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Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Request explain the steps marked in red
![THEOREM 20.1. The adjoint A* of an arbitrary function, A, XDY,
where X and Y are inner product spaces and D₁ is dense in X, is a closed linear
transformation.
Our next theorem appeals to one's love of symmetry.
THEOREM 20.2. Let A, X, and Y be as in the preceding theorem. In addition,
suppose A-¹ exists and that D₁-1 is dense in Y. Then
(A-¹)=(A) ¹.
Proof. To show the equality of these two functions, we must show that their
domains are the same and that they agree on every vector in the domain. First,
suppose x € D₁ and that y € D-1).. We have
(x, y) = (A¹Ax, y) = (Ax, (A¹)*y).
Therefore [see Eqs. (20.6) and (20.7)], since x is any vector in D₁,
(A-¹)*y € D₁.
and
A*(A¹)*y = y.
On the other hand, suppose x € D₁-1 and ye D₁.. In this case we have
(x, y) = (AA¯¹x, y) = (A¯¹x, A*y),
A*y € D₁A-1).
(A¹)*A*y = y.
which implies
(20.8)
Req explain on
(20.9)
and
(20.10)
(20.11)
the mapping A* being called the adjoint of A. In summary if, for ye Y, there exists
ze Y such that
(Ax, y) = (x, z)
(20.6)
for all x € D₁, we say that
yεD₁.
and
A" y = 2.
(20.7)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F9af0d132-d473-413c-8b6f-3ba7075bc4d8%2Fgjhni6g_processed.png&w=3840&q=75)
Transcribed Image Text:THEOREM 20.1. The adjoint A* of an arbitrary function, A, XDY,
where X and Y are inner product spaces and D₁ is dense in X, is a closed linear
transformation.
Our next theorem appeals to one's love of symmetry.
THEOREM 20.2. Let A, X, and Y be as in the preceding theorem. In addition,
suppose A-¹ exists and that D₁-1 is dense in Y. Then
(A-¹)=(A) ¹.
Proof. To show the equality of these two functions, we must show that their
domains are the same and that they agree on every vector in the domain. First,
suppose x € D₁ and that y € D-1).. We have
(x, y) = (A¹Ax, y) = (Ax, (A¹)*y).
Therefore [see Eqs. (20.6) and (20.7)], since x is any vector in D₁,
(A-¹)*y € D₁.
and
A*(A¹)*y = y.
On the other hand, suppose x € D₁-1 and ye D₁.. In this case we have
(x, y) = (AA¯¹x, y) = (A¯¹x, A*y),
A*y € D₁A-1).
(A¹)*A*y = y.
which implies
(20.8)
Req explain on
(20.9)
and
(20.10)
(20.11)
the mapping A* being called the adjoint of A. In summary if, for ye Y, there exists
ze Y such that
(Ax, y) = (x, z)
(20.6)
for all x € D₁, we say that
yεD₁.
and
A" y = 2.
(20.7)
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