This means that each linear combination of functions in the family will be a different member of the family. Furthermore, we will not discuss the elementary consequences of the axioms such as A L� Xi = L� AXi.
Section 1.1 Definitions and Examples 3 symbol C denotes the set of complex numbers.) Other fields can be employed
In this space, vectors are effectively equivalence classes of functions, where two functions are considered equivalent if they differ only on a set of measure o. Recall that a subset S in a linear space is linearly independent if no finite, nonempty set of distinct vectors XI, X2, can be found.
A sequence [xnl in a normed linear space X is said to have the Cauchy property or to be a Cauchy sequence if. A compact set in a normed linear space must be bounded; i.e. contained in some multiple of the unit sphere.
Any normed linear space X can be embedded as a dense subspace in a complete normed linear space X. A real-valued continuous function whose domain is a compact set in a normed linear space attains its maximum and infi.
The following is a generalization for normed linear spaces of a theorem that should be known from elementary calculus.
Continuity, Open Sets, Closed Sets
Theorem 4. The inverse image of a closed set by a continuous map is closed
By the compactness of the interval [-c, cl, there exists an increasing sequence li of N which has the property that lim [xle(l) : k E lt ]. Then, there exists another increasing sequence h C h such that there exists lim [xd2) : k E 12l. A subgroup S in a normed linear space is said to be bounded if there is a constant c such that IIxll � c for all x E S.
To show that AI is compact, we can use Lemmas 1 and 2 above if we can show that for some c,.
Corollary 2. Every finite-dimensional subspace in a normed linear space is closed
Prove that the collection of open sets (as we defined them) in a normed linear space satisfies the topology axioms. Prove that if the unit sphere in normed linear space is complete, then the space is complete.
A linear functional on a normed space is continuous if and only if its kernel ("zero space") is closed. These points belong to the kernel f and converge to x, which is not in the kernel, so the latter does not exist.
Corollary 1. Every linear functional on a finite-dimensional normed linear space is continuous
Corollary 2. Every linear transformation from a finite-dimensional IIormed space to another normed space is continuous
If AA is well defined (eg, the range A is contained in its domain), then we write it as A2. Since the space of all bounded linear operators on X in X is complete (Theorem 4), the sequence [Bn] converges to the bounded linear operator B. Neumann's theorem is a powerful tool that has applications to many practical problems, such as the integral of equations and the solution of large systems of linear equations.
The Axiom of Choice is an axiom that most mathematicians use unre. served, but still controversial. In 1963, Paul Cohen [Coh] proved that the Axiom of Choice is independent of the remaining axioms in the Zermelo-Fraenkel system.
Theorem 2. Every nontrivial vector space has a Hamel base
An element m in a partially ordered set X is said to be a maximal element if every x in X satisfying the condition m � x also satisfies x � m. A partially ordered array contains a maximal element if every fully ordered subset has an upper bound.
Using Problem 33 and Problem 1.5.3, page 28, prove that every infinite-dimensional normed linear space is the union of a disjoint pair of dense convex sets. The corollary implies that if X is a complete metric space, then X is of the second category in X. is a family of continuous linear transformations defined on the Banach space X and taking values in the normed linear space. the previous corollary implies that F is of second category in X. F is of second category in X. Since F is of second category, every Fn is a closed subset. of X, the definition of the second category category implies that some Fm contains a ball.
7 Bair's Theorem and the Uniform Boundedness 43 It can be shown that the norm An treated as a mapping of C21r to itself, It can be shown that the norm of An treated as a mapping of C21r to itself is approximately log n . Since sUPn IIt< 0 Anll = +00, the set x in C21r whose Fourier series diverges at a certain point t is a set of the second category.
Is every set of the second category the complement of a set of the first category? Prove that in a complete metric space the complement of a set of the first category is dense and of the second category.
A normed linear space that is the image of a Banach space by a bounded, linear inner map is also a Banach space. By problem 1 .2.38 (page 14), it suffices to prove that every absolutely convergent series in Y is convergent.
Prove that if L is a linear operator with a closed domain and acts between normed linear spaces, then the equation Lx = y is solvable for x if and only if y E [N(L·)J 1. Describe the domain of L and prove that no contains functions f(x) = et. Continued) Draw the same conclusion as in Exercise 18 by referring to the Closed Area Theorem.
Section 1.9 Weak Convergence 53
The space fp is defined as the set of all real sequences x for which L�=I Ix(n)iP < 00.
Section 1.9 Weak Convergence
James exhibits an X that is isometrically isomorphic to X" , but the isometry is not the canonical map J, and indeed the canonical image of J( X ) is a proper subspace of X" in the example. One of the problems asks for a proof of the fact that B* is an isometric isomer.
Section 1.10 Reflexive Spaces
The Lebesgue spaces Lp[a, bj can be defined without knowing anything about Lebesgue measure or integration. Thus, if J is the natural map of Cp[a, bj into its second conjugate space, then.
Chapter 2
- Geometry 61
- Orthogonality and Bases 70
- Sturm-Liouville Theory 105
Item e in Theorem 1 is called the parallelogram equality (or "law") because it states that the sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of the two diagonals.
The notation £2 (S) then denotes the space of measurable complex functions on S such that J l!(s)12 dlL <. If K is a closed, convex, nonempty set in a Hilbert space X, then for each x in X there corresponds a unique point Y in K that is closest to X, then for each x in X there corresponds a unique point Y in K that is closest to x; that is,.
Geometry 67
Let x and y be points in a real inner product space such that IIx + Yll2 == IIxll2 + IIYIl2. Find all solutions of the equation (x, a)c == b, assuming that a, b, and c are given vectors in an inner product space.
Let F and G be two maps (not supposed to be linear or continuous) of an inner product space X into itself. What should we mean by the sum of the elements in an arbitrary subset A in X.
Section 2.2 Orthogonality and Bases 71
Section 2.2 Orthogonality and Bases
Every nontrivial inner-product space has an orthonor
We introduce an inner product in .
Section 2.2 Orthogonality and Bases We should be very cautious about writing
Prove that any orthonormal set in an inner product space can be enlarged to form an orthonormal basis.
An indexed set lUi : i E 1) in a Hilbert space is said to be stable if there exist positive constants A and B such that. Assume that it is orthogonal in this sense: If Xi f. Show by example that the law of Pythagoras in Theorem 1 can fail.
Section 2.3 Linear Functionals and Operators 81
The orthogonal projection P of a Hilbert space X onto a closed subspace Y is a bounded linear operator from X to X. If A is a bounded Jinear operator on a Hilbert space X (so A X -t X), then there exists a uniquely defined bounded linear operator A' such that.
Section 2.3 Linear Functiona/s and Operators For this it will be sufficient to prove
It is obvious that For each y in the unit sphere, define a functional ¢y is linear, and we also see that it is bounded since ¢y by writing ¢y(x).
IIxll=l sup I(Ax,x)1
- Section 2.3 Linear Functionals and Operators From the definit ion of m A III and a homogeneity argument, we obtain
- If A is Hermitian, then IIAII = m AI l !
- Lemma 3. Every continuous linear operator (from one normed linear space into another) having finite-dimensional range is compact
- Linear Functionals and Operators 87 Equation (3) suggests truncating the series that defines T in order to obtain
- Theorem 6. The null space of a bounded linear operator on a Hilbert space is the orthogonal complement of the range of its adjoint
- Lemma 4. A weakly Cauchy sequence in a Hilbert space is weakly convergent to a point in the Hilbert space
- Section 2.3 Linear Functionals and Operators
- Section 2.4 Spectral Theory 91
- Section 2.4 Spectral Theory
- Spectral Theory
- Let A be an operator on a Hilbert space such that A x =
- Section 2.4 Spectral Theory 101
The null space of a bounded linear operator in a Hilbert space is the orthogonal complement of its adjacent range. What are the eigenvalues of P. Give the spectral shape of P and 1 -P. Let A be a bounded linear operator in a Hilbert space.
B = P-IAP
Therefore, we often try to recast a differential equation as an equivalent integral equation in the hope that the transformed problem will be less difficult. Let Ax = -x" as before, but let the inner product space be the subspace of L2[O, 1TJ consisting of twice continuously differentiable functions satisfying x(O) = X(1T) = o.
If the homogeneous boundary value problem is known to have only the trivial solution, then B is also a left inverse of A.
Section 2.5 Sturm-Liouville Theory 109
The use of this formula is similar to the traditional method of solving the boundary value problem. Note that Theorem 2 has given us an alternative method to solve the inhomogeneous boundary value problem.
Section 2.5 Sturm-Liouville Theory
The Green's function for the above problem is charac- terized by these five properties
The previous theorem states that gt must solve the homogeneous differential equation in the intervals 0 < s < t < 1 and 0 < t < s < 1.
Section 2.5 Sturm-Liouville Theory 113
Show that the Wronskian for any two solutions of the equation (px' )' - qx = 0 is a scalar multiple of lip, and is therefore identically zero or never zero. Assume that the domain of A is the set of twice continuously differentiable functions on [0, IJ that have limits x(O) = x(l) = o.
Chapter 3
- The Frechet Derivative 1 15
- The Chain Rule and Mean Value Theorems 121 3.3 Newton's Method 125 3.3 Newton's Method 125
- Implicit Function Theorems 135
- Extremum Problems and Lagrange Multipliers 1 4 5
- The Calculus of Variations 152
If J is bounded in a neighborhood of x and if a linear map A has the property in equation (1), then A is a bounded linear.
Thus we set VO = x and Vi = Vi-I + hiei, where ei is the ith standard unit vector. Informally, we say that the Frechet derivative of f is given by the Jacobian matrix J of f at x: Jij = (DjJ;(x)).
Prove that the relation holds if the norms in X and Y are made equal. is a unique affine mapping tangent to I at xo. Prove that if I Fnkhet is differentiable at x, then Gateaux is differentiable at x and both derivatives are equal.
Section 3.2 The Chain Rule and Mean-Value Theorems 121
Let J be a continuous map over a compact interval [a, b] of the real line in a normed linear space Y.
Section 3.2 The Chain Rule and Mean-Value Theorems 123
Section 3.3 Newton's Method 125
For an iterative process, this is an extremely favorable condition, as it implies a doubling of the number of significant figures in the numerical solution at each step. To see how well it works, we can use a computer system such as Mathematica, Maple, or Matlab to obtain the Newton approximation to \1'2.
Section 3.3 Newton's Method 127
III )
Section 3.3 Newton '8 Method 129
Section 3.3 Newton's Method
Theorem 4. There is a neighborhood of x· such that the iteration sequence defined in Equation (7) converges to x' for arbitrary starting
Section 3.3 Newton's Method 133
From these equations it is clear that r(s, t)/ st is on the one hand a function of.
Section 3.4 Implicit FUnction Theorems
If F(xo, Yo) = 0 and F2(xo, Yo) I-0, then there is a continuously differentiable function J defined in a neighborhood of Xo such that. According to the implicit function theorem (with the roles of x and y reversed!), there exists a neighborhood N of Yo and.
Implicit Function Theorems
Let f be defined on an open set n in the direct sum space X = L�=I GlXj and take values in a normed space Y. Let f : n --t Y be a continuous differentiable map, where n is an open set in a Banach space, and Y is a normed linear space.
Extremum Problems and Lagrange Multipliers 147
In the next theorem, X and Y are normed linear spaces and Y is an ordered vector space. If Xo is a local maximum point of f on the set {x : G(x) � O} interior point of the positive cone, then there is a non-negative functional and if there is a h E X such that G(xo) + G '(xo)h is a G(x) � O} interior point of the positive cone, then there is a non-negative functional and if there is an h E X such that G(xo) + G'( xo)h is on.
152 Chapter 3 Calculus in Banach Spaces Now we conclude that. a) Use Lagrange multipliers to find the maximum of xy subject to x + Y = c. Use the method of Lagrange multipliers to find a point on the surface (x - y)2 - z2 = 1 as close as possible to the origin in IRJ .
Calculus of Variations 153 problems are not simple numbers but functions. We begin with some classical
But when he recognized the problem, he found he couldn't sleep until he discovered it, and after he did, he published the solution anonymously. However, Bernoulli knew immediately that the leader of the solution was Newton, and in a sense.
Calculus of Variations
If this superline must pass through the other given endpoint (b, j3), then A will have to satisfy the equation.
- If the terminal point is above the envelope, two catenaries of the family (6) pass through it. One of these is a local minimum in the problem but not
- For terminal points sufficiently far above the envelope, the upper catenary of the two passing through the point is the solution to the problem
This is the Brachistochrone Problem, except that the endpoint is allowed to be anywhere on a given vertical line.
Fermat's principle states that a light ray passing between two points will follow a path that minimizes the elapsed time. What is the path of a light ray from a point in the first medium to a point in the second.
Section 3.6 Calculus of Variations
1 1 , but there is no smaller length, since the infimum of acceptable lengths is 1, but it is not reached with an acceptable y. Methods based only on the use of necessary conditions never prove the existence of the solution.).
Chapter 4
It is clear that the coefficient matrix for this system is tridiagonal, because the equation contains only the three unknowns Vi-I, Vi and Vi+I. We write System (5) in the form Av = c, where A is the tridiagonal matrix and c now denotes the vector having Ci components.
This is a system of n linear equations in the unknowns x(Sj); it can be solved using standard methods. Two-stage approximations were made and the resulting function x is not the solution to the original problem.
Section 4.2 The Method of Iteration 177
It is assumed that w is continuous and that K(s, t, r) is continuous on the domain in 1R3 defined by the inequalities.
Section 4.2 The Method of Iteration 1 79
If f is continuous but does not satisfy the Lipschitz condition in Theorem 3, the conclusions of the theorem may not be valid. The correct way to look at this is as a single equation involving the function x : S � JRn where S is an interval of the form [0, bJ.
Section 4.2 The Method of Iteration 181
Thus F has at least one fixed point (namely �), and � can be obtained by iteration using the function Fm. Thus, it remains to be proved that the sequence x is a fixed point of Fm, and x = �. by the first part of the proof.
Section 4.2 The Method of Iteration
For what values of A can we be sure that the integral equation x(t) = A 11 est cosx(s) ds + tan t has a continuous solution in [O,I]. Prove that there is no contraction of X to X if X is a compact metric space that has at least two points.
Section 4.2 The Method of Itera.tion 185
Recall Neumann's theorem in section 1.5, page 28, which states that if a linear operator on a Banach space, A : X � For linear operator equations in a Banach space, one should not overlook the possibility of a Neumann series solution.
Methods Based on the Neumann Series
If the hypothesis 111 - ABII < 1 is satisfied, the Neumann series converges, partial sums in equation (8) converge, and by the theorem the sequence [xnl. Thus (since every linear operator on Rn is affected by a matrix) we have Xo = Bv, where B is a particular N x N matrix.
Section 4.3 Methods Based on the Neumann Series 189
Observe that the set of non-singular n x n matrices is open and dense in the set of all n x n. Prove that this statement can be false if the upper limit in the integral is replaced by 1.
Section 4.4 Projections and Projection Methods
The adjoint of a projection is also a projection
