Advanced Calculus (I)

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Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

WEN-CHINGLIEN Advanced Calculus (I)

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5.5 Functions of bounded variation

Let φ : [a, b] →R. To measure how much φ wiggles on an interval [a,b], set

V (φ, P) =

n

X

j=1

|φ(x_j) − φ(xj−1)|

for each partition P = {x0,x1, · · · ,xn} of [a,b]. The variation of φ is defined by

Var (φ) := sup{V (φ, P) : P is a partition of [a, b]}.

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5.5 Functions of bounded variation

Let φ : [a, b] →R. To measure how much φ wiggles on an interval [a,b], set

V (φ, P) =

n

X

j=1

|φ(x_j) − φ(xj−1)|

for each partition P = {x0,x1, · · · ,xn} of [a,b]. The variation of φ is defined by

Var (φ) := sup{V (φ, P) : P is a partition of [a, b]}.

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Definition

Let [a,b] be a closed, nondegenerate interval and φ : [a, b] →R. Then φ is said to be of bounded variation on [a,b] if and only if Var (φ) < ∞.

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Definition

Let [a,b] be a closed, nondegenerate interval and φ : [a, b] →R. Then φ is said to be of bounded variation on [a,b] if and only if Var (φ) < ∞.

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Remark:

If φ ∈ C¹[a, b], then φ is of bounded variation on [a,b].

However, there exist functions of bounded variation that are not continuously differentiable.

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Remark:

If φ ∈ C¹[a, b], then φ is of bounded variation on [a,b].

However, there exist functions of bounded variation that are not continuously differentiable.

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Proof:

Let P = {x0,x1, · · · ,xn} be a partition of [a,b]. By the Extreme Value Theorem,there is an M > 0 such that

|φ⁰(x )| ≤ M for all x ∈ [a, b]. Therefore, it follows from the Mean Value Theorem that for each k between 1 and n, there is a point ck between xk −1and xk such that

|φ(xk) − φ(xk −1)| = |φ⁰(ck)|(xk − xk −1) ≤M(xk − xk −1).

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Proof:

Let P = {x0,x1, · · · ,xn} be a partition of [a,b]. By the Extreme Value Theorem, there is an M > 0 such that

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Proof:

Let P = {x0,x1, · · · ,xn} be a partition of [a,b]. By the Extreme Value Theorem,there is an M > 0 such that

|φ⁰(x )| ≤ M for all x ∈ [a, b]. Therefore,it follows from the Mean Value Theorem that for each k between 1 and n, there is a point ck between xk −1and xk such that

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Proof:

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Proof:

|φ⁰(x )| ≤ M for all x ∈ [a, b]. Therefore,it follows from the Mean Value Theorem that for each k between 1 and n, there is a point ck between xk −1and xk such that

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Proof:

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Proof:

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By telescoping, we obtain V (φ, P) ≤ M(b − a) for any partition P of [a,b]. Therefore,

Var (φ) ≤ M(b − a).

On the other hand, x²sin(1/x ) is of bounded variation on [0,1] (see Exercise 2) but does not belong to C¹[0, 1] (see Example 4.8)2

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By telescoping,we obtain V (φ, P) ≤ M(b − a) for any partition P of [a,b]. Therefore,

Var (φ) ≤ M(b − a).

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Var (φ) ≤ M(b − a).

On the other hand,x²sin(1/x ) is of bounded variation on [0,1] (see Exercise 2) but does not belong to C¹[0, 1] (see Example 4.8)2

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Var (φ) ≤ M(b − a).

On the other hand, x²sin(1/x ) is of bounded variation on [0,1] (see Exercise 2)but does not belong to C¹[0, 1] (see Example 4.8)2

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Var (φ) ≤ M(b − a).

On the other hand,x²sin(1/x ) is of bounded variation on [0,1] (see Exercise 2) but does not belong to C¹[0, 1] (see Example 4.8)2

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Var (φ) ≤ M(b − a).

On the other hand, x²sin(1/x ) is of bounded variation on [0,1] (see Exercise 2)but does not belong to C¹[0, 1] (see Example 4.8)2

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Var (φ) ≤ M(b − a).

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Remark:

If φ is of bounded variation on [a,b], then φ is bounded on [a,b]. However, there exist bounded functions that are not of bounded variation.

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Remark:

If φ is of bounded variation on [a,b], then φ is bounded on [a,b]. However, there exist bounded functions that are not of bounded variation.

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Proof:

Let x ∈ [a, b] and note by definition that

Hence,by the triangle inequality,

|φ(x)| ≤ |φ(a)| + Var (φ).

To find a bounded function that is not of bounded variation, consider

φ(x ) :=

( sin(1

x) x 6= 0

0 x = 0

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Proof:

Hence, by the triangle inequality,

|φ(x)| ≤ |φ(a)| + Var (φ).

φ(x ) :=

( sin(1

x) x 6= 0

0 x = 0

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Proof:

Hence,by the triangle inequality,

|φ(x)| ≤ |φ(a)| + Var (φ).

To find a bounded function that is not of bounded variation,consider

φ(x ) :=

( sin(1

x) x 6= 0

0 x = 0

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Proof:

|φ(x)| ≤ |φ(a)| + Var (φ).

φ(x ) :=

( sin(1

x) x 6= 0

0 x = 0

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Proof:

|φ(x)| ≤ |φ(a)| + Var (φ).

To find a bounded function that is not of bounded variation,consider

φ(x ) :=

( sin(1

x) x 6= 0

0 x = 0

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Proof:

|φ(x)| ≤ |φ(a)| + Var (φ).

φ(x ) :=

( sin(1

x) x 6= 0

0 x = 0

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Clearly, φ is bounded by 1. On the other hand, if

xj =







0 j = 0

2

(n − j)π 0 < j < n,

then n

X

j=1

|φ(x_j) − φ(xj−1)| =2n → ∞

as n → ∞. Thus φ is not of bounded variation on [0,2 π]. 2

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Clearly, φ is bounded by 1. On the other hand, if

xj =







0 j = 0

2

(n − j)π 0 < j < n,

then n

X

j=1

|φ(x_j) − φ(xj−1)| =2n → ∞

as n → ∞. Thus φ is not of bounded variation on [0,2 π]. 2

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Theorem

If φ and ψ are of bounded variation on a closed interval [a,b], then so are φ + ψ and φ − ψ.

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Theorem

If φ and ψ are of bounded variation on a closed interval [a,b], then so are φ + ψ and φ − ψ.

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Definition

Let φ be of bounded variation on a closed interval [a,b].

The total variation of φ is the function Φ deifined on [a,b]

by

Φ(x ) := sup







k

X

j=1

|φ(x_j) − φ(xj−1)| : {x0,x1, · · · ,xk}

is a partition of [a, x ]}

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Definition

Let φ be of bounded variation on a closed interval [a,b].

The total variation of φ is the function Φ deifined on [a,b]

by

Φ(x ) := sup







k

X

j=1

|φ(x_j) − φ(xj−1)| : {x0,x1, · · · ,xk}

is a partition of [a, x ]}

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Theorem

Let φ be of bounded variation on [a,b] and Φ be its total variation. Then

(i) |φ(y ) − φ(x )| ≤ Φ(y ) − Φ(x ) for all a ≤ x < y ≤ b, (ii)Φ and Φ − φ are increasing on [a,b], and

(iii)Var (φ) ≤ Var (Φ).

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Theorem

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Theorem

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Theorem

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Theorem

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Proof:

(i)

Let x < y belong to [a,b] and {x0,x1, · · · ,xk} be a partition of [a,x]. Then {x0,x1, · · · ,xk,y } is a partition [a,y], and we have by Definition 5.55 that

k

X

j=1

|φ(x)−φ(xj−1)| ≤

k

X

j=1

|φ(xj)−φ(xj−1)|+|φ(y )−φ(x )| ≤ Φ(y ).

Taking the supremum of this inequality over all partitions {x₀,x1, · · · ,xk} of [a,x], we obtain

Φ(x ) ≤ Φ(x ) + |Φ(y ) − φ(x )| ≤ Φ(y ).

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Proof:

(i)

Let x < y belong to [a,b] and {x0,x1, · · · ,xk} be a partition of [a,x]. Then {x0,x1, · · · ,xk,y } is a partition [a,y],and we have by Definition 5.55 that

k

X

j=1

|φ(x)−φ(xj−1)| ≤

k

X

j=1

|φ(xj)−φ(xj−1)|+|φ(y )−φ(x )| ≤ Φ(y ).

Φ(x ) ≤ Φ(x ) + |Φ(y ) − φ(x )| ≤ Φ(y ).

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Proof:

(i)

k

X

j=1

|φ(x)−φ(xj−1)| ≤

k

X

j=1

|φ(xj)−φ(xj−1)|+|φ(y )−φ(x )| ≤ Φ(y ).

Φ(x ) ≤ Φ(x ) + |Φ(y ) − φ(x )| ≤ Φ(y ).

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Proof:

(i)

Let x < y belong to [a,b] and {x0,x1, · · · ,xk} be a partition of [a,x]. Then {x0,x1, · · · ,xk,y } is a partition [a,y],and we have by Definition 5.55 that

k

X

j=1

|φ(x)−φ(xj−1)| ≤

k

X

j=1

|φ(xj)−φ(xj−1)|+|φ(y )−φ(x )| ≤ Φ(y ).

Taking the supremum of this inequality over all partitions {x₀,x1, · · · ,xk} of [a,x],we obtain

Φ(x ) ≤ Φ(x ) + |Φ(y ) − φ(x )| ≤ Φ(y ).

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Proof:

(i)

k

X

j=1

|φ(x)−φ(xj−1)| ≤

k

X

j=1

|φ(xj)−φ(xj−1)|+|φ(y )−φ(x )| ≤ Φ(y ).

Φ(x ) ≤ Φ(x ) + |Φ(y ) − φ(x )| ≤ Φ(y ).

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Proof:

(i)

k

X

j=1

|φ(x)−φ(xj−1)| ≤

k

X

j=1

|φ(xj)−φ(xj−1)|+|φ(y )−φ(x )| ≤ Φ(y ).

Taking the supremum of this inequality over all partitions {x₀,x1, · · · ,xk} of [a,x],we obtain

Φ(x ) ≤ Φ(x ) + |Φ(y ) − φ(x )| ≤ Φ(y ).

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Proof:

(i)

k

X

j=1

|φ(x)−φ(xj−1)| ≤

k

X

j=1

|φ(xj)−φ(xj−1)|+|φ(y )−φ(x )| ≤ Φ(y ).

Φ(x ) ≤ Φ(x ) + |Φ(y ) − φ(x )| ≤ Φ(y ).

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(ii)

By the Monotone Property of Suprema,Φis increasing on [a,b]. To show that Φ − φ is increasing, suppose that a ≤ x < y ≤ b. By part (i),