Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral...
Transcript of Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral...
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Integral Transforms - p. 1/38
Integral TransformsMath 2025
Gestur OlafssonMathematics Department
Louisiana State University
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Vector Spaces over R Integral Transforms - p. 2/38
Chapter 1
Vector Spaces over R
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Vector Spaces over R
Definition. vector space over R is a set V with operations ofaddition + and scalar multiplication · satisfying the followingproperties:
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Vector Spaces over R
Definition. vector space over R is a set V with operations ofaddition + and scalar multiplication · satisfying the followingproperties:■ A1 (Closure of addition)
For all u, v ∈ V, u + v is defined and u + v ∈ V .
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Vector Spaces over R Integral Transforms - p. 3/38
Vector Spaces over R
Definition. vector space over R is a set V with operations ofaddition + and scalar multiplication · satisfying the followingproperties:■ A1 (Closure of addition)
For all u, v ∈ V, u + v is defined and u + v ∈ V .
■ A2 (Commutativity for addition)u + v = v + u for all u, v ∈ V .
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Vector SpacesDefinitionImmediate resultsExamplesR
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Vector Spaces over R Integral Transforms - p. 3/38
Vector Spaces over R
Definition. vector space over R is a set V with operations ofaddition + and scalar multiplication · satisfying the followingproperties:■ A1 (Closure of addition)
For all u, v ∈ V, u + v is defined and u + v ∈ V .
■ A2 (Commutativity for addition)u + v = v + u for all u, v ∈ V .
■ A3 (Associativity for addition)u + (v + w) = (u + v) + w for all u, v, w ∈ V .
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Vector SpacesDefinitionImmediate resultsExamplesR
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Vector Spaces over R Integral Transforms - p. 3/38
Vector Spaces over R
Definition. vector space over R is a set V with operations ofaddition + and scalar multiplication · satisfying the followingproperties:■ A1 (Closure of addition)
For all u, v ∈ V, u + v is defined and u + v ∈ V .
■ A2 (Commutativity for addition)u + v = v + u for all u, v ∈ V .
■ A3 (Associativity for addition)u + (v + w) = (u + v) + w for all u, v, w ∈ V .
■ A4 (Existence of additive identity)
There exists an element ~0 such that u +~0 = u for allu ∈ V .
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Vector SpacesDefinitionImmediate resultsExamplesR
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Vector Spaces over R Integral Transforms - p. 3/38
Vector Spaces over R
Definition. vector space over R is a set V with operations ofaddition + and scalar multiplication · satisfying the followingproperties:■ A1 (Closure of addition)
For all u, v ∈ V, u + v is defined and u + v ∈ V .
■ A2 (Commutativity for addition)u + v = v + u for all u, v ∈ V .
■ A3 (Associativity for addition)u + (v + w) = (u + v) + w for all u, v, w ∈ V .
■ A4 (Existence of additive identity)
There exists an element ~0 such that u +~0 = u for allu ∈ V .
■ A5 (Existence of additive inverse)For each u ∈ V , there exists an element -denoted by −u-
such that u + (−u) = ~0.
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Vector Spaces over R Integral Transforms - p. 4/38
Vector Spaces over R
Definition. A vector space over R is a set V with operationsof addition + and scalar multiplication · satisfying thefollowing properties:
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Vector Spaces over R
Definition. A vector space over R is a set V with operationsof addition + and scalar multiplication · satisfying thefollowing properties:■ M1 (Closure for scalar multiplication)
For each number r and each u ∈ V , r · u is defined andr · u ∈ V .
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Vector Spaces over R Integral Transforms - p. 4/38
Vector Spaces over R
Definition. A vector space over R is a set V with operationsof addition + and scalar multiplication · satisfying thefollowing properties:■ M1 (Closure for scalar multiplication)
For each number r and each u ∈ V , r · u is defined andr · u ∈ V .
■ M2 (Multiplication by 1)1 · u = u for all u ∈ V .
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Vector Spaces over R Integral Transforms - p. 4/38
Vector Spaces over R
Definition. A vector space over R is a set V with operationsof addition + and scalar multiplication · satisfying thefollowing properties:■ M1 (Closure for scalar multiplication)
For each number r and each u ∈ V , r · u is defined andr · u ∈ V .
■ M2 (Multiplication by 1)1 · u = u for all u ∈ V .
■ M3 (Associativity for multiplication)r · (s · u) = (r · s) · u for r, s ∈ R and all u ∈ V .
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Vector SpacesDefinitionImmediate resultsExamplesR
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Vector Spaces over R Integral Transforms - p. 4/38
Vector Spaces over R
Definition. A vector space over R is a set V with operationsof addition + and scalar multiplication · satisfying thefollowing properties:■ M1 (Closure for scalar multiplication)
For each number r and each u ∈ V , r · u is defined andr · u ∈ V .
■ M2 (Multiplication by 1)1 · u = u for all u ∈ V .
■ M3 (Associativity for multiplication)r · (s · u) = (r · s) · u for r, s ∈ R and all u ∈ V .
■ D1 (First distributive property)r · (u + v) = r · u + r · v for all r ∈ R and all u, v ∈ V .
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Vector Spaces over R Integral Transforms - p. 4/38
Vector Spaces over R
Definition. A vector space over R is a set V with operationsof addition + and scalar multiplication · satisfying thefollowing properties:■ M1 (Closure for scalar multiplication)
For each number r and each u ∈ V , r · u is defined andr · u ∈ V .
■ M2 (Multiplication by 1)1 · u = u for all u ∈ V .
■ M3 (Associativity for multiplication)r · (s · u) = (r · s) · u for r, s ∈ R and all u ∈ V .
■ D1 (First distributive property)r · (u + v) = r · u + r · v for all r ∈ R and all u, v ∈ V .
■ D2 (Second distributive property)(r + s) · u = r · u + s · u for all r, s ∈ R and all u ∈ V .
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Some immediate results
Remark. The zero element ~0 is unique, i.e., if ~01, ~02 ∈ V aresuch that
u + ~01 = u + ~02 = u,∀u ∈ V
then ~01 = ~02.
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Some immediate results
Remark. The zero element ~0 is unique, i.e., if ~01, ~02 ∈ V aresuch that
u + ~01 = u + ~02 = u,∀u ∈ V
then ~01 = ~02.
Proof. We have ~01 = ~01 + ~02 = ~02 + ~01 = ~02
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Some immediate results
Remark. The zero element ~0 is unique, i.e., if ~01, ~02 ∈ V aresuch that
u + ~01 = u + ~02 = u,∀u ∈ V
then ~01 = ~02.
Proof. We have ~01 = ~01 + ~02 = ~02 + ~01 = ~02
Lemma. Let u ∈ V , then 0 · u = ~0.
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Some immediate results
Remark. The zero element ~0 is unique, i.e., if ~01, ~02 ∈ V aresuch that
u + ~01 = u + ~02 = u,∀u ∈ V
then ~01 = ~02.
Proof. We have ~01 = ~01 + ~02 = ~02 + ~01 = ~02
Lemma. Let u ∈ V , then 0 · u = ~0.
Proof.
u + 0 · u = 1 · u + 0 · u
= (1 + 0) · u
= 1 · u
= u
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Some immediate results
Remark. The zero element ~0 is unique, i.e., if ~01, ~02 ∈ V aresuch that
u + ~01 = u + ~02 = u,∀u ∈ V
then ~01 = ~02.
Proof. We have ~01 = ~01 + ~02 = ~02 + ~01 = ~02
Lemma. Let u ∈ V , then 0 · u = ~0.
Proof.
Thus ~0 = u + (−u) = (0 · u + u) + (−u)
= 0 · u + (u + (−u))
= 0 · u +~0
= 0 · u
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Lemma. a) The element −u is unique.
b) −u = (−1) · u.
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Lemma. a) The element −u is unique.
b) −u = (−1) · u.
Proof of part (b).
u + (−1) · u = 1 · u + (−1) · u
= (1 + (−1)) · u
= 0 · u
= ~0
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Examples
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Rn as column vectors
Before examining the axioms in more detail, let usdiscuss two examples.
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Rn as column vectors
Example. Let V = Rn ,considered as column vectors
Rn = {
x1
x2
...
xn
|x1, x2, . . . , xn ∈ R}
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Rn as column vectors
Example. Let V = Rn .Then for
u =
x1
...
xn
, v =
y1
...
yn
∈ Rn and r ∈ R :
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Rn as column vectors
Example. Let V = Rn .Then for
u =
x1
...
xn
, v =
y1
...
yn
∈ Rn and r ∈ R :
Define
u + v =
x1 + y1
...
xn + yn
and r · u =
rx1
...
rxn
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Rn as column vectors
Example. Let V = Rn .Then for
u =
x1
...
xn
, v =
y1
...
yn
∈ Rn and r ∈ R :
Note that the zero vector and the additive inverse of u aregiven by:
~0 =
0...
0
, −u =
−x1
...
−x2
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Rn as row vectors
Remark. Rn can be considered as the space of all rowvectors.
Rn = {(x1, . . . , xn) | x1, . . . , xn ∈ R}
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Rn as row vectors
Remark. Rn can be considered as the space of all rowvectors.
Rn = {(x1, . . . , xn) | x1, . . . , xn ∈ R}
The addition and scalar multiplication is again givencoordinate wise
(x1, . . . , xn) + (y1, . . . , yn) = (x1 + y1, . . . , xn + yn)
r · (x1, . . . , xn) = (rx1, . . . , rxn)
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Example. If ~x = (2, 1, 3), ~y = (−1, 2,−2) and r = −4find ~x + ~y and r · ~x.
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Example. If ~x = (2, 1, 3), ~y = (−1, 2,−2) and r = −4find ~x + ~y and r · ~x.
Solution.
~x + ~y = (2, 1, 3) + (−1, 2,−2)
= (2 − 1, 1 + 2, 3 − 2)
= (1, 3, 1)
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Example. If ~x = (2, 1, 3), ~y = (−1, 2,−2) and r = −4find ~x + ~y and r · ~x.
Solution.
~x + ~y = (2, 1, 3) + (−1, 2,−2)
= (2 − 1, 1 + 2, 3 − 2)
= (1, 3, 1)
r · ~x = −4 · (2, 1, 3) = (−8,−4,−12).
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Remark.
(x1, . . . , xn) + (0, . . . , 0) = (x1 + 0, . . . , xn + 0)
= (x1, . . . , xn)
So the additive identity is ~0 = (0, . . . , 0).
Note also that
0 · (x1, . . . , xn) = (0x1, . . . , 0xn)
= (0, . . . , 0)
for all (x1, . . . , xn) ∈ Rn.
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Vector space of real-valued functions
Example. Let A be the interval [0, 1) and V be the space offunctions f : A −→ R,i.e.,
V = {f : [0, 1) −→ R}
Define addition and scalar multiplication by
(f + g)(x) = f(x) + g(x)
(r · f)(x) = rf(x)
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Vector space of real-valued functions
Example. Let A be the interval [0, 1) and V be the space offunctions f : A −→ R,i.e.,
V = {f : [0, 1) −→ R}
Define addition and scalar multiplication by
(f + g)(x) = f(x) + g(x)
(r · f)(x) = rf(x)
For instance, the function f(x) = x4 is an element of V andso are
g(x) = x + 2x2, h(x) = cos x, k(x) = ex
We have (f + g)(x) = x + 2x2 + x4.
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Vector SpacesDefinitionImmediate resultsExamplesR
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Vector Spaces over R Integral Transforms - p. 12/38
Vector space of real-valued functions
Example. Let A be the interval [0, 1) and V be the space offunctions f : A −→ R,i.e.,
V = {f : [0, 1) −→ R}
Define addition and scalar multiplication by
(f + g)(x) = f(x) + g(x)
(r · f)(x) = rf(x)
Remark. (a) The zero element is the function ~0 whichassociates to each x the number 0:
~0(x) = 0 for all x ∈ [0, 1)
Proof.(f +~0)(x) = f(x) +~0(x) = f(x) + 0 = f(x).
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Vector SpacesDefinitionImmediate resultsExamplesR
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Vector Spaces over R Integral Transforms - p. 12/38
Vector space of real-valued functions
Example. Let A be the interval [0, 1) and V be the space offunctions f : A −→ R,i.e.,
V = {f : [0, 1) −→ R}
Define addition and scalar multiplication by
(f + g)(x) = f(x) + g(x)
(r · f)(x) = rf(x)
Remark. (b) The additive inverse is the function−f : x 7→ −f(x).
Proof. (f + (−f))(x) = f(x) − f(x) = 0 for all x.
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The vector space V A
Example. Instead of A = [0, 1) we can take any set A 6= ∅,and we can replace R by any vector space V . We set
V A = {f : A −→ V }
and set
(f + g)(x) = f(x)+g(x)
(r · f)(x) = r·f(x)
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The vector space V A
Example. Instead of A = [0, 1) we can take any set A 6= ∅,and we can replace R by any vector space V . We set
V A = {f : A −→ V }
and set
(f + g)(x) = f(x)+g(x)
(r · f)(x) = r·f(x)
multiplication in V addition in V
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The vector space V A
Example. Instead of A = [0, 1) we can take any set A 6= ∅,and we can replace R by any vector space V . We set
V A = {f : A −→ V }
and set
(f + g)(x) = f(x)+g(x)
(r · f)(x) = r·f(x)
Remark. (a) The zero element is the function which
associates to each x the vector ~0:
0 : x 7→ ~0Proof
(f + 0)(x) = f(x) + 0(x)
= f(x) +~0 = f(x)
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Remark.
(b) Here we prove that + is associative:
Proof. Let f, g, h ∈ V A. Then
[(f + g) + h](x) = (f + g)(x) + h(x)
= (f(x) + g(x)) + h(x)
= f(x) + (g(x) + h(x)) associativity inV
= f(x) + (g + h)(x)
= [f + (g + h)](x)
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Exercises
Let V = R4. Evaluate the following:
a) (2,−1, 3, 1) + (3,−1, 1,−1).b) (2, 1, 5,−1) − (3, 1, 2,−2).c) 10 · (2, 0,−1, 1).d) (1,−2, 3, 1) + 10 · (1,−1, 0, 1) − 3 · (0, 2, 1,−2).e) x1 · (1, 0, 0, 0) + x2 · (0, 1, 0, 0) + x3 · (0, 0, 1, 0) +
x4 · (0, 0, 0, 1).
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SubspacesDefinitionExamples
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Chapter 2
Subspaces
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SubspacesDefinitionExamples
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Subspace of a vector space
In most applications we will be working with asubset W of a vector space V such that W itself isa vector space.
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Subspaces Integral Transforms - p. 17/38
Subspace of a vector space
In most applications we will be working with asubset W of a vector space V such that W itself isa vector space.
Question: Do we have to test all the axioms to findout if W is a vector space?
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Subspaces Integral Transforms - p. 17/38
Subspace of a vector space
In most applications we will be working with asubset W of a vector space V such that W itself isa vector space.
Question: Do we have to test all the axioms to findout if W is a vector space?
The answer is NO.
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Subspaces Integral Transforms - p. 17/38
Subspace of a vector space
In most applications we will be working with asubset W of a vector space V such that W itself isa vector space.
Question: Do we have to test all the axioms to findout if W is a vector space?
The answer is NO.
Theorem. Let W 6= ∅ be a subset of a vector space V .Then W , with the addition and scalar multiplication as V , is avector space if and only if:■ u + v ∈ W for all u, v ∈ W (or W + W ⊆ W )
■ r · u ∈ W for all r ∈ R and all u ∈ W (or RW ⊆ W ).
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Subspaces of R2
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Subspaces Integral Transforms - p. 17/38
Subspace of a vector space
In most applications we will be working with asubset W of a vector space V such that W itself isa vector space.
Question: Do we have to test all the axioms to findout if W is a vector space?
The answer is NO.
Theorem. Let W 6= ∅ be a subset of a vector space V .Then W , with the addition and scalar multiplication as V , is avector space if and only if:■ u + v ∈ W for all u, v ∈ W (or W + W ⊆ W )
■ r · u ∈ W for all r ∈ R and all u ∈ W (or RW ⊆ W ).
In this case we say that W is a subspace of V .
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Proof. Assume that W + W ⊆ W and RW ⊆ W .To show that W is a vector space we have to show that all the10 axioms hold for W . But that follows because the axiomshold for V and W is a subset of V :
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Proof. Assume that W + W ⊆ W and RW ⊆ W .To show that W is a vector space we have to show that all the10 axioms hold for W . But that follows because the axiomshold for V and W is a subset of V :■ A1 (Commutativity of addition)
For u, v ∈ W , we have u + v = v + u. This is becauseu, v are also in V and commutativity holds in V .
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Subspaces Integral Transforms - p. 18/38
Proof. Assume that W + W ⊆ W and RW ⊆ W .To show that W is a vector space we have to show that all the10 axioms hold for W . But that follows because the axiomshold for V and W is a subset of V :■ A1 (Commutativity of addition)
For u, v ∈ W , we have u + v = v + u. This is becauseu, v are also in V and commutativity holds in V .
■ A4 (Existence of additive identity)Take any vector u ∈ W . Then by assumption
0 · u = ~0 ∈ W . Hence ~0 ∈ W .
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Proof. Assume that W + W ⊆ W and RW ⊆ W .To show that W is a vector space we have to show that all the10 axioms hold for W . But that follows because the axiomshold for V and W is a subset of V :■ A1 (Commutativity of addition)
For u, v ∈ W , we have u + v = v + u. This is becauseu, v are also in V and commutativity holds in V .
■ A4 (Existence of additive identity)Take any vector u ∈ W . Then by assumption
0 · u = ~0 ∈ W . Hence ~0 ∈ W .
■ A5 (Existence of additive inverse)If u ∈ W then −u = (−1) · u ∈ W .
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Proof. Assume that W + W ⊆ W and RW ⊆ W .To show that W is a vector space we have to show that all the10 axioms hold for W . But that follows because the axiomshold for V and W is a subset of V :■ A1 (Commutativity of addition)
For u, v ∈ W , we have u + v = v + u. This is becauseu, v are also in V and commutativity holds in V .
■ A4 (Existence of additive identity)Take any vector u ∈ W . Then by assumption
0 · u = ~0 ∈ W . Hence ~0 ∈ W .
■ A5 (Existence of additive inverse)If u ∈ W then −u = (−1) · u ∈ W .
■ One can check that the other axioms follow in the same way.
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Examples
Usually the situation is that we are given a vectorspace V and a subset of vectors W satisfyingsome conditions and we need to see if W is asubspace of V .
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Examples
Usually the situation is that we are given a vectorspace V and a subset of vectors W satisfyingsome conditions and we need to see if W is asubspace of V .
W = {v ∈ V : some conditions on v}
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Examples
Usually the situation is that we are given a vectorspace V and a subset of vectors W satisfyingsome conditions and we need to see if W is asubspace of V .
W = {v ∈ V : some conditions on v}
We will then have to show that
u, v ∈ W u + v
r ∈ R r · u
}
Satisfy the same conditions.
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Lines through the origin as subspaces of R2
Example.
V = R2,
W = {(x, y)|y = kx} for a given k
= line through (0, 0) with slope k.
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Lines through the origin as subspaces of R2
Example.
V = R2,
W = {(x, y)|y = kx} for a given k
= line through (0, 0) with slope k.
To see that W is in fact a subspace of R2:Let u = (x1, y1), v = (x2, y2) ∈ W . Then y1 = kx1 andy2 = kx2
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Lines through the origin as subspaces of R2
Example.
V = R2,
W = {(x, y)|y = kx} for a given k
= line through (0, 0) with slope k.
To see that W is in fact a subspace of R2:Let u = (x1, y1), v = (x2, y2) ∈ W . Then y1 = kx1 andy2 = kx2
and
u + v = (x1 + x2, y1 + y2)
= (x1 + x2, kx1 + kx2)
= (x1 + x2, k(x1 + x2)) ∈ W
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Subspaces Integral Transforms - p. 20/38
Lines through the origin as subspaces of R2
Example.
V = R2,
W = {(x, y)|y = kx} for a given k
= line through (0, 0) with slope k.
To see that W is in fact a subspace of R2:Let u = (x1, y1), v = (x2, y2) ∈ W . Then y1 = kx1 andy2 = kx2
and
u + v = (x1 + x2, y1 + y2)
= (x1 + x2, kx1 + kx2)
= (x1 + x2, k(x1 + x2)) ∈ W
Similarly, r · u = (rx1, ry1) = (rx1, krx1) ∈ W
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Subspaces of R2
So what are the subspaces of R2?
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 21/38
Subspaces of R2
So what are the subspaces of R2?1. {0}
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 21/38
Subspaces of R2
So what are the subspaces of R2?1. {0}2. Lines. But only those that contain (0, 0). Why?
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 21/38
Subspaces of R2
So what are the subspaces of R2?1. {0}2. Lines. But only those that contain (0, 0). Why?3. R2
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 21/38
Subspaces of R2
So what are the subspaces of R2?1. {0}2. Lines. But only those that contain (0, 0). Why?3. R2
Remark (First test). If W is a subspace, then ~0 ∈ W .
Thus: If ~0 /∈ W , then W is not a subspace.
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 21/38
Subspaces of R2
So what are the subspaces of R2?1. {0}2. Lines. But only those that contain (0, 0). Why?3. R2
Remark (First test). If W is a subspace, then ~0 ∈ W .
Thus: If ~0 /∈ W , then W is not a subspace.
This is why a line not passing through (0, 0) cannot be a subspace of R2.
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 22/38
A subset of R2 that is not a subspace
Warning. We can not conclude from the fact that ~0 ∈ W , thatW is a subspace.
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 22/38
A subset of R2 that is not a subspace
Warning. We can not conclude from the fact that ~0 ∈ W , thatW is a subspace.
Example. Lets consider the following subset of R2:
W = {(x, y)|x2 − y2 = 0}
Is W a subspace of R2? Why?
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 22/38
A subset of R2 that is not a subspace
Warning. We can not conclude from the fact that ~0 ∈ W , thatW is a subspace.
Example. Lets consider the following subset of R2:
W = {(x, y)|x2 − y2 = 0}
Is W a subspace of R2? Why?
The answer is NO.
We have (1, 1) and (1,−1) ∈ W but(1, 1) + (1,−1) = (2, 0) /∈ W . i.e., W is not closed underaddition.
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 22/38
A subset of R2 that is not a subspace
Warning. We can not conclude from the fact that ~0 ∈ W , thatW is a subspace.
Example. Lets consider the following subset of R2:
W = {(x, y)|x2 − y2 = 0}
Is W a subspace of R2? Why?
The answer is NO.
We have (1, 1) and (1,−1) ∈ W but(1, 1) + (1,−1) = (2, 0) /∈ W . i.e., W is not closed underaddition.
Notice that (0, 0) ∈ W and W is closed under multiplicationby scalars.
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 23/38
Subspaces of R3
What are the subspaces of R3?
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 23/38
Subspaces of R3
What are the subspaces of R3?1. {0} and R3.
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 23/38
Subspaces of R3
What are the subspaces of R3?1. {0} and R3.2. Planes: A plane W ⊆ R3 is given by a normal
vector (a, b, c) and its distance from (0, 0, 0) or
W = {(x, y, z)| ax + by + cz = p︸ ︷︷ ︸
}
condition on (x, y, z)
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 23/38
Subspaces of R3
What are the subspaces of R3?1. {0} and R3.2. Planes: A plane W ⊆ R3 is given by a normal
vector (a, b, c) and its distance from (0, 0, 0) or
W = {(x, y, z)| ax + by + cz = p︸ ︷︷ ︸
}
condition on (x, y, z)
For W to be a subspace, (0, 0, 0) must be in W
by the first test. Thus
p = a · 0 + b · 0 + c · 0 = 0
or
p = 0
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 24/38
Planes containing the origin
A plane containing (0, 0, 0) is indeed a subspace ofR3.
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 24/38
Planes containing the origin
A plane containing (0, 0, 0) is indeed a subspace ofR3.
Proof. Let (x1, y1, z1) and (x2, y2, z2) ∈ W . Then
ax1 + by1 + cz1 = 0
ax2 + by2 + cz2 = 0
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 24/38
Planes containing the origin
A plane containing (0, 0, 0) is indeed a subspace ofR3.
Proof. Let (x1, y1, z1) and (x2, y2, z2) ∈ W . Then
ax1 + by1 + cz1 = 0
ax2 + by2 + cz2 = 0
Then we havea(x1 + x2) + b(y1 + y2) + c(z1 + z2)
= (ax1 + by1 + cz1)︸ ︷︷ ︸
+(ax2 + by2 + cz2)︸ ︷︷ ︸
0 0= 0
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 24/38
Planes containing the origin
A plane containing (0, 0, 0) is indeed a subspace ofR3.
Proof. Let (x1, y1, z1) and (x2, y2, z2) ∈ W . Then
ax1 + by1 + cz1 = 0
ax2 + by2 + cz2 = 0
Then we havea(x1 + x2) + b(y1 + y2) + c(z1 + z2)
= (ax1 + by1 + cz1)︸ ︷︷ ︸
+(ax2 + by2 + cz2)︸ ︷︷ ︸
0 0= 0
and a(rx1) + b(ry1) + c(rz1) = r(ax1 + by1 + cz1)
= 0
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 25/38
Summary of subspaces of R3
1. {0} and R3.
2. Planes containing (0, 0, 0).
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 25/38
Summary of subspaces of R3
1. {0} and R3.
2. Planes containing (0, 0, 0).
3. Lines containing (0, 0, 0).(Intersection of two planes containing (0, 0, 0))
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 26/38
Exercises
Determine whether the given subset of Rn is asubspace or not (Explain):
a) W = {(x, y) ∈ R2| xy = 0}.b) W = {(x, y, z) ∈ R3| 3x + 2y2 + z = 0}.c) W = {(x, y, z) ∈ R3| 2x + 3y − z = 0}.
d) The set of all vectors (x1, x2, x3)satisfying
2x3 = x1 − 10x2
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SubspacesDefinitionExamples
Subspaces of R2
Subspaces of R3
Exercises
Subspaces Integral Transforms - p. 26/38
Exercises
Determine whether the given subset of Rn is asubspace or not (Explain):
e) The set of all vectors in R4 satisfying the systemof linear equations
2x1 + 3x2 + 5x4 = 0
x1 + x2 − 3x3 = 0
f) The set of all points (x1, x2, x3, x4) ∈ R4
satisfying
x1 + 2x2 + 3x3 + x4 = −1
![Page 83: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/83.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 27/38
Chapter 3
Vector Spaces of Functions
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 28/38
Space of Continuous Functions
Let I ⊆ R be an interval.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 28/38
Space of Continuous Functions
Let I ⊆ R be an interval. Then I is of the form (forsome a < b)
I =
{x ∈ R| a < x < b}, an open interval;{x ∈ R| a ≤ x ≤ b}, a closed interval;{x ∈ R| a ≤ x < b}
{x ∈ R| a < x ≤ b}.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 28/38
Space of Continuous Functions
Let I ⊆ R be an interval. Then recall that thespace of all functions f : I −→ R is a vector space.We will now list some important subspaces:
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 28/38
Space of Continuous Functions
Let I ⊆ R be an interval. Then recall that thespace of all functions f : I −→ R is a vector space.We will now list some important subspaces:
Example (1). Let C(I) be the space of continuous functions.If f and g are continuous, so are the functions f + g andrf (r ∈ R). Hence C(I) is a vector space.
![Page 88: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/88.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 28/38
Space of Continuous Functions
Let I ⊆ R be an interval. Then recall that thespace of all functions f : I −→ R is a vector space.We will now list some important subspaces:
Example (1). Let C(I) be the space of continuous functions.If f and g are continuous, so are the functions f + g andrf (r ∈ R). Hence C(I) is a vector space.
Recall, that a function is continuous, if the graph has no gaps.This can be formulated in different ways:
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 28/38
Space of Continuous Functions
Let I ⊆ R be an interval. Then recall that thespace of all functions f : I −→ R is a vector space.We will now list some important subspaces:
Example (1). Let C(I) be the space of continuous functions.If f and g are continuous, so are the functions f + g andrf (r ∈ R). Hence C(I) is a vector space.
Recall, that a function is continuous, if the graph has no gaps.This can be formulated in different ways:
a) Let x0 ∈ I and let ǫ > 0 . Then there exists a δ > 0 suchthat for all x ∈ I ∩ (x0 − δ, x0 + δ) we have
|f(x) − f(x0)| < ǫ
This tells us that the value of f at nearby points isarbitrarily close to the value of f at x0.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 28/38
Space of Continuous Functions
Let I ⊆ R be an interval. Then recall that thespace of all functions f : I −→ R is a vector space.We will now list some important subspaces:
Example (1). Let C(I) be the space of continuous functions.If f and g are continuous, so are the functions f + g andrf (r ∈ R). Hence C(I) is a vector space.
Recall, that a function is continuous, if the graph has no gaps.This can be formulated in different ways:
b) A reformulation of (a) is:
limx→x0
f(x) = f(x0)
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 29/38
Space of continuously differentiable functs.
Example (2). The space C1(I). Here we assume that I isopen.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 29/38
Space of continuously differentiable functs.
Example (2). The space C1(I). Here we assume that I isopen. Recall that f is differentiable at x0 if
limx→x0
f(x) − f(x0)
x − x0
= limh→0
f(x0 + h) − f(x0)
h=: f ′(x0)
exists.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 29/38
Space of continuously differentiable functs.
Example (2). The space C1(I). Here we assume that I isopen. Recall that f is differentiable at x0 if
limx→x0
f(x) − f(x0)
x − x0
= limh→0
f(x0 + h) − f(x0)
h=: f ′(x0)
exists.If f ′(x0) exists for all x0 ∈ I , then we say that f isdifferentiable on I . In this case we get a new function on I
x 7→ f ′(x)
We say that f is continuously differentiable on I if f ′ exists
and is continuous on I .
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 29/38
Space of continuously differentiable functs.
Example (2). The space C1(I). Here we assume that I isopen. Recall that f is differentiable at x0 if
limx→x0
f(x) − f(x0)
x − x0
= limh→0
f(x0 + h) − f(x0)
h=: f ′(x0)
exists. We say that f is continuously differentiable on I if f ′
exists and is continuous on I . Recall that if f and g aredifferentiable, then so are
f + g and rf (r ∈ R)moreover
(f + g)′ = f ′ + g′ ; (rf)′ = rf ′
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 29/38
Space of continuously differentiable functs.
Example (2). The space C1(I). Here we assume that I isopen. Recall that f is differentiable at x0 if
limx→x0
f(x) − f(x0)
x − x0
= limh→0
f(x0 + h) − f(x0)
h=: f ′(x0)
exists. We say that f is continuously differentiable on I if f ′
exists and is continuous on I . Recall that if f and g aredifferentiable, then so are
f + g and rf (r ∈ R)moreover
(f + g)′ = f ′ + g′ ; (rf)′ = rf ′
As f ′ + g′ and rf ′ are continuous by Example (1), it followsthat C1(I) is a vector space.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 30/38
A continuous but not differentiable function
Let f(x) = |x| for x ∈ R. Then f is continuous on Rbut it is not differentiable on R.
![Page 97: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/97.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 30/38
A continuous but not differentiable function
Let f(x) = |x| for x ∈ R. Then f is continuous on Rbut it is not differentiable on R. We show that f isnot differentiable at x0 = 0.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 30/38
A continuous but not differentiable function
Let f(x) = |x| for x ∈ R. Then f is continuous on Rbut it is not differentiable on R. We show that f isnot differentiable at x0 = 0. For h > 0 we have
f(x0 + h) − f(x0)
h=
|h| − 0
h=
h
h= 1
hence
limh→0+
f(x0 + h) − f(x0)
h= 1
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 30/38
A continuous but not differentiable function
Let f(x) = |x| for x ∈ R. Then f is continuous on Rbut it is not differentiable on R. We show that f isnot differentiable at x0 = 0. For h > 0 we have
f(x0 + h) − f(x0)
h=
|h| − 0
h=
h
h= 1
hence
limh→0+
f(x0 + h) − f(x0)
h= 1
But if h < 0, then
f(x0 + h) − f(x0)
h=
|h| − 0
h=
−h
h= −1
hence
limh→0−
f(x0 + h) − f(x0)
h= −1
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 30/38
A continuous but not differentiable function
Let f(x) = |x| for x ∈ R. Then f is continuous on Rbut it is not differentiable on R. We show that f isnot differentiable at x0 = 0.Therefore,
limh→0
f(x0 + h) − f(x0)
h
does not exist.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 31/38
Space of r-times continuously diff. functs.
Example (3). The space Cr(I)Let I = (a, b) be an open interval. and letr ∈ N = {1, 2, 3, · · · }.
Definition. The function f : I −→ R is said to ber-times continuously differentiable if all the derivatives
f ′, f ′′, · · · , f (r) exist and f (r) : I −→ R is continuous.
We denote by Cr(I) the space of r-times continuouslydifferentiable functions on I . Cr(I) is a subspace of C(I).
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 31/38
Space of r-times continuously diff. functs.
Example (3). The space Cr(I)Let I = (a, b) be an open interval. and letr ∈ N = {1, 2, 3, · · · }.
Definition. The function f : I −→ R is said to ber-times continuously differentiable if all the derivatives
f ′, f ′′, · · · , f (r) exist and f (r) : I −→ R is continuous.
We denote by Cr(I) the space of r-times continuouslydifferentiable functions on I . Cr(I) is a subspace of C(I).
We have
Cr(I) $ Cr−1(I) ( · · · $ C1(I) $ C(I).
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 32/38
Cr(I) 6= Cr−1(I)
We have seen that C1(I) 6= C(I). Let us try to finda function that is in C1(I) but not in C2(I).
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 32/38
Cr(I) 6= Cr−1(I)
We have seen that C1(I) 6= C(I). Let us try to finda function that is in C1(I) but not in C2(I).
Assume 0 ∈ I and let f(x) = x5
3 . Then f isdifferentiable and
f ′(x) = 53x
2
3
which is continuous.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 32/38
Cr(I) 6= Cr−1(I)
We have seen that C1(I) 6= C(I). Let us try to finda function that is in C1(I) but not in C2(I).
Assume 0 ∈ I and let f(x) = x5
3 . Then f isdifferentiable and
f ′(x) = 53x
2
3
which is continuous.
If x 6= 0, then f ′ is differentiable and
f ′′(x) = 103x−
1
3
![Page 106: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/106.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 33/38
Cr(I) 6= Cr−1(I)
But for x = 0 we have
limh→0
f ′(h) − 0
h= lim
h→0
5
3
h2
3
h= lim
h→0
5
3h−
1
3
which does not exist.
![Page 107: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/107.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 33/38
Cr(I) 6= Cr−1(I)
But for x = 0 we have
limh→0
f ′(h) − 0
h= lim
h→0
5
3
h2
3
h= lim
h→0
5
3h−
1
3
which does not exist.
Remark. One can show that the function
f(x) = x3r−1
3
is in Cr−1(R), but not in Cr(R).
![Page 108: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/108.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 33/38
Cr(I) 6= Cr−1(I)
But for x = 0 we have
limh→0
f ′(h) − 0
h= lim
h→0
5
3
h2
3
h= lim
h→0
5
3h−
1
3
which does not exist.
Remark. One can show that the function
f(x) = x3r−1
3
is in Cr−1(R), but not in Cr(R).
Thus, as stated before, we have
Cr(I) $ Cr−1(I) · · · $ C1(I) $ C(I).
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 34/38
Piecewise-continuous functions
Example (4). Piecewise-continuous functions
Definition. Let I = [a, b). A function f : I −→ R is calledpiecewise-continuous if there exists finitely many points
a = x0 < x1 < · · · < xn = b
such that f is continuous on each of the sub-intervals(xi, xi+1) for i = 0, 1, · · · , n − 1.
![Page 110: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/110.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 34/38
Piecewise-continuous functions
Example (4). Piecewise-continuous functions
Definition. Let I = [a, b). A function f : I −→ R is calledpiecewise-continuous if there exists finitely many points
a = x0 < x1 < · · · < xn = b
such that f is continuous on each of the sub-intervals(xi, xi+1) for i = 0, 1, · · · , n − 1.
Remark. If f and g are both piecewise-continuous, then
f + g and rf (r ∈ R)
are piecewise-continuous.
![Page 111: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/111.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 34/38
Piecewise-continuous functions
Example (4). Piecewise-continuous functions
Definition. Let I = [a, b). A function f : I −→ R is calledpiecewise-continuous if there exists finitely many points
a = x0 < x1 < · · · < xn = b
such that f is continuous on each of the sub-intervals(xi, xi+1) for i = 0, 1, · · · , n − 1.
Remark. If f and g are both piecewise-continuous, then
f + g and rf (r ∈ R)
are piecewise-continuous.
Hence the space of piecewise-continuous functions is a vectorspace. Denote this vector space by PC(I).
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 35/38
The indicator function χA
Important elements of PC(I) are the indicatorfunctions χA, where A ⊆ I a sub-interval.
![Page 113: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/113.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 35/38
The indicator function χA
Important elements of PC(I) are the indicatorfunctions χA, where A ⊆ I a sub-interval.
Let A ⊆ R be a set. Define
χA(x) =
{
1, if x ∈ A
0, if x /∈ A
![Page 114: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/114.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 35/38
The indicator function χA
Important elements of PC(I) are the indicatorfunctions χA, where A ⊆ I a sub-interval.
Let A ⊆ R be a set. Define
χA(x) =
{
1, if x ∈ A
0, if x /∈ A
So the values of χA tell us whether x is in A ornot.If x ∈ A, then χA(x) = 1 and if x /∈ A, thenχA(x) = 0.
![Page 115: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/115.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 35/38
The indicator function χA
Important elements of PC(I) are the indicatorfunctions χA, where A ⊆ I a sub-interval.
Let A ⊆ R be a set. Define
χA(x) =
{
1, if x ∈ A
0, if x /∈ A
So the values of χA tell us whether x is in A ornot.If x ∈ A, then χA(x) = 1 and if x /∈ A, thenχA(x) = 0.
We will work a lot with indicator functions so let uslook at some of their properties.
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Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 36/38
Some properties of χA
Lemma. Let A, B ⊆ I . Then
χA∩B(x) = χA(x) χB(x) (∗)
![Page 117: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/117.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 36/38
Some properties of χA
Lemma. Let A, B ⊆ I . Then
χA∩B(x) = χA(x) χB(x) (∗)
Proof. We have to show that the two functions
x 7→ χA∩B(x) and x 7→ χA(x)χB(x)
take the same values at every point x ∈ I . So lets evaluateboth functions:
![Page 118: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/118.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 36/38
Some properties of χA
Lemma. Let A, B ⊆ I . Then
χA∩B(x) = χA(x) χB(x) (∗)
Proof. We have to show that the two functions
x 7→ χA∩B(x) and x 7→ χA(x)χB(x)
take the same values at every point x ∈ I . So lets evaluateboth functions:
If x ∈ A and x ∈ B, that is x ∈ A ∩ B, thenχA∩B(x) = 1 and,
![Page 119: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/119.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 36/38
Some properties of χA
Lemma. Let A, B ⊆ I . Then
χA∩B(x) = χA(x) χB(x) (∗)
Proof. We have to show that the two functions
x 7→ χA∩B(x) and x 7→ χA(x)χB(x)
take the same values at every point x ∈ I . So lets evaluateboth functions:
If x ∈ A and x ∈ B, that is x ∈ A ∩ B, thenχA∩B(x) = 1 and,
since χA(x) = 1 and χB(x) = 1, we also haveχA(x)χB(x) = 1.
Thus, the left and the right hand sides of (∗) agree.
![Page 120: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/120.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 36/38
Some properties of χA
Lemma. Let A, B ⊆ I . Then
χA∩B(x) = χA(x) χB(x) (∗)
Proof. We have to show that the two functions
x 7→ χA∩B(x) and x 7→ χA(x)χB(x)
take the same values at every point x ∈ I . So lets evaluateboth functions:
On the other hand, if x /∈ A ∩ B, then there are twopossibilities:
![Page 121: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/121.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 36/38
Some properties of χA
Lemma. Let A, B ⊆ I . Then
χA∩B(x) = χA(x) χB(x) (∗)
Proof. We have to show that the two functions
x 7→ χA∩B(x) and x 7→ χA(x)χB(x)
take the same values at every point x ∈ I . So lets evaluateboth functions:
On the other hand, if x /∈ A ∩ B, then there are twopossibilities:■ x /∈ A then χA(x) = 0, so χA(x)χB(x) = 0.
![Page 122: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/122.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 36/38
Some properties of χA
Lemma. Let A, B ⊆ I . Then
χA∩B(x) = χA(x) χB(x) (∗)
Proof. We have to show that the two functions
x 7→ χA∩B(x) and x 7→ χA(x)χB(x)
take the same values at every point x ∈ I . So lets evaluateboth functions:
On the other hand, if x /∈ A ∩ B, then there are twopossibilities:■ x /∈ A then χA(x) = 0, so χA(x)χB(x) = 0.
■ x /∈ B then χB(x) = 0, so χA(x)χB(x) = 0.
It follows that
0 = χA∩B(x) = χA(x)χB(x)
![Page 123: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/123.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 37/38
Some properties of χA
What about χA∪B? Can we express it in terms ofχA, χB?
![Page 124: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/124.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 37/38
Some properties of χA
What about χA∪B? Can we express it in terms ofχA, χB?If A and B are disjoint, that is A ∩ B = ∅ then
χA∪B(x) = χA(x) + χB(x).
![Page 125: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/125.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 37/38
Some properties of χA
What about χA∪B? Can we express it in terms ofχA, χB?If A and B are disjoint, that is A ∩ B = ∅ then
χA∪B(x) = χA(x) + χB(x).
Let us prove this:■ If x /∈ A ∪ B, then x /∈ A and x /∈ B. Thus the
LHS (left hand side) and the RHS (right handside) are both zero.
![Page 126: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/126.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 37/38
Some properties of χA
What about χA∪B? Can we express it in terms ofχA, χB?If A and B are disjoint, that is A ∩ B = ∅ then
χA∪B(x) = χA(x) + χB(x).
Let us prove this:■ If x /∈ A ∪ B, then x /∈ A and x /∈ B. Thus the
LHS (left hand side) and the RHS (right handside) are both zero.
■ If x ∈ A ∪ B then either
![Page 127: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/127.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 37/38
Some properties of χA
What about χA∪B? Can we express it in terms ofχA, χB?If A and B are disjoint, that is A ∩ B = ∅ then
χA∪B(x) = χA(x) + χB(x).
Let us prove this:■ If x /∈ A ∪ B, then x /∈ A and x /∈ B. Thus the
LHS (left hand side) and the RHS (right handside) are both zero.
■ If x ∈ A ∪ B then either
◆ x is in A but not in B. In this caseχA∪B(x) = 1 and χA(x) + χB(x) = 1 + 0 = 1
![Page 128: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/128.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 37/38
Some properties of χA
What about χA∪B? Can we express it in terms ofχA, χB?If A and B are disjoint, that is A ∩ B = ∅ then
χA∪B(x) = χA(x) + χB(x).
Let us prove this:■ If x /∈ A ∪ B, then x /∈ A and x /∈ B. Thus the
LHS (left hand side) and the RHS (right handside) are both zero.
■ If x ∈ A ∪ B then either
◆ x is in A but not in B. In this caseχA∪B(x) = 1 and χA(x) + χB(x) = 1 + 0 = 1
or◆ x is in B but not in A. In this case
χA∪B(x) = 1 and χA(x) + χB(x) = 0 + 1 = 1
![Page 129: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/129.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
![Page 130: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/130.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
Now, what if A ∩ B 6= ∅?
![Page 131: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/131.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
Now, what if A ∩ B 6= ∅?
Lemma. χA∪B(x) = χA(x) + χB(x) − χA∩B(x).
![Page 132: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/132.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
Now, what if A ∩ B 6= ∅?
Lemma. χA∪B(x) = χA(x) + χB(x) − χA∩B(x).
Proof.■ If x /∈ A ∪ B, then both of the LHS and the RHS take the
value 0.
![Page 133: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/133.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
Now, what if A ∩ B 6= ∅?
Lemma. χA∪B(x) = χA(x) + χB(x) − χA∩B(x).
Proof.■ If x /∈ A ∪ B, then both of the LHS and the RHS take the
value 0.
■ If x ∈ A ∪ B, then we have the following possibilities:
![Page 134: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/134.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
Now, what if A ∩ B 6= ∅?
Lemma. χA∪B(x) = χA(x) + χB(x) − χA∩B(x).
Proof.■ If x /∈ A ∪ B, then both of the LHS and the RHS take the
value 0.
■ If x ∈ A ∪ B, then we have the following possibilities:1. If x ∈ A, x /∈ B, then
χA∪B(x) = 1
χA(x) + χB(x) − χA∩B = 1 + 0 − 0 = 1
![Page 135: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/135.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
Now, what if A ∩ B 6= ∅?
Lemma. χA∪B(x) = χA(x) + χB(x) − χA∩B(x).
Proof.■ If x /∈ A ∪ B, then both of the LHS and the RHS take the
value 0.
■ If x ∈ A ∪ B, then we have the following possibilities:1. If x ∈ A, x /∈ B, then
χA∪B(x) = 1
χA(x) + χB(x) − χA∩B = 1 + 0 − 0 = 1
2. Similarly for the case x ∈ B, x /∈ A: LHS equals theRHS.
![Page 136: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/136.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
Now, what if A ∩ B 6= ∅?
Lemma. χA∪B(x) = χA(x) + χB(x) − χA∩B(x).
Proof.■ If x /∈ A ∪ B, then both of the LHS and the RHS take the
value 0.
■ If x ∈ A ∪ B, then we have the following possibilities:3. If x ∈ A ∩ B, then
χA∪B(x) = 1
χA(x) + χB(x) − χA∩B = 1 + 1 − 1 = 1
![Page 137: Integral Transforms Math 2025 - LSU Mathematicsolafsson/pdf_files/ch1_2_3.pdf · Integral Transforms Math 2025 Gestur Olafsson ... Vector Spaces over R Integral Transforms - p. 3/38](https://reader034.fdocuments.in/reader034/viewer/2022052419/5a714c9b7f8b9a93538cce06/html5/thumbnails/137.jpg)
Spaces of FunctionsC(I)
C1(I)Cr(I)PC(I)Indicator functionsχA∩B
χA∪B (disjoint)χA∪B
Vector Spaces of Functions Integral Transforms - p. 38/38
Some properties of χA
Thus we have,If A ∩ B = ∅, then χA∪B(x) = χA(x) + χB(x).
Now, what if A ∩ B 6= ∅?
Lemma. χA∪B(x) = χA(x) + χB(x) − χA∩B(x).
Proof.■ If x /∈ A ∪ B, then both of the LHS and the RHS take the
value 0.
■ If x ∈ A ∪ B, then we have the following possibilities:3. If x ∈ A ∩ B, then
χA∪B(x) = 1
χA(x) + χB(x) − χA∩B = 1 + 1 − 1 = 1
As we have checked all possibilities, we have shown that thestatement in the lemma is correct