7.1 Lemma Let X be a linear space over C. Regarding X as a linear space over R., consider a real-linear functional u: X → R. Define f(x) = u(r)-iu(ir), x € X. Then f is a complex-linear functional on X. Proof: As u is real-linear, it is easy to see that f is also real-linear. In fact, f is complex-linear, since f(ix) = u(ix) - iu(-x) = u(ir) + iu(r) =i[u(r) - iu(ix)] = if(x) for all z € X. The following result is of crucial importance. 7.2 Lemma Let X be a linear space over K and Y be a subspace of X which is not a hyperspace in X. If z₁ and ₂ are in X but not in Y, then there is some x in X such that for all t = [0, 1], ta₁ + (1-t)x Y and tr₂+(1-t)x & Y. If X is a normed space, then the complement Y of Y is connected. Proof: If tæ₁ + (1 - t)x₂ # Y for all t € (0, 1), then clearly we can let x = 1₂. Assume now that there is some s € (0, 1) such that sa₁ + (1 - s)x₂ € Y. Then it is easy to see that span {Y, ₁} = span {Y, 2₂}. Since Y is not a hyperspace in X, span {Y, 2₁} # X. Let z X be such that x span {Y, ₁}. Let, if possible, y = tx₁ + (1 – t)x belong to Y for some t€ (0,1). Thenx=(y-tr₁)/(1 t) would belong to span {Y, ₁}, contrary to our choice of z. Hence tx₁ + (1-t)x & Y Raquest explain th part.
7.1 Lemma Let X be a linear space over C. Regarding X as a linear space over R., consider a real-linear functional u: X → R. Define f(x) = u(r)-iu(ir), x € X. Then f is a complex-linear functional on X. Proof: As u is real-linear, it is easy to see that f is also real-linear. In fact, f is complex-linear, since f(ix) = u(ix) - iu(-x) = u(ir) + iu(r) =i[u(r) - iu(ix)] = if(x) for all z € X. The following result is of crucial importance. 7.2 Lemma Let X be a linear space over K and Y be a subspace of X which is not a hyperspace in X. If z₁ and ₂ are in X but not in Y, then there is some x in X such that for all t = [0, 1], ta₁ + (1-t)x Y and tr₂+(1-t)x & Y. If X is a normed space, then the complement Y of Y is connected. Proof: If tæ₁ + (1 - t)x₂ # Y for all t € (0, 1), then clearly we can let x = 1₂. Assume now that there is some s € (0, 1) such that sa₁ + (1 - s)x₂ € Y. Then it is easy to see that span {Y, ₁} = span {Y, 2₂}. Since Y is not a hyperspace in X, span {Y, 2₁} # X. Let z X be such that x span {Y, ₁}. Let, if possible, y = tx₁ + (1 – t)x belong to Y for some t€ (0,1). Thenx=(y-tr₁)/(1 t) would belong to span {Y, ₁}, contrary to our choice of z. Hence tx₁ + (1-t)x & Y Raquest explain th part.
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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![7.1 Lemma
Let X be a linear space over C. Regarding X as a linear space over
R, consider a real-linear functional u: X → R. Define
f(x) = u(x) - iu(ix), x € X.
Then f is a complex-linear functional on X.
Proof:
As u is real-linear, it is easy to see that f is also real-linear. In fact,
f is complex-linear, since
f(ix) = u(ix) - iu(-x) = u(ix) + iu(r) =i[u(r) - iu(ix)] = if(x)
for all z € X.
The following result is of crucial importance.
0
7.2 Lemma
Let X be a linear space over K and Y be a subspace of X which is
not a hyperspace in X. If ₁ and ₂ are in X but not in Y, then there
is some x in X such that for all t = [0, 1],
ta₁ + (1 - t)x Y and tx₂+(1-t)x & Y.
If X is a normed space, then the complement Y of Y is connected.
Proof:
If tx₁ + (1 -t)x₂ Y for all t € (0, 1), then clearly we can let x = I2.
Assume now that there is some s € (0, 1) such that sx₁ + (1 - s)x₂ € Y.
Then it is easy to see that span {Y,₁} = span {Y, 2₂}. Since Y is
not a hyperspace in X, span {Y, 2₁} # X. Let 2 € X be such that
x span {Y, ₁}. Let, if possible, y ta₁ + (1 t)x belong to Y
for some t€ (0,1). Then x = (y tr₁)/(1 t) would belong to
span {Y, ₁}, contrary to our choice of z. Hence tx₁ + (1 - t)a Y
-
Regreest explain this
part.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F7ea5806e-65a0-4d46-8891-4f3dfc1f4684%2Fb75xx2_processed.png&w=3840&q=75)
Transcribed Image Text:7.1 Lemma
Let X be a linear space over C. Regarding X as a linear space over
R, consider a real-linear functional u: X → R. Define
f(x) = u(x) - iu(ix), x € X.
Then f is a complex-linear functional on X.
Proof:
As u is real-linear, it is easy to see that f is also real-linear. In fact,
f is complex-linear, since
f(ix) = u(ix) - iu(-x) = u(ix) + iu(r) =i[u(r) - iu(ix)] = if(x)
for all z € X.
The following result is of crucial importance.
0
7.2 Lemma
Let X be a linear space over K and Y be a subspace of X which is
not a hyperspace in X. If ₁ and ₂ are in X but not in Y, then there
is some x in X such that for all t = [0, 1],
ta₁ + (1 - t)x Y and tx₂+(1-t)x & Y.
If X is a normed space, then the complement Y of Y is connected.
Proof:
If tx₁ + (1 -t)x₂ Y for all t € (0, 1), then clearly we can let x = I2.
Assume now that there is some s € (0, 1) such that sx₁ + (1 - s)x₂ € Y.
Then it is easy to see that span {Y,₁} = span {Y, 2₂}. Since Y is
not a hyperspace in X, span {Y, 2₁} # X. Let 2 € X be such that
x span {Y, ₁}. Let, if possible, y ta₁ + (1 t)x belong to Y
for some t€ (0,1). Then x = (y tr₁)/(1 t) would belong to
span {Y, ₁}, contrary to our choice of z. Hence tx₁ + (1 - t)a Y
-
Regreest explain this
part.
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