is left as a problem. | No |
Find the function \( y \) that minimizes an integral\n\n\[ \n{\int }_{a}^{b}F\left( {x, y\left( x\right) ,{y}^{\prime }\left( x\right) }\right) {dx} \n\]\n\nsubject to constraints that \( y \) belong to \( {C}^{1}\left\lbrack {a, b}\right\rbrack \) and\n\n\[ \n{\int }_{a}^{b}G\left( {x, y\left( x\right) ,{y}^{\prime }\... | Theorem 4. If \( F \) and \( G \) map \( {\mathbb{R}}^{3} \) to \( \mathbb{R} \) and have continuous partial derivatives of the second order, and if \( y \) is an element of \( {C}^{2}\left\lbrack {a, b}\right\rbrack \) that minimizes \( {\int }_{a}^{b}F\left( {x, y\left( x\right) ,{y}^{\prime }\left( x\right) }\right)... | Yes |
Theorem 4. If \( F \) and \( G \) map \( {\mathbb{R}}^{3} \) to \( \mathbb{R} \) and have continuous partial derivatives of the second order, and if \( y \) is an element of \( {C}^{2}\left\lbrack {a, b}\right\rbrack \) that minimizes \( {\int }_{a}^{b}F\left( {x, y\left( x\right) ,{y}^{\prime }\left( x\right) }\right)... | Proof. As in previous problems of this section, we try to obtain a necessary condition for a solution by perturbing the solution in such a way that the constraints are not violated. Suppose that \( y \) is a solution in \( {C}^{1}\left\lbrack {a, b}\right\rbrack \) . Let \( {\eta }_{1} \) and \( {\eta }_{2} \) be two f... | Yes |
It is required to find the curve of given length \( \ell \) joining the point \( \left( {-1,0}\right) \) to the point \( \left( {1,0}\right) \) that, together with the interval \( \left\lbrack {-1,1}\right\rbrack \) on the horizontal axis, encloses the greatest possible area. | We assume that \( 2 < \ell < \pi \) . Let the curve be given by \( y = y\left( x\right) \), where \( y \) belongs to \( {C}^{1}\left\lbrack {-1,1}\right\rbrack \) . The area to be maximized is then\n\n\[ \n{\int }_{-1}^{1}y\left( x\right) {dx} \n\]\n\nand the constraints are\n\n\[ \n{\int }_{-1}^{1}\sqrt{1 + {y}^{\prim... | Yes |
Among all the plane curves having a prescribed length, find one enclosing the greatest area. We assume a parametric representation \( x = x\\left( t\\right) \) and \( y = y\\left( t\\right) \) with continuously differentiable functions. We can also assume that \( 0 \\leq t \\leq b \) and that \( x\\left( 0\\right) = x\... | \[ \n{\\int }_{\\Gamma }\\left( {{Pdx} + {Qdy}}\\right) = {\\iint }_{R}\\left\\lbrack {{Q}_{1}\\left( {x, y}\\right) - {P}_{2}\\left( {x, y}\\right) }\\right\\rbrack {dxdy} \]\n\nwhere \( R \) is the region enclosed by the curve \( \\Gamma \) and the subscripts denote partial derivatives. A special case of Green's Theo... | Yes |
Example 8. What is the path of a light beam if the velocity of light in the medium is \( c = {\alpha y} \) (where \( \alpha \) is a constant)? | Solution. The path is the graph of a function \( y \) such that\n\n\[ x = \int \frac{k\alpha y}{\sqrt{1 - {k}^{2}{\alpha }^{2}{y}^{2}}}{dy} \]\n\nThe integration produces\n\n\[ x = \frac{-1}{k\alpha }\sqrt{1 - {k}^{2}{\alpha }^{2}{y}^{2}} + A \]\n\nHere \( A \) and \( k \) are constants that can be adjusted so that the... | Yes |
Theorem 5. Suppose that \( {y}_{1},\ldots ,{y}_{n} \) are functions (of \( t \) ) in \( {C}^{2}\left\lbrack {a, b}\right\rbrack \)\n\nthat minimize the integral\n\n\[ \n{\int }_{a}^{b}F\left( {{y}_{1},\ldots ,{y}_{n},{y}_{1}^{\prime },\ldots ,{y}_{n}^{\prime }}\right) {dt} \n\]\n\nsubject to endpoint constraints that p... | Proof. Take functions \( {\eta }_{1},\ldots ,{\eta }_{n} \) in \( {C}^{2}\left\lbrack {a, b}\right\rbrack \) that vanish at the endpoints. The expression\n\n\[ \n{\int }_{a}^{b}F\left( {{y}_{1} + {\theta }_{1}{\eta }_{1},\ldots ,{y}_{n} + {\theta }_{n}{\eta }_{n}}\right) {dt} \n\]\n\nwill have a minimum when \( \left( ... | Yes |
We search for geodesics on a cylinder. Let the surface be the cylinder \( {x}^{2} + {z}^{2} = 1 \), or \( z = {\left( 1 - {x}^{2}\right) }^{1/2} \) (upper-half cylinder). In the general theory, \( F\left( {x, y,{x}^{\prime },{y}^{\prime }}\right) = \sqrt{{x}^{\prime 2} + {y}^{\prime 2} + {\left( {z}_{x}{x}^{\prime } + ... | Then computations show that \[ \frac{\partial F}{\partial x} = \frac{x{x}^{\prime 2}}{{\left( 1 - {x}^{2}\right) }^{2}F}\;\frac{\partial F}{\partial {x}^{\prime }} = \frac{{x}^{\prime }}{\left( {1 - {x}^{2}}\right) F}\;\frac{\partial F}{\partial y} = 0\;\frac{\partial F}{\partial {y}^{\prime }} = \frac{{y}^{\prime }}{F... | Yes |
We wish to minimize the expression \( {\int }_{0}^{b}{\int }_{0}^{a}\left( {{\phi }_{x}^{2} + {\phi }_{y}^{2}}\right) {dxdy} \) subject to the constraints that \( \phi \) be a continuously differentiable function on the rectangle \( R = \{ \left( {x, y}\right) : 0 \leq x \leq a,0 \leq y \leq b\} \), that \( \phi = 0 \)... | A suitable set of base functions for this problem is the doubly indexed sequence\n\n\[ \n{u}_{nm}\left( {x, y}\right) = \frac{2}{\sqrt{ab}}\sin \frac{n\pi x}{a}\sin \frac{m\pi y}{b}\;\left( {n, m \geq 1}\right) \n\]\n\nIt turns out that this is an orthonormal set with respect to the inner product \( \langle u, v\rangle... | Yes |
\[ \left\{ \begin{array}{ll} {u}^{\prime \prime } + a{u}^{\prime } + {bu} = c & 0 < t < 1 \\ u\left( 0\right) = 0\;u\left( 1\right) = 0 & \end{array}\right. \] | \[ \left\{ \begin{array}{l} \frac{{v}_{i + 1} - 2{v}_{i} + {v}_{i - 1}}{{h}^{2}} + {a}_{i}\frac{{v}_{i + 1} - {v}_{i - 1}}{2h} + {b}_{i}{v}_{i} = {c}_{i}\;\left( {1 \leq i \leq n}\right) \\ {v}_{0} = {v}_{n + 1} = 0 \end{array}\right. \] | No |
Lemma 1. If an \( n \times n \) matrix \( A \) is diagonally dominant, then it is nonsingular, and\n\n\[ \n{\begin{Vmatrix}{A}^{-1}\end{Vmatrix}}_{\infty } \leq \mathop{\max }\limits_{i}{\left\{ \left| {a}_{ii}\right| - \mathop{\sum }\limits_{\substack{{j = 1} \\ {j \neq i} }}^{n}\left| {a}_{ij}\right| \right\} }^{-1} ... | Proof. Let \( x \) be any nonzero vector, and let \( y = {Ax} \) . Select \( i \) so that \( \left| {x}_{i}\right| = \) \( \parallel x{\parallel }_{\infty } \) . Then\n\n\[ \n{a}_{ii}{x}_{i} + \mathop{\sum }\limits_{\substack{{j = 1} \\ {j \neq i} }}^{n}{a}_{ij}{x}_{j} = {y}_{i} \n\]\n\n\[ \n\left| {{a}_{ii}{x}_{i}}\ri... | Yes |
Lemma 2. If \( {f}^{\left( 4\right) } \) is continuous on \( \left( {t - h, t + h}\right) \), then\n\n\[ \n{f}^{\prime \prime }\left( t\right) = {h}^{-2}\left\lbrack {f\left( {t + h}\right) - {2f}\left( t\right) + f\left( {t - h}\right) }\right\rbrack - \frac{1}{12}{h}^{2}{f}^{\left( 4\right) }\left( \xi \right) \n\] | Proof. We derive the first formula and leave the second as a problem. By Taylor's Theorem we have\n\n\[ \nf\left( {t + h}\right) = f\left( t\right) + h{f}^{\prime }\left( t\right) + \frac{1}{2}{h}^{2}{f}^{\prime \prime }\left( t\right) + \frac{1}{6}{h}^{3}{f}^{\prime \prime \prime }\left( t\right) + \frac{1}{24}{h}^{4}... | No |
Consider a linear integral equation, such as\n\n\\[ \n{\\int }_{a}^{b}k\\left( {s, t}\\right) x\\left( s\\right) {ds} = v\\left( t\\right) \\;\\left( {a \\leq t \\leq b}\\right) \n\\]\n\nIn this equation, the kernel \\( k \\) and the function \\( v \\) are prescribed. We seek the unknown function \\( x \\). | Suppose that a quadrature formula of the type\n\n\\[ \n{\\int }_{a}^{b}f\\left( s\\right) {ds} \\approx \\mathop{\\sum }\\limits_{{j = 1}}^{n}{c}_{j}f\\left( {s}_{j}\\right) \n\\]\n\nis available. (The points \\( {s}_{j} \\) need not be equally spaced.) Taking \\( t = {s}_{i} \\) in the integral equation, we have\n\n\\... | Yes |
Theorem 1. Contraction Mapping Theorem. If \( F \) is a contraction on a complete metric space \( X \), then \( F \) has a unique fixed point \( \xi \) . The point \( \xi \) is the limit of every sequence generated from an arbitrary point \( x \) by iteration\n\n\[ \left\lbrack {x,{Fx},{F}^{2}x,\ldots }\right\rbrack \;... | Proof. Reverting to the previous notation, we select \( {x}_{0} \) arbitrarily in \( X \) and define \( {x}_{n + 1} = F{x}_{n} \) for \( n = 0,1,2,\ldots \) . We have\n\n\[ d\left( {{x}_{n},{x}_{n - 1}}\right) = d\left( {F{x}_{n - 1}, F{x}_{n - 2}}\right) \leq {\theta d}\left( {{x}_{n - 1},{x}_{n - 2}}\right) \]\n\nThi... | Yes |
Consider the nonlinear Fredholm equation\n\n\[ \nx\left( t\right) = \frac{1}{2}{\int }_{0}^{1}\cos \left( {{stx}\left( s\right) }\right) {ds} \]\n | By the mean value theorem,\n\n\[ \n\left| {\cos \left( {st\xi }\right) - \cos \left( {st\eta }\right) }\right| = \left| {\sin \left( {st\zeta }\right) }\right| \;\left| {{st\xi } - {st\eta }}\right| \leq \left| {\xi - \eta }\right| \]\n\nThus the preceding theory is applicable with \( \theta = \frac{1}{2} \) . If the i... | Yes |
Theorem 3. Let \( S \) be an interval of the form \( S = \left\lbrack {0, b}\right\rbrack \) . Let \( f \) be a continuous map of \( S \times \mathbb{R} \) to \( \mathbb{R} \) . Assume a Lipschitz condition in the second argument:\n\n\[ \left| {f\left( {s,{t}_{1}}\right) - f\left( {s,{t}_{2}}\right) }\right| \leq \lamb... | Proof. We introduce a new norm in \( C\left( S\right) \) by defining\n\n\[ \parallel x{\parallel }_{w} = \mathop{\sup }\limits_{{s \in S}}\left| {x\left( s\right) }\right| {e}^{-{2\lambda s}} \]\n\nThe space \( C\left( S\right) \), accompanied by this norm, is complete. Since the initial-value problem is equivalent to ... | Yes |
Does the following initial value problem have a solution in the space \( C\left\lbrack {0,{10}}\right\rbrack \) ?\n\n\[ \n{x}^{\prime } = \cos \left( {x{e}^{s}}\right) \;x\left( 0\right) = 0 \n\] | This is an illustration of the general theory in which \( f\left( {s, t}\right) = \cos \left( {t{e}^{s}}\right) \) . By the mean value theorem,\n\n\[ \n\left| {f\left( {s,{t}_{1}}\right) - f\left( {s,{t}_{2}}\right) }\right| = \left| {\frac{\partial f}{\partial t}\left( {s,\tau }\right) }\right| \left| {{t}_{1} - {t}_{... | Yes |
If \( f \) is continuous but does not satisfy the Lipschitz condition in Theorem 3, the conclusions of the theorem may fail. For example, the problem \( {x}^{\prime } = {x}^{2/3}, x\left( 0\right) = 0 \) has two solutions, \( x\left( s\right) = 0 \) and \( x\left( s\right) = {s}^{3}/{27} \) . There is no Lipschitz cond... | \[ \left| {{t}_{1}^{2/3} - {t}_{2}^{2/3}}\right| \leq \lambda \left| {{t}_{1} - {t}_{2}}\right| \] (Consider the implications of this inequality when \( {t}_{2} = 0 \) .) | Yes |
\[ {x}^{\prime } = {2t}\left( {1 + x}\right) \;x\left( 0\right) = 0 \] | The formula for the Picard iteration in this example is\n\n\[ {x}_{n + 1}\left( t\right) = {\int }_{0}^{t}{2s}\left( {1 + {x}_{n}\left( s\right) }\right) {ds} = {t}^{2} + {\int }_{0}^{t}{2s}{x}_{n}\left( s\right) {ds} \]\n\nIf \( {x}_{0} = 0 \), then successive computations yield\n\n\[ {x}_{1}\left( t\right) = {t}^{2}\... | Yes |
Theorem 4. Let \( F \) be a mapping of a complete metric space into itself such that for some \( m,{F}^{m} \) is contractive. Then \( F \) has a unique fixed point. It is the limit of every sequence \( \left\lbrack {{F}^{k}x}}\right\rbrack \), for arbitrary \( x \) . | Proof. Since \( {F}^{m} \) is contractive, it has a unique fixed point \( \xi \) by Theorem 1.\n\nThen\n\n\[ \n{F\xi } = F\left( {{F}^{m}\xi }}\right) = {F}^{m + 1}\xi = {F}^{m}\left( {F\xi }}\right) \n\]\n\nThis shows that \( {F\xi } \) is also a fixed point of \( {F}^{m} \). By the uniqueness of \( \xi ,{F\xi } = \xi... | Yes |
Theorem 6. Let \( F \) be a mapping of a Hilbert space into itself such that (a) \( \langle {Fx} - {Fy}, x - y\rangle \geq \alpha \parallel x - y{\parallel }^{2}\;\left( {\alpha > 0}\right) \) (b) \( \parallel {Fx} - {Fy}\parallel \leq \beta \parallel x - y\parallel \) Then \( F \) is surjective and injective. Conseque... | Proof. The injectivity follows at once from (a): If \( x \neq y \), then \( {Fx} \neq {Fy} \) . For the surjectivity, let \( w \) be any point in the Hilbert space. It is to be proved that, for some \( x,{Fx} = w \) . It is equivalent to prove, for any \( \lambda > 0 \), that an \( x \) exists satisfying \( x - \lambda... | Yes |
First, consider an integral equation of the form\n\n\[ x\left( t\right) - \lambda {\int }_{0}^{1}K\left( {s, t}\right) x\left( s\right) {ds} = v\left( t\right) \;\left( {0 \leq t \leq 1}\right) \] | Write the equation in the form\n\n\[ \left( {I - {\lambda A}}\right) x = v \] \nin which \( A \) is the integral operator in Equation (4). If we have chosen a suitable Banach space and if \( \parallel {\lambda A}\parallel < 1 \), then the Neumann series gives a formula for the solution:\n\n\[ x = {\left( I - \lambda A\... | Yes |
Example 2. For a concrete example of this, consider\n\n\[ x\left( t\right) = \lambda {\int }_{0}^{1}{e}^{t - s}x\left( s\right) {ds} + v\left( t\right) \]\n\nHere, we use an operator \( A \) defined by\n\n\[ \left( {Ax}\right) \left( t\right) = {\int }_{0}^{1}{e}^{t - s}x\left( s\right) {ds} \] | If we compute \( {A}^{2}x \), the result is\n\n\[ \left( {{A}^{2}x}\right) \left( t\right) = {\int }_{0}^{1}{e}^{t - s}\left( {Ax}\right) \left( s\right) {ds} \]\n\n\[ = {\int }_{0}^{1}{e}^{t - s}{\int }_{0}^{1}{e}^{s - \sigma }x\left( \sigma \right) {d\sigma ds} \]\n\n\[ = {\int }_{0}^{1}\left\lbrack {{\int }_{0}^{1}{... | Yes |
Can we use an operator \( B \) such that \( \parallel I - {AB}\parallel < 1 \) to solve the problem \( {Ax} = v \)? | Obviously, \( {x}_{0} = {Bv} \) is a first approximation to \( x \) . By the Neumann theorem, we know that \( {AB} \) is invertible and that\n\n\[ {\left( AB\right) }^{-1} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{\left( I - AB\right) }^{k} \]\n\nIt is clear that the vector \( x = B{\left( AB\right) }^{-1}v \) is a s... | Yes |
Theorem 1. The range of a projection is identical with the set of its fixed points. | Proof. Let \( P : X \rightarrow X \) be a projection and \( V \) its range. If \( v \in V \), then \( v = {Px} \) for some \( x \), and consequently,\n\n\[ \n{Pv} = {P}^{2}x = {Px} = v \n\]\n\nThus \( v \) is a fixed point of \( P \) . The reverse inclusion is obvious: If \( x = {Px} \), then \( x \) is in the range of... | Yes |
Theorem 2. If \( P \) is a projection, then so is \( I - P \) . The range of each is the null space of the other. | \[ {\left( I - P\right) }^{2} = \left( {I - P}\right) \left( {I - P}\right) = I - {2P} + {P}^{2} = I - P \] \n\nUse \( \mathcal{R} \) and \( \mathcal{N} \) for \ | No |
Theorem 3. The adjoint of a projection is also a projection. | Proof. Let \( P \) be a projection defined on the normed space \( X \) . Recall that its adjoint \( {P}^{ * } \) maps \( {X}^{ * } \) to \( {X}^{ * } \) and is defined by the equation\n\n\[ \n{P}^{ * }\phi = \phi \circ P\;\phi \in {X}^{ * } \n\]\n\nThus it follows that\n\n\[ \n{\left( {P}^{ * }\right) }^{2}\phi = {P}^{... | Yes |
Theorem 4. Let \( P \) be a projection of a normed space \( X \) onto a finite-dimensional subspace \( V \) . If \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) is any basis for \( V \) then there exist functionals \( {\phi }_{i} \) in \( {X}^{ * } \) such that\n\n(4)\n\n\[{\phi }_{i}\left( {v}_{j}\right) = {\delta }_{... | Proof. Select \( {\psi }_{i} \in {V}^{ * } \) such that for any \( v \in V \),\n\n\[v = \mathop{\sum }\limits_{{i = 1}}^{n}{\psi }_{i}\left( v\right) {v}_{i}\]\n\nThe functionals \( {\psi }_{i} \) are linear and continuous, by Corollary 1 in Section 1.5, page 26. (See also the proof of Corollary 2 in the same section.)... | Yes |
Theorem 5. If \( P \) is a projection of \( X \) onto a subspace \( V \), then for all \( x \in X \) ,\n\n\[ \parallel x - {Px}\parallel \leq \parallel I - P\parallel \operatorname{dist}\left( {x, V}\right) \] | Proof. For any \( v \in V,{Pv} = v \) . Hence\n\n\[ \parallel x - {Px}\parallel = \parallel \left( {x - v}\right) - P\left( {x - v}\right) \parallel = \parallel \left( {I - P}\right) \left( {x - v}\right) \parallel \leq \parallel I - P\parallel \parallel x - v\parallel \]\n\nNow take an infimum as \( v \) ranges over \... | Yes |
Theorem 6. Let \( \left\lbrack {P}_{n}\right\rbrack \) be a sequence of projections on a Banach space \( X \), and assume that \( {P}_{n}y \rightarrow y \) for each \( y \) in \( X \) . Let \( b \in X \) and \( A \in \mathcal{L}\left( {X, X}\right) \) . For each \( n \) let \( {x}_{n} \) be a point such that \( {P}_{n}... | Proof. Since \( {P}_{n}y \rightarrow y \), we also have \( \begin{Vmatrix}{{P}_{n}y}\end{Vmatrix} \rightarrow \parallel y\parallel \), and therefore \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{{P}_{n}y}\end{Vmatrix} < \) \( \infty \) for each \( y \) . Since \( X \) is complete, we may apply the Uniform Boundedness Pri... | Yes |
Theorem 7. Let \( P \) be a projection of the normed space \( X \) onto a subspace \( V \). Suppose that \( x \in X, x - {Ax} - b = 0,\widetilde{x} \in V \), and \( P\left( {\widetilde{x} - A\widetilde{x} - b}\right) = 0 \). If \( I - {PA} \) is invertible, then\n\n\[ \parallel x - \widetilde{x}\parallel \leq \begin{Vm... | Proof. From Equation (15), \( {PAx} = {Px} - {Pb} \). Hence\n\n\[ x - {PAx} = x - {Px} + {Pb} \]\n\nSince \( \widetilde{x} \in V \), it follows that \( P\widetilde{x} = \widetilde{x} \). Consequently,\n\n\[ \widetilde{x} - {PA}\widetilde{x} = {Pb} \]\n\nSubtraction between Equations (17) and (18) gives\n\n\[ x - \widet... | Yes |
Theorem 8. Let \( \\left\\lbrack {{v}_{1},{v}_{2},\\ldots }\\right\\rbrack \) be an orthonormal basis in a Hilbert space \( X \) . Let \( A \\in \\mathcal{L}\\left( {X, X}\\right) \) and \( \\parallel A\\parallel < 1 \) . For each \( n \), let \( {x}_{n} \) be a linear combination of \( {v}_{1},\\ldots ,{v}_{n} \) chos... | Of course, the Neumann Theorem can be used to solve the equation \( \\left( {I - A}\\right) x = b \) . It gives \( x = {\\left( I - A\\right) }^{-1}b = \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }{A}^{n}b \) . There seems to be no obvious connection between this solution and the one provided by Theorem 8. | No |
Theorem 1. If \( \Omega \) is a region to which Green’s Theorem applies, then the Laplacian is self-adjoint with respect to the inner product (9) when applied to functions vanishing on \( \partial \Omega \) . | Proof. Using subscripts to denote partial derivatives, we write Green's Theorem in the form\n\[ \n{\int }_{\Omega }\left( {{Q}_{x} - {P}_{y}}\right) = {\int }_{\partial \Omega }\left( {{Pdx} + {Qdy}}\right) \]\n\nApplying this to the functions \( Q = u{v}_{x} - v{u}_{x} \) and \( P = v{u}_{y} - u{v}_{y} \), we obtain\n... | Yes |
Under the hypotheses listed above we have the following: For each \( z \) in \( V \) there is a unique \( w \) in \( U \) such that \[ B\left( {w, v}\right) = \langle z, v\rangle \;\text{ for all }v\text{ in }V \] Furthermore, \( w \) depends linearly and continuously on \( z \) . | Proof. As usual, we define the \( u \) -sections of \( B \) by \( {B}_{u}\left( v\right) = B\left( {u, v}\right) \) . Then each \( {B}_{u} \) is a continuous linear functional on \( V \) . Indeed, \[ \begin{Vmatrix}{B}_{u}\end{Vmatrix} = \sup \{ \left| {B\left( {u, v}\right) }\right| : v \in V,\parallel v\parallel = 1\... | Yes |
Consider the two-point boundary-value problem\n\n\[ \n{\left( p{u}^{\prime }\right) }^{\prime } - {qu} = f\;u\left( a\right) = 0\;u\left( b\right) = 0 \n\] | This is a Sturm-Liouville problem, the subject of Theorem 1 in the next section (page 206) as well as Section 2.5, pages 105ff. In order to apply Theorem 3, one requires the bilinear form and linear functional appropriate to the problem. They are revealed by a standard procedure: Multiply the differential equation by a... | Yes |
The steady-state distribution of heat in a two-dimensional domain \( \Omega \) is governed by Poisson’s Equation:\n\n\[ \n{\nabla }^{2}u = f\;\text{ in }\Omega \n\] \n\nHere, \( u\left( {x, y}\right) \) is the temperature at the location \( \left( {x, y}\right) \) in \( {\mathbb{R}}^{2} \), and \( f \) is the heat-sour... | To arrive at this form of the problem, first write the equivalent equation\n\n\[ \n\left\langle {{\nabla }^{2}u, v}\right\rangle = \langle f, v\rangle \;\text{ for all }v \n\] \n\nThe integral form of this is\n\n\[ \n{\int }_{\Omega }v{\nabla }^{2}u = {\int }_{\Omega }{vf}\;\text{ for all }v \n\] \n\nThe integral on th... | Yes |
Theorem 1. If \( x \) is a function in \( {C}^{2}\left\lbrack {a, b}\right\rbrack \) that minimizes \( \Phi \left( x\right) \) locally subject to the constraints \( x\left( a\right) = \alpha \) and \( x\left( b\right) = \beta \), then \( x \) solves the problem in (5). | Proof. Assume the hypotheses, and let \( y \) be any element of \( {C}^{2}\left\lbrack {a, b}\right\rbrack \) such that \( y\left( a\right) = y\left( b\right) = 0 \) . We use what is known as a variational argument. For each real number \( \lambda, x + {\lambda y} \) is a competitor of \( x \) in the minimization of \(... | Yes |
Theorem 2. Assume that \( p\left( t\right) > 0 \) and \( q\left( t\right) \geq 0 \) on \( \left\lbrack {a, b}\right\rbrack \) . If \( x \) is a function in \( {C}^{2}\left\lbrack {a, b}\right\rbrack \) that solves the boundary-value problem (5) then \( x \) is the unique local minimizer of \( \Phi \) subject to the bou... | Proof. Let \( z \in {C}^{2}\left\lbrack {a, b}\right\rbrack, z \neq x, z\left( a\right) = \alpha \), and \( z\left( b\right) = \beta \) . Then the function \( y = z - x \) satisfies 0 -boundary conditions but is not 0 . By calculations like those in the preceding proof,\n\n(9)\n\n\[ \Phi \left( {x + y}\right) = \Phi \l... | Yes |
Theorem 3. Let \( x \) denote the solution of the boundary-value problem \( \left( {{11},{12}}\right) \), and let \( {x}_{n} \) be a point in \( {U}_{n} \) that minimizes \( \Phi \) on \( {U}_{n} \) . Assume \( q \geq 0 \) and \( p > 0 \) on \( \left\lbrack {0,1}\right\rbrack \) . Then \( {x}_{n}\left( t\right) \righta... | Proof. Since we have assumed that \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{U}_{n} \) is dense in \( X \), the preceding lemma and Theorem 2 imply that \( \Phi \left( {x}_{n}\right) \downarrow \Phi \left( x\right) \) . Notice that our choice of norm on \( X \) guarantees the continuity of \( \Phi \) . In the foll... | Yes |
Theorem 1. Each solution of the boundary-value problem (4) solves the integral equation (6) and conversely. | Proof. Let \( x \) be any function in \( C\left\lbrack {0,1}\right\rbrack \), and define \( y \) by writing\n\n\[ y\left( t\right) = - {\int }_{0}^{1}G\left( {t, s}\right) f\left( {s, x\left( s\right) }\right) {ds} \]\n\n\[ = - {\int }_{0}^{t}G\left( {t, s}\right) f\left( {s, x\left( s\right) }\right) {ds} + {\int }_{1... | Yes |
Theorem 2. Let \( f\left( {s, t}\right) \) be continuous in the domain defined by the inequalities \( 0 \leq s \leq 1, - \infty < t < \infty \) . Assume also that \( f \) satisfies a Lipschitz condition in this domain:\n\n(8)\n\n\[ \left| {f\left( {s,{t}_{1}}\right) - f\left( {s,{t}_{2}}\right) }\right| \leq k\left| {{... | Proof. Consider the nonlinear mapping \( F : C\left\lbrack {0,1}\right\rbrack \rightarrow C\left\lbrack {0,1}\right\rbrack \) defined by\n\n\[ \left( {Fx}\right) \left( t\right) = - {\int }_{0}^{1}G\left( {t, s}\right) f\left( {s, x\left( s\right) }\right) {ds}\;x \in C\left\lbrack {0,1}\right\rbrack \]\n\nWe shall pro... | Yes |
Consider the two-point boundary-value problem\n\n\[ \left\{ \begin{array}{l} {x}^{\prime \prime }\left( t\right) = \frac{1}{2}\exp \left\{ {\frac{1}{2}\left( {t + 1}\right) \cos \left\lbrack {x\left( t\right) + 7 - {3t}}\right\rbrack }\right\} \; - 1 \leq t \leq 1 \\ x\left( {-1}\right) = - {10}\;x\left( 1\right) = - 4... | Our existence theorem does not apply to this directly, and some changes of variables are called for. We set\n\n\[ z\left( t\right) = x\left( t\right) - {3t} + 7 \]\n\nand find that \( z \) should solve this problem:\n\n\[ \left\{ \begin{array}{l} {z}^{\prime \prime }\left( t\right) = \frac{1}{2}\exp \left\{ {\frac{1}{2... | Yes |
Theorem 3. Problems (14) and (15) are equivalent. | Proof. We proceed as in the proof of Theorem 1, which concerns the \ | No |
Theorem 4. If \( k\left( {1 + 8\parallel v{\parallel }_{\infty }}\right) < 8 \), and if the weights \( {A}_{j} \) in Equation (19) are all positive, then for \( i = 1,2,\ldots, n \), \[ \left| {x\left( {s}_{i}\right) - {y}_{i}}\right| \leq \lambda \parallel u{\parallel }_{\infty }\;\text{ where }\lambda = {\left\lbrack... | Proof. Let \( {\varepsilon }_{i} = \left| {x\left( {s}_{i}\right) - {y}_{i}}\right| \) and \( \varepsilon = \max {\varepsilon }_{i} \). Then for each \( i \) we have \[ {\varepsilon }_{i} = \left| {{\int }_{0}^{1}G\left( {{s}_{i}, s}\right) f\left( {s, x\left( s\right) }\right) {ds} - \mathop{\sum }\limits_{{j = 1}}^{n... | Yes |
Theorem 6. If the nodes and the function \( w \) are prescribed, then there exists a formula of the type displayed in Equation (26) that is exact for all polynomials of degree at most \( n - 1 \) . | Proof. Recall the Lagrange interpolation operator described in Example 1 of Section 4.4, page 193. Its formula is\n\n(27)\n\n\[ \n{Lx} = \mathop{\sum }\limits_{{i = 1}}^{n}x\left( {t}_{i}\right) {\ell }_{i} \n\]\n\nSince \( L \) is a projection of \( C\left\lbrack {a, b}\right\rbrack \) onto \( \mathop{\prod }\limits_{... | Yes |
Example 2. If \( \left( {{t}_{1},{t}_{2},{t}_{3}}\right) = \left( {-1,0, + 1}\right) \) and \( \left\lbrack {a, b}\right\rbrack = \left\lbrack {-1,1}\right\rbrack \), what is the quadrature formula produced by the preceding method when \( w\left( t\right) = 1 \) ? | We follow the prescription and begin with the functions \( {\ell }_{i} \) :\n\n\[ \n{\ell }_{1}\left( t\right) = \left( {t - {t}_{2}}\right) \left( {t - {t}_{3}}\right) {\left( {t}_{1} - {t}_{2}\right) }^{-1}{\left( {t}_{1} - {t}_{3}\right) }^{-1} = t\left( {t - 1}\right) /2 \n\]\n\n\[ \n{\ell }_{2}\left( t\right) = \l... | Yes |
Theorem 7. Gaussian Quadrature. For appropriate nodes and weights, the quadrature formula (26) is exact on \( \mathop{\prod }\limits_{{{2n} - 1}} \) . | Proof. Define an inner product on \( C\left\lbrack {a, b}\right\rbrack \) by putting\n\n(30)\n\n\[ \langle x, y\rangle = {\int }_{a}^{b}x\left( t\right) y\left( t\right) w\left( t\right) {dt} \]\n\nLet \( p \) be the unique monic polynomial in \( \mathop{\prod }\limits_{n} \) that is orthogonal to \( \mathop{\prod }\li... | Yes |
Theorem 8. The weights in a Gaussian quadrature formula are positive. | Proof. Suppose that Formula (26) is exact on \( {\Pi }_{{2n} - 1} \) . Then it will integrate \( {\ell }_{j}^{2} \) exactly:\n\n\[ 0 < {\int }_{a}^{b}{\ell }_{j}^{2}\left( t\right) w\left( t\right) {dt} = \mathop{\sum }\limits_{{i = 1}}^{n}{A}_{i}{\ell }_{j}^{2}\left( {t}_{i}\right) = {A}_{j} \]\n | Yes |
Theorem 1. A lower semicontinuous functional on a compact set attains its infimum. | Proof. Recall that lower semicontinuity of \( \Phi \) means that each set of the form\n\n\[ \n{K}_{\lambda } = \{ x \in K : \Phi \left( x\right) \leq \lambda \} \n\]\n\nis closed. If \( \lambda > \rho \), then \( {K}_{\lambda } \) is nonempty. The family of closed sets \( \left\{ {K}_{\lambda }\right. \) : \( \lambda >... | Yes |
Theorem 2. Under the hypotheses above, a point \( y \) satisfies the equation \( {Ay} = b \) if and only if \( y \) is a global minimum point of \( \Phi \) . | Proof. Let \( x \) be an arbitrary point and \( v \) any nonzero vector. Then\n\n\[ \Phi \left( {x + {tv}}\right) = \langle {Ax} + {tAv} - {2b}, x + {tv}\rangle \]\n\n(10)\n\n\[ = \langle {Ax} - {2b}, x\rangle + t\langle {Ax} - {2b}, v\rangle + t\langle {Av}, x\rangle + {t}^{2}\langle {Av}, v\rangle \]\n\n\[ = \Phi \le... | Yes |
Theorem 3. If \( A \) is self-adjoint and satisfies\n\n\[ \mathop{\inf }\limits_{{\parallel x\parallel = 1}}\langle {Ax}, x\rangle > 0 \]\n\nthen the steepest-descent sequence in Equation (15) converges to the solution of the equation \( {Ax} = b \) . | There is more to this theorem than meets the eye, because the hypotheses on \( A \) imply its invertibility, and consequently the equation \( {Ax} = b \) has a unique solution for each \( b \) in the Hilbert space. See the lemma in Section 4.9, page 234, for the appropriate formal result. | No |
Theorem 1. In the conjugate direction algorithm (3), using an A-orthonormal set of direction vectors, each residual \( {r}_{n} = b - A{x}_{n} \) is orthogonal to all the previous search directions \( {v}_{1},\ldots ,{v}_{n - 1} \) . | Proof. Let \( {r}_{n} = b - A{x}_{n} \) . We wish to prove that\n\n(5)\n\n\[ {r}_{n} \bot {v}_{1},{v}_{2},\ldots ,{v}_{n - 1}\;\left( {n = 2,3,\ldots }\right) \]\n\nFirst we observe that\n\n(6)\n\n\[ {r}_{n + 1} = b - A{x}_{n + 1} = b - A\left( {{x}_{n} + {\alpha }_{n}{v}_{n}}\right) = {r}_{n} - {\alpha }_{n}A{v}_{n} \... | Yes |
Theorem 2. Let \( A \) be a self-adjoint operator on a Hilbert space \( X \) . Assume that\n\n\[ m\parallel x{\parallel }^{2} \leq \langle x,{Ax}\rangle \leq M\parallel x{\parallel }^{2}\;\left( {m > 0}\right) \]\n\nLet \( {v}_{1},{v}_{2},\ldots \) be an \( A \) -orthonormal sequence whose linear span is dense in \( X ... | Proof. (After [Lue2]) Putting \( {\alpha }_{n} = \left\langle {{v}_{n}, b - A{x}_{n}}\right\rangle \), we have, from Equation\n\n\[ {x}_{2} = {x}_{1} + {\alpha }_{1}{v}_{1} \]\n\n\[ {x}_{3} = {x}_{2} + {\alpha }_{2}{v}_{2} = {x}_{1} + {\alpha }_{1}{v}_{1} + {\alpha }_{2}{v}_{2} \]\n\nand so on. Thus, in general,\n\n\[ ... | Yes |
Example 1. Let \( X = Y = {\mathbb{R}}^{2} \), and define\n\n\[ f\left( x\right) = \left\lbrack \begin{matrix} \sin {\xi }_{1} + {e}^{{\xi }_{2}} - 3 \\ {\left( {\xi }_{2} + 3\right) }^{2} - {\xi }_{1} - 4 \end{matrix}\right\rbrack \;x = \left( {{\xi }_{1},{\xi }_{2}}\right) \in {\mathbb{R}}^{2} \] | A convenient homotopy is defined by Equation (4), and we select the starting point \( {x}_{0} = \left( {5,3}\right) \) . The derivatives on the right side of Equation (8) are computed to be\n\n\[ {h}_{2} = {f}^{\prime }\left( x\right) = \left\lbrack \begin{matrix} \cos {\xi }_{1} & {e}^{{\xi }_{2}} \\ - 1 & 2{\xi }_{2}... | Yes |
Example 2. Let \( f \) be the mapping\n\n\[ f\left( x\right) = \left\lbrack \begin{matrix} {\xi }_{1}^{2} - 3{\xi }_{2}^{2} + 3 \\ {\xi }_{1}{\xi }_{2} + 6 \end{matrix}\right\rbrack \;x = \left( {{\xi }_{1},{\xi }_{2}}\right) \in {\mathbb{R}}^{2} \]\n\nWe take the starting point \( {x}_{0} = \left( {1,1}\right) \) and ... | Then\n\n\[ h\left( {t, x}\right) = \left\lbrack \begin{matrix} {\xi }_{1}^{2} - 3{\xi }_{2}^{2} + 2 + t \\ {\xi }_{1}{\xi }_{2} - 1 + {7t} \end{matrix}\right\rbrack \]\n\nThe differential equation (9) is given by\n\n(11)\n\n\[ \left\lbrack \begin{matrix} 1 & 2{\xi }_{1} & - 6{\xi }_{2} \\ 7 & {\xi }_{2} & {\xi }_{1} \e... | Yes |
For any polynomial \( P \), the function \( f : \mathbb{R} \rightarrow \mathbb{R} \) defined by \[ f\left( x\right) = \left\{ \begin{array}{ll} P\left( {1/x}\right) {e}^{-1/x} & x > 0 \\ 0 & x \leq 0 \end{array}\right. \] is in \( {C}^{\infty }\left( \mathbb{R}\right) \) . | First we show that \( f \) is continuous. The only questionable point is \( x = 0 \) . We have \[ \mathop{\lim }\limits_{{x \downarrow 0}}f\left( x\right) = \mathop{\lim }\limits_{{x \downarrow 0}}\frac{P\left( {1/x}\right) }{\exp \left( {1/x}\right) } = \mathop{\lim }\limits_{{t \uparrow \infty }}\frac{P\left( t\right... | Yes |
Lemma 2. The function \( \rho \) defined in Equation (1) belongs to \( \mathbf{D} \) . | Proof. The function \( f \) in the preceding lemma (with \( P\left( x\right) = 1 \) ) has the property that \( \rho \left( x\right) = {cf}\left( {1 - {\left| x\right| }^{2}}\right) \) . Thus \( \rho = {cf} \circ g \), where \( g\left( x\right) = 1 - {\left| x\right| }^{2} \) and belongs to \( {C}^{\infty }\left( {\math... | Yes |
Theorem 1. For every multi-index \( \alpha ,{D}^{\alpha } \) is a continuous linear transformation of \( \mathbf{D} \) into \( \mathbf{D} \) . | Proof. The linearity is a familiar feature of differentiation. For the continuity, it suffices to prove continuity at 0 because \( {D}^{\alpha } \) is linear. Thus, suppose that \( {\phi }_{j} \in \mathcal{D} \) and \( {\phi }_{j} \rightarrow 0 \) . Let \( K \) be a compact set containing the supports of all the functi... | Yes |
Example 1. A Dirac distribution \( {\delta }_{\xi } \) is defined by selecting \( \xi \in {\mathbb{R}}^{n} \) and writing\n\n(3)\n\n\[ \n{\delta }_{\xi }\left( \phi \right) = \phi \left( \xi \right) \;\left( {\phi \in \mathfrak{D}}\right) \n\] | It is a distribution, because firstly, it is linear:\n\n\[ \n{\delta }_{\xi }\left( {{\lambda }_{1}{\phi }_{1} + {\lambda }_{2}{\phi }_{2}}\right) = \left( {{\lambda }_{1}{\phi }_{1} + {\lambda }_{2}{\phi }_{2}}\right) \left( \xi \right) = {\lambda }_{1}{\phi }_{1}\left( \xi \right) + {\lambda }_{2}{\phi }_{2}\left( \x... | Yes |
Example 3. Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be continuous. With \( f \) we associate a distribution \( \widetilde{f} \) by means of the definition\n\n(5)\n\n\[ \widetilde{f}\left( \phi \right) = \int f\left( x\right) \phi \left( x\right) {dx}\;\left( {\phi \in \mathcal{D}}\right) \] | The linearity of \( \widetilde{f} \) is obvious. For the continuity, we observe that if \( {\phi }_{j} \rightarrow 0 \), then there is a compact \( K \) containing the supports of the \( {\phi }_{j} \) . Then we have\n\n\[ \left| {\widetilde{f}\left( {\phi }_{j}\right) }\right| = \left| {{\int }_{K}f\left( x\right) {\p... | Yes |
Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be continuous. With \( f \) we associate a distribution \( \widetilde{f} \) by means of the definition\n\n\[ \widetilde{f}\left( \phi \right) = \int f\left( x\right) \phi \left( x\right) {dx}\;\left( {\phi \in \mathcal{D}}\right) \] | The linearity of \( \widetilde{f} \) is obvious. For the continuity, we observe that if \( {\phi }_{j} \rightarrow 0 \), then there is a compact \( K \) containing the supports of the \( {\phi }_{j} \) . Then we have\n\n\[ \left| {\widetilde{f}\left( {\phi }_{j}\right) }\right| = \left| {{\int }_{K}f\left( x\right) {\p... | Yes |
Theorem 2. If \( f \in C\left( {\mathbb{R}}^{n}\right) \), then \( \widetilde{f} \), as defined in Equation (5), is a distribution. The map \( f \mapsto \widetilde{f} \) is linear and injective from \( C\left( {\mathbb{R}}^{n}\right) \) into \( {\mathcal{D}}^{\prime } \) . | Proof. We have already seen that \( \widetilde{f} \) is a distribution. The linearity of the mapping \( f \mapsto \widetilde{f} \) follows from the equation\n\n\[{\left( {\alpha }_{1}{f}_{1} + {\alpha }_{2}{f}_{2}\right) }^{ \sim }\left( \phi \right) = \int \left( {{\alpha }_{1}{f}_{1} + {\alpha }_{2}{f}_{2}}\right) \p... | Yes |
Theorem 2. If \( f \in C\left( {\mathbb{R}}^{n}\right) \), then \( \widetilde{f} \), as defined in Equation (5), is a distribution. The map \( f \mapsto \widetilde{f} \) is linear and injective from \( C\left( {\mathbb{R}}^{n}\right) \) into \( {\mathcal{D}}^{\prime } \) . | Proof. We have already seen that \( \widetilde{f} \) is a distribution. The linearity of the mapping \( f \mapsto \widetilde{f} \) follows from the equation\n\n\[ \n{\left( {\alpha }_{1}{f}_{1} + {\alpha }_{2}{f}_{2}\right) }^{ \sim }\left( \phi \right) = \int \left( {{\alpha }_{1}{f}_{1} + {\alpha }_{2}{f}_{2}}\right)... | Yes |
Theorem 4. Let \( \mu \) be a positive Borel measure on \( {\mathbb{R}}^{n} \) such that \( \mu \left( K\right) < \infty \) for each compact set \( K \) in \( {\mathbb{R}}^{n} \) . Then \( \mu \) induces a distribution \( T \) by the formula\n\n\[ T\left( \phi \right) = {\int }_{{\mathbb{R}}^{n}}\phi \left( x\right) {d... | Proof. The linearity is obvious. For the continuity of \( T \), let \( {\phi }_{j} \in \mathcal{D} \) and \( {\phi }_{j} \rightarrow 0 \) . Then there is a compact set \( K \) containing the supports of all \( {\phi }_{j} \) . Consequently,\n\n\[ \left| {T\left( {\phi }_{j}\right) }\right| \leq {\int }_{K}\left| {{\phi... | Yes |
Let \( \widetilde{H} \) be the Heaviside distribution (Example 2, page 250), and let \( \delta \) be the Dirac distribution at 0 (Example 1, page 250). Then with \( n = 1 \) and \( \alpha = \left( 1\right) \), we have \( \partial \widetilde{H} = \delta \) . | Indeed, for any test function \( \phi \) ,\n\n\[ \left( {\partial \widetilde{H}}\right) \left( \phi \right) = - \widetilde{H}\left( {D\phi }\right) = - {\int }_{0}^{\infty }{\phi }^{\prime } = \phi \left( 0\right) - \phi \left( \infty \right) = \phi \left( 0\right) = \delta \left( \phi \right) \] | Yes |
Example 3. What is the distribution derivative of the function \( f\left( x\right) = \left| x\right| \) ? | It is a distribution \( \widetilde{g} \), where \( g \) is a function such that for all test functions \( \phi \) ,\n\n\[ \n\int {g\phi } = - \int f{\phi }^{\prime } = - {\int }_{-\infty }^{0}\left( {-x}\right) {\phi }^{\prime }\left( x\right) {dx} - {\int }_{0}^{\infty }x{\phi }^{\prime }\left( x\right) {dx} \n\]\n\n\... | Yes |
What is the distribution derivative \( {f}^{\prime \prime } \) when \( f\left( x\right) = \left| x\right| \) ? | If we blindly use the techniques of classical calculus, we have from Examples 1 and \( 3,{f}^{\prime } = {2H} - 1 \) and \( {f}^{\prime \prime } = {2\delta } \) . This procedure is justified by the next theorem. | No |
Theorem 1. The operators \( {\partial }^{\alpha } \) are linear from \( {\mathcal{D}}^{\prime } \) into \( {\mathcal{D}}^{\prime } \) . Furthermore, \( {\partial }^{\alpha }{\partial }^{\beta } = {\partial }^{\beta }{\partial }^{\alpha } = {\partial }^{\alpha + \beta } \) for any pair of multi-indices. | Proof. The linearity of \( {\partial }^{\alpha } \) is obvious from the definition, Equation (1). The commutative property rests upon a theorem of classical calculus that states that for any function \( f \) of two variables, if \( \frac{{\partial }^{2}f}{\partial x\partial y} \) and \( \frac{{\partial }^{2}f}{\partial... | Yes |
Theorem 3. Let \( n = 1 \), and let \( T \) be a distribution for which \( \partial T = 0 \). Then \( T \) is \( \widetilde{c} \) for some constant \( c \). | Proof. Adopt the notation of the preceding proof. The familiar equation\n\n\[ \phi \left( x\right) = \frac{d}{dx}{\int }_{-\infty }^{x}\phi \left( y\right) {dy} \]\n\nsays that \( \phi = {DB\phi } \), and this is valid for all \( \phi \in M \). Since \( {A\phi } \in M \) for all \( \phi \in \mathfrak{D} \), we have \( ... | Yes |
Theorem 4. If \( T \) is a distribution and \( K \) is a compact set in \( {\mathbb{R}}^{n} \) , then there exists an \( f \in C\left( {\mathbb{R}}^{n}\right) \) and a multi-index \( \alpha \) such that for all \( \phi \in \mathcal{D} \) whose supports are in \( K \) ,\n\n\[ T\left( \phi \right) = \left( {{\partial }^{... | Proof. For the proof, consult [Ru1], page 152. | No |
Theorem 1. For every multi-index \( \alpha ,{\partial }^{\alpha } \) is a continuous linear map of \( {\mathcal{D}}^{\prime } \) into \( {\mathcal{D}}^{\prime } \) . | Proof. Let \( {T}_{j} \rightarrow 0 \) . In order to prove that \( {\partial }^{\alpha }{T}_{j} \rightarrow 0 \), we select an arbitrary test function \( \phi \), and attempt to prove that \( \left( {{\partial }^{\alpha }{T}_{j}}\right) \left( \phi \right) \rightarrow 0 \) . This means that \( {\left( -1\right) }^{\lef... | No |
Theorem 2. If a sequence of distributions \( \left\lbrack {T}_{j}\right\rbrack \) has the property that \( \left\lbrack {{T}_{j}\left( \phi \right) }\right\rbrack \) is convergent for each test function \( \phi \), then the equation \( T\left( \phi \right) = \mathop{\lim }\limits_{j}{T}_{j}\left( \phi \right) \) define... | The theorem asserts that for a sequence of distributions \( {T}_{j} \) if \( \mathop{\lim }\limits_{j}{T}_{j}\left( \phi \right) \) exists in \( \mathbb{R} \) for every test function \( \phi \), then the equation \[ T\left( \phi \right) = \mathop{\lim }\limits_{j}{T}_{j}\left( \phi \right) \] defines a distribution. Th... | "No" |
Corollary 2. If \( \sum {T}_{j} \) is a convergent series of distributions, then for any multi-index \( \alpha ,{\partial }^{\alpha }\sum {T}_{j} = \sum {\partial }^{\alpha }{T}_{j} \) . | Proof. By Theorem \( 1,{\partial }^{\alpha } \) is continuous. Hence\n\n\[ \n{\partial }^{\alpha }\left( {\mathop{\sum }\limits_{{j = 1}}^{\infty }{T}_{j}}\right) = {\partial }^{\alpha }\left( {\mathop{\lim }\limits_{{m \rightarrow \infty }}\mathop{\sum }\limits_{{j = 1}}^{m}{T}_{j}}\right) = \mathop{\lim }\limits_{{m ... | Yes |
Corollary 2. If \( \sum {T}_{j} \) is a convergent series of distributions, then for any multi-index \( \alpha ,{\partial }^{\alpha }\sum {T}_{j} = \sum {\partial }^{\alpha }{T}_{j} \) . | Proof. By Theorem \( 1,{\partial }^{\alpha } \) is continuous. Hence\n\n\[ \n{\partial }^{\alpha }\left( {\mathop{\sum }\limits_{{j = 1}}^{\infty }{T}_{j}}\right) = {\partial }^{\alpha }\left( {\mathop{\lim }\limits_{{m \rightarrow \infty }}\mathop{\sum }\limits_{{j = 1}}^{m}{T}_{j}}\right) = \mathop{\lim }\limits_{{m ... | Yes |
Theorem 3. Let \( f,{f}_{1},{f}_{2},\ldots \) belong to \( {L}_{\text{loc }}^{1}\left( {\mathbb{R}}^{n}\right) \), and suppose that \( {f}_{j} \rightarrow f \) pointwise almost everywhere. If there is an element \( g \in {L}_{\text{loc }}^{1}\left( {\mathbb{R}}^{n}\right) \) such that \( \left| {f}_{j}\right| \leq g \)... | Proof. The question is whether\n\n(1)\n\n\[\n\int {f}_{j}\phi \rightarrow \int {f\phi }\n\]\n\nfor all test functions \( \phi \) . We have \( {f}_{j}\phi \in {L}^{1}\left( K\right) \) if \( K \) is the support of \( \phi \) . Furthermore, \( \left| {{f}_{j}\phi }\right| \leq g\left| \phi \right| \) and \( \left( {{f}_{... | Yes |
Theorem 4. Let \( \\left\\lbrack {f}_{j}\\right\\rbrack \) be a sequence of nonnegative functions in\n\n\( {L}_{\\text{loc }}^{1}\\left( {\\mathbb{R}}^{n}\\right) \) such that \( \\int {f}_{j} = 1 \) for each \( j \) and such that\n\n\\[ \n\\mathop{\\lim }\\limits_{{j \\rightarrow \\infty }}{\\int }_{\\left| x\\right| ... | Proof. Let \( \\phi \\in \\mathcal{D} \) and put \( \\psi = \\phi - \\phi \\left( 0\\right) \) . Let \( \\varepsilon > 0 \), and select \( r > 0 \) so that \( \\left| {\\psi \\left( x\\right) }\\right| < \\varepsilon \) when \( \\left| x\\right| < r \) . Then\n\n\\[\n\\left| {\\int {f}_{j}\\phi - \\phi \\left( 0\\right... | Yes |
Theorem 1 The set of monomials \( x \mapsto {x}^{\alpha } \) on \( {\mathbb{R}}^{n} \) is linearly independent. | Proof. If \( n = 1 \), the monomials are the elementary functions \( {x}_{1} \mapsto {x}_{1}^{j} \) for \( j = 0,1,2,\ldots \) They form a linearly independent set, because a nontrivial linear combination \( \mathop{\sum }\limits_{{j = 0}}^{m}{c}_{j}{x}_{1}^{j} \) cannot vanish as a function. (Indeed, it can have at mo... | Yes |
Theorem 2 The dimension of \( {\Pi }_{m}\left( {\mathbb{R}}^{n}\right) \) is \( \left( \begin{matrix} m + n \\ n \end{matrix}\right) \) . | Proof. The preceding theorem asserts that a basis for \( {\Pi }_{m}\left( {\mathbb{R}}^{n}\right) \) is \( \left\{ {x \mapsto {x}^{\alpha }}\right. \) : \( \left| \alpha \right| \leq m\} \) . Here \( x \in {\mathbb{R}}^{n} \) . Using #to denote the number of elements in a set, we have only to prove\n\n\[ \n\# \left\{ {... | Yes |
Theorem 4. Let \( D \) be a simple partial derivative, say \( D = \frac{\partial }{\partial {x}_{i}} \) , and let \( \partial \) be the corresponding distribution derivative. If \( T \in {\mathcal{D}}^{\prime } \) and \( f \in {C}^{\infty }\left( {\mathbb{R}}^{n}\right) \), then \[ \partial \left( {fT}\right) = {Df} \c... | Proof. \[ \left( {{Df} \cdot T + f \cdot \partial T}\right) \left( \phi \right) = T\left( {{Df} \cdot \phi }\right) + \partial T\left( {f \cdot \phi }\right) = T\left( {{Df} \cdot \phi }\right) - T\left( {D\left( {f\phi }\right) }\right) \] \[ = T\left( {{Df} \cdot \phi }\right) - T\left( {{Df} \cdot \phi + f \cdot {D\... | Yes |
Theorem 5. Let \( n = 1 \), let \( T \) be a distribution, and let \( u \) be an element of \( {C}^{\infty }\left( \mathbb{R}\right) \) . If \( \partial T + {uT} = \widetilde{f} \), for some \( f \) in \( C\left( \mathbb{R}\right) \), then \( T = \widetilde{g} \) for some \( g \) in \( {C}^{1}\left( \mathbb{R}\right) \... | Proof. If \( u = 0 \), then \( \partial T = \widetilde{f} \) . Write \( f = {h}^{\prime } \), where\n\n\[ h\left( x\right) = {\int }_{a}^{x}f\left( y\right) {dy} \]\n\nThen \( h \in {C}^{1}\left( \mathbb{R}\right) \) . From the equation\n\n\[ \partial \left( {T - \widetilde{h}}\right) = \partial T - \widetilde{f} = 0 \... | Yes |
Theorem 6. If \( \phi \in \mathcal{D} \), then\n\n\[ \n{\int }_{{\mathbb{R}}^{n}}\phi \left( x\right) {dx} = \mathop{\lim }\limits_{{h \downarrow 0}}{h}^{n}\mathop{\sum }\limits_{{\alpha \in {\mathbb{Z}}^{n}}}\phi \left( {h\alpha }\right)\n\] | Proof. The right side is just the limit of Riemann sums for the integral. In the case \( n = 2 \), we set up a lattice of points in \( {\mathbb{R}}^{2} \) . These points are of the form \( \left( {{ih},{jh}}\right) = h\left( {i, j}\right) = {h\alpha } \), where \( \alpha \) runs over the set of all multi-integers, havi... | No |
Lemma 1. For \( T \in {\mathcal{D}}^{\prime } \) and \( \phi \in \mathcal{D} \), \[ {E}_{x}\left( {T * \phi }\right) = \left( {{E}_{x}T}\right) * \phi = T * {E}_{x}\phi \] | Proof. Straightforward calculation, using some results in Problem 1, gives us \[ \left\lbrack {{E}_{x}\left( {T * \phi }\right) }\right\rbrack \left( y\right) = \left( {T * \phi }\right) \left( {y - x}\right) = T\left( {{E}_{y - x}{B\phi }}\right) \] \[ \left\lbrack {\left( {{E}_{x}T}\right) * \phi }\right\rbrack \left... | Yes |
Lemma 2. If \( T \) is a distribution and if \( {\phi }_{j} \rightarrow \phi \) in \( \mathcal{D} \), then \( T * {\phi }_{i} \rightarrow \) \( T * \phi \) pointwise. | Proof. By linearity (see Problem 3), it suffices to consider the case when \( \phi = 0 \) . If \( {\phi }_{j} \rightarrow 0 \) in \( \mathcal{D} \), then for all \( x \) ,\n\n\[ \n\left( {T * {\phi }_{j}}\right) \left( x\right) = T\left( {{E}_{x}B{\phi }_{j}}\right) \rightarrow 0 \n\]\n\nby the continuity of \( B,{E}_{... | No |
Lemma 3. Let \( \left\lbrack {x}_{j}\right\rbrack \) be a sequence of points in \( {\mathbb{R}}^{n} \) converging to \( x \) . For each \( \phi \in \mathfrak{D} \), (7) \[ {E}_{{x}_{j}}\phi \rightarrow {E}_{x}\phi \;\text{ (convergence in }\;\mathfrak{D}\text{ ) } \] | Proof. If \( {K}_{1} = \left\{ {x,{x}_{1},{x}_{2},\ldots }\right\} \) and if \( {K}_{2} \) is the support of \( \phi \), then (as is easily verified) the supports of \( {E}_{{x}_{j}}\phi \) are contained in the compact set \[ {K}_{1} + {K}_{2} = \left\{ {u + v : u \in {K}_{1}, v \in {K}_{2}}\right\} \] Now we observe t... | Yes |
Lemma 4. Let \( e = \left( {1,0,\ldots ,0}\right) ,0 < \left| t\right| < 1 \), and \( {F}_{t} = {t}^{-1}\left( {{E}_{0} - {E}_{te}}\right) \). Then for each test function \( \phi ,{F}_{t}\phi \rightarrow \frac{\partial \phi }{\partial {x}_{1}} \) as \( t \rightarrow 0 \) . (This convergence is in the topology of \( \ma... | Proof. Since \( \left| t\right| < 1 \), there is a single compact set \( K \) containing the supports of \( {F}_{t}\phi \) and \( \frac{\partial \phi }{\partial {x}_{1}} \). By the mean value theorem (used twice) we have (for \( 0 < \theta ,{\theta }^{\prime } < 1 \) )\n\n\[ \left| {\frac{\partial \phi }{\partial {x}_{... | Yes |
What are the fundamental solutions of the operator \( D \) in the case \( n = 1 \) ? \( \left( {D = \frac{d}{dx}}\right) \) . We seek all the distributions \( T \) that satisfy \( \partial T = \delta \) . | We saw in Example 1 of Section 5.2 (page 254) that \( \partial \widetilde{H} = \delta \), where \( H \) is the Heaviside function. Thus \( \widetilde{H} \) is one of the fundamental solutions. Since the distributions sought are exactly those for which \( \partial T = \partial \widetilde{H} \), we see by Theorem 3 in Se... | Yes |
Theorem 1. The Malgrange-Ehrenpreis Theorem. Every operator \( \sum {c}_{\alpha }{D}^{\alpha } \) has a fundamental solution. | For the proof of this basic theorem, consult [Ho] page 189, or [Ru1] page 195. | No |
Theorem 2. Let \( A \) be a linear differential operator with constant coefficients, and let \( T \) be a distribution that is a fundamental solution of \( A \) . Then for each test function \( \phi, A\left( {T * \phi }\right) = \phi \) . | Proof. Let \( A = \sum {c}_{\alpha }{D}^{\alpha } \) . Then \( \sum {c}_{\alpha }{\partial }^{\alpha }T = \delta \) . The basic formula (the theorem of Section 5, page 271) states that\n\n\[ \n{D}^{\alpha }\left( {T * \phi }\right) = {\partial }^{\alpha }T * \phi \n\]\n\nFrom this we conclude that\n\n\[ \nA\left( {T * ... | Yes |
Example 2. We use the theory of distributions to find a solution of the differential equation \( \frac{du}{dx} = \phi \), where \( \phi \) is a test function. | By Example 1, one fundamental solution is the distribution \( \widetilde{H} \) . By the preceding theorem, \( \widetilde{H} * \phi \) will solve the differential equation. We have, with a simple change of variable,\n\n\[ u\left( x\right) = \left( {\widetilde{H} * \phi }\right) \left( x\right) = {\int }_{-\infty }^{\inf... | Yes |
Let us search for a solution of the differential equation\n\n\\[ \n{u}^{\prime } + {au} = \phi \n\\]\n\nusing distribution theory. First, we try to discover a fundamental solution, i.e., a distribution \\( T \\) such that \\( \\partial T + {aT} = \\delta \\) . | If \\( T \\) is such a distribution and if \\( v\\left( x\\right) = {e}^{ax} \\), then\n\n\\[ \n\\partial \\left( {v \\cdot T}\\right) = {Dv} \\cdot T + v \\cdot \\partial T = {av} \\cdot T + v \\cdot \\left( {\\delta - {aT}}\\right) = v \\cdot \\delta = \\delta \n\\]\n\nConsequently, by Example 1,\n\n\\[ \nv \\cdot T ... | Yes |
Lemma 1. For \( x \neq 0,\frac{\partial }{\partial {x}_{j}}\left| x\right| = {x}_{j}{\left| x\right| }^{-1} \) . | \[ \frac{\partial }{\partial {x}_{j}}\left| x\right| = \frac{\partial }{\partial {x}_{j}}{\left( {x}_{1}^{2} + \cdots + {x}_{n}^{2}\right) }^{1/2} = \frac{1}{2}{\left( {x}_{1}^{2} + \cdots + {x}_{n}^{2}\right) }^{-1/2}\left( {2{x}_{j}}\right) = {x}_{j}{\left| x\right| }^{-1} \] | Yes |
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