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Material Type: Assignment; Class: Sptp:Optimization; Subject: Mathematics; University: Duquesne University; Term: Fall 2008;
Typology: Assignments
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Math 491/CPMA 591 Inverse Problems Fall, 2008
Problems 3
Problem 1 The following bounds are not necessarily best possible; your answers may be different.
a) ‖h‖ 1 ≤ 2 π, ‖h‖ 2 ≤
2 π, ‖h‖∞ ≤ 1. b) ‖h‖ 1 = 2
2 π, ‖h‖ 2 = ∞, ‖h‖∞ = ∞. c) ‖h‖ 1 = 16 [(3/4)− 3 − 1], ‖h‖ 2 <
47 / 38 , ‖h‖∞ = 2^7 / 34. d) ‖h‖ 11 /e, ‖h‖ 2 = 1 sqrte, ‖h‖∞ = e−^1.
Problem 2 For one norms, if ‖K‖∞ < ∞, the basic calculation is the following:
‖h‖ 1 =
∫ (^) b
a
∫ (^) b
a
K(x, y)h(y)
dx
∫ (^) b
a
∫ (^) b
a
|K(x, y)h(y)| dydx
≤ ‖h‖∞‖K‖∞
∫ (^) b
a
∫ (^) b
a
dydx
= ‖h‖∞‖K‖∞(b − a)^2.
If ‖K‖∞ is not finite but ‖K(·, y)‖ 1 is, the basic calculation for one-norms becomes:
‖h‖ 1 =
∫ (^) b
a
∫ (^) b
a
K(x, y)h(y)
dx
∫ (^) b
a
∫ (^) b
a
|K(x, y)h(y)| dydx
≤ ‖h‖∞
∫ (^) b
a
‖K‖ 1 dx
= ‖h‖∞‖K‖ 1 (b − a).
The calculations for two norms and sup norms are similar. Using these techniques, you can show:
a) ‖q‖ 1 ≤ ‖h‖∞, ‖q‖ 2 ≤ ‖h‖∞, ‖q‖∞ ≤ ‖h‖∞. b) Similar. (More generally: whenever K is bounded and the region is finite, q will have 1,2, and sup norms. c) Here the kernel is not bounded. You can still that show that ‖q‖ 1 , ‖q‖ 2 , and ‖q‖∞ exist, and are bounded by some multiple of ‖h‖∞.
Problem 3 Calculations similar to the above yield:
a) ‖q‖ 1 , ‖q‖ 2 , and ‖q‖∞ exist and are bounded by some multiple of ‖h‖ 1. b) Ditto. c) Here, neither ‖q‖ 1 , ‖q‖ 2 , nor ‖q‖∞ need exist.
Problem 4 Recall that a norm is a positive function which satisfies the triangle inequality and is equal to zero if and only if its argument is zero. Thus:
a) not a norm (has non-zero kernel) b) norm (verify) c) not a norm (could be negative)
Problem 5 Note the kernel is separable, and thus any solution must lie in the span of the αi(x)’s. In this case, α 1 (x) = x and α 2 (x) = 1. Since x^2 + 1 does not lie in the span of the 1 and x, we conclude there is no solution.
Problem 6 Set ψˆi := y100+1^ for i = 1, 2 , · · ·. Note that ψˆi does not lie in the span of the βi (since it is impossible to write a high order polynomial as a sum of low order polynomials.) If we let P denote projection on the span of the βi and define
ψi(x) = ψ(ˆx) − P ( ψ(ˆx)),
then ψi is orthogonal to the βi for all i. Concretely, we have:
ψi(x) = ψˆ(x) −
j=
ψ(ˆx), βi(x)
βi(x)/‖βi(x)‖^2.
Problem 7 The kernel is separable, with α 1 (x) = sin x, α 2 (x) = sin 2x, and α 3 (x) = cos 2x. The left hand side of the equation is a linear combination of these elements, with coefficients 1,3, and 1, respectively. Thus the equation might have a solution. To find out, suppose f is a solution. Note that f must be a linear combination of the βi’s, where in this case β 1 = cos y, β 2 = cos 2y, and β 3 = sin 2y. Let Γ be the matrix whose (i, j)’th entry is 〈βi, βj 〉. Then
π/ 2 0 0 0 π/ 4 0 0 0 π/ 4
and the equation has a solution if and only if the matrix equation
f 1 f 2 f 3
has a solution. Clearly, the solution is f 1 = 2/π, f 2 = 12/π, and f 3 = 4/π.
Problem 8 As above, form the matrix Γ, this time with the (i, j)th entry equal to 〈βi, αj 〉. We are seeking a solution for some scalar λ (the eigenvalue) and vector y = [y 1 , y 2 ] (the eigenvector). The system has a solution if and only if the matrix equation
Γy = λy.
Since Γ =
we see from linear algebra that the eigenvalues will be ±8. The eigenvectors are v 1 = [−. 71 , .71] and v 2 = [. 71 , .71].