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About 3307 results. 1294 free access solutions
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free/or 0.5$ |
| m82177 | Let X = [0, 2]. Show that the correspondence |
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| m82178 | Let X and Θ be metric spaces, and let f: X × Θ → X where
• X is complete
• For every θ ∊ Θ, the function fθ(x) = f (x, θ) is contraction mapping on X with modulus β
• f is continuous in θ, that is for every θ0 ∊ Θ, limθ→ θ0 fθ(x) = fθ0(x) for every x ∊ X
Then fθ has a unique fixed point xθ for every θ ∊ Θ and limθ→θ0 xθ = xθ0.
Although there are many direct methods for solving systems of linear equations, iterative methods are sometimes used in practice. The following exercise outlines one such method and devises a sufficient condition for convergence. |
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| m82179 | Let X and Y be Banach spaces. Any linear function f: X → Y is continuous if and only if its graph
graph (f) = {(x, y) : y = f(x), x ∊ X}
is a closed subset of X × Y. |
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| m82180 | Let X and Y be compact subsets of a finite-dimensional normed linear space, and let f be a continuous functional on X × Y. Then
if and only if there exists a point (x*, y*) ∈ X × Y such that
for every x ∈ X and y ∈ Y |
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| m82182 | Let X be a compact metric space. A closed subspace of C(X) is compact if and only if it is bounded and equicontinuous. |
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| m82183 | Let x be a feasible solution to the linear system
Ax = c, x ≥ 0
(23)
where A is an m × n matrix and c ∈ ℜm. Then
c = x1A1 + x2A2 + ... + xnAn
where Ai ∈ ℜm are the columns of A and and xi ≥ 0 for every i. Without loss of generality, assume that the first k components are positive and the rest are zero, that is,
c = x1A1 + x2A2 + ... + xkAk
with k ≤ n and xi > 0 for every a = 1, 2, . . . k.
1. The columns {A1, A2,..., Ak} are vectors in ℜm. If the columns {A1, A2,..., Ak} are linearly independent, then k < m and there exists a basic feasible solution.
2. If the vectors {A1, A2,..., Ak} are linearly dependent,
a. There exists a nonzero solution to the homogeneous system
y1A1 + y2A2 + ... + ykAk = 0
b. For t ∈ ℜ define = x - ty. is a solution to the non homogeneous system
Ax = c
c. Let
Then = x - tx is a feasible solution, that is, ≥ 0.
d. There exists h such that
is a feasible solution with one less positive component.
3. If there exists a feasible solution, there exists a basic feasible solution. |
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| m82184 | Let X be a finite-dimensional normed linear space, and {x1; x2,..., xn} any basis for X. The function f: ℜn → X defined by
is a linear homeomorphism (remark 2.12). That is,
• f is linear
• f is one-to-one and onto
• f and f-1 are continuous |
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| m82185 | Let X be a linear space and |
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| m82186 | Let X be a linear space. Two affine subsets S and T are parallel if one is a translate of the other, that is,
S = T + x for some x ∈ X
Show that the relation S is parallel to T is an equivalence relation in the set of affine subsets of X. |
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| m82187 | Let X be a partially ordered set which has a best element x*. If every nonempty subset S of a X has a greatest lower bound, then X is a complete lattice. |
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| m82190 | Let X be any set. Let F(X) denote the set of all functions on X. Show that F(X) is a linear space. |
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| m82191 | Let X be ordered by 7. |
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| m82193 | Let X be the product of n sets Xi each of which is ordered by 7i. Show that the lexicographic order is complete if and only if the component orders are complete. |
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| m82194 | Let X = C[0; 1] be the space of all continuous functions x(t) on the interval 0, 1. Show that the functional defined by
f(x) = x(1/2)
is a linear functional on C[0; 1]. |
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| m82195 | Let |
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| m82196 | Let |
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| m82197 | Let X, Y be Banach spaces. Their product X × Y with norm
||(x,y)|| = max {||x||, ||y||}
Is also a Banach space.
The natural space of economic models is Rn, the home space of consumption and production sets, which is a typical finite-dimensional normed linear space. In these spaces the interaction between linearity and topology is most acute, and many of the results obtained above can be sharpened. The most important results are summarized in the following proposition |
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| m82198 | Let |
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| m82199 | Let |
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| m82200 | Let X, Y, Z be linear spaces. The set BL(Y, Z) of all bounded linear functions from Y to Z is a linear space (exercise 3.33). Let BL(X, BL(Y, Z)) denote the set of bounded linear functions from X to the set BL(Y, Z). Show that
1. BL(X, BL(Y, Z)) is a linear space.
2. Let |
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