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(1)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

(2)

## Ch11: Differentiability on R

n

### 11.1: Partial Derivatives and Partial Integrals

Notation:

(1) E1× E2× . . . × En

:= {(x1,x2, . . . ,xn) :xj ∈ Ej, for j = 1, . . . , n}

(2) The partial derivative fxj exists at a pointa if and only if the limit

∂f

∂xj

(a) = lim

h→0

f (a + hej) −f (a) h

exists.

(3)

fxjxk := ∂2f

∂xk∂xj

:= ∂

∂xk

 ∂f

∂xj

 .

(3)

Let V be a nonempty, open subset ofRn, let f : V →Rm, and let p ∈N.

(i) f is said to be Cp on V if and only if each partial derivative of f of order k ≤ p exists and is continuous on V .

(ii) f is said to be Con V if and only if f is Cp on V for all p ∈N.

(4)

Theorem (11.2)

Suppose that V is open inR2, that (a, b) ∈ V , and that f : V →R. If f is C1on V , and if one of the mixed second partial derivatives of f exists on V and is continuous at the point (a, b), then the other mixed second partial derivative exists at (a, b) and

2f

∂y ∂x(a, b) = ∂2f

∂x ∂y(a, b).

Note: These hypotheses are met if f ∈ C2(H).

(5)

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b), define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V .Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f2

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b), we have

(1) lim lim ∆(h, k )

= ∂2f

(a, b).

(6)

Proof.

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b),define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V . Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f2

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b), we have

(1) lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(7)

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b), define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V .Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f2

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b), we have

(1) lim lim ∆(h, k )

= ∂2f

(a, b).

(8)

Proof.

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b), define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V . Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f2

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b), we have

(1) lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(9)

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b), define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V . Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f2

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b),we have

(1) lim lim ∆(h, k )

= ∂2f

(a, b).

(10)

Proof.

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b), define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V . Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f2

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b), we have

(1) lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(11)

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b), define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V . Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f2

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b),we have

(1) lim lim ∆(h, k )

= ∂2f

(a, b).

(12)

Proof.

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b), define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V . Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f2

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b), we have

(1) lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(13)

On the other hand, the Mean Value Theorem also implies that there is a scalar u ∈ (0, 1) such that

∆(h, k ) = f (a + h, b + k ) − f (a, b + k ) − f (a + h, b) + f (a, b)

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

Hence, it follows from (1) that

k →0lim lim

h→0

1 k

 ∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)



= lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(14)

Proof.

On the other hand, the Mean Value Theorem also implies that there is a scalar u ∈ (0, 1) such that

∆(h, k ) = f (a + h, b + k ) − f (a, b + k ) − f (a + h, b) + f (a, b)

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

Hence, it follows from (1) that

k →0lim lim

h→0

1 k

 ∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)



= lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(15)

On the other hand, the Mean Value Theorem also implies that there is a scalar u ∈ (0, 1) such that

∆(h, k ) = f (a + h, b + k ) − f (a, b + k ) − f (a + h, b) + f (a, b)

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

Hence, it follows from (1) that

k →0lim lim

h→0

1 k

 ∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)



= lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(16)

Proof.

On the other hand, the Mean Value Theorem also implies that there is a scalar u ∈ (0, 1) such that

∆(h, k ) = f (a + h, b + k ) − f (a, b + k ) − f (a + h, b) + f (a, b)

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

Hence, it follows from (1) that

k →0lim lim

h→0

1 k

 ∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)



= lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(17)

On the other hand, the Mean Value Theorem also implies that there is a scalar u ∈ (0, 1) such that

∆(h, k ) = f (a + h, b + k ) − f (a, b + k ) − f (a + h, b) + f (a, b)

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

Hence, it follows from (1) that

k →0lim lim

h→0

1 k

 ∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)



= lim

k →0lim

h→0

∆(h, k )

hk = ∂2f

∂x ∂y(a, b).

(18)

Proof.

Since fx is continuous on Br(a, b), we can let h = 0 in the first expression. We conclude by definition that

2f

∂y ∂x(a, b)= lim

k →0

1 k

 ∂f

∂x(a, b+k )−∂f

∂x(a, b)



= ∂2f

∂x ∂y(a, b).

(19)

Since fx is continuous on Br(a, b), we can let h = 0 in the first expression.We conclude by definition that

2f

∂y ∂x(a, b) = lim

k →0

1 k

 ∂f

∂x(a, b+k )−∂f

∂x(a, b)



= ∂2f

∂x ∂y(a, b).

(20)

Proof.

Since fx is continuous on Br(a, b), we can let h = 0 in the first expression. We conclude by definition that

2f

∂y ∂x(a, b)= lim

k →0

1 k

 ∂f

∂x(a, b+k )−∂f

∂x(a, b)



= ∂2f

∂x ∂y(a, b).

(21)

Since fx is continuous on Br(a, b), we can let h = 0 in the first expression. We conclude by definition that

2f

∂y ∂x(a, b) = lim

k →0

1 k

 ∂f

∂x(a, b+k )−∂f

∂x(a, b)



= ∂2f

∂x ∂y(a, b).

(22)

Proof.

Since fx is continuous on Br(a, b), we can let h = 0 in the first expression. We conclude by definition that

2f

∂y ∂x(a, b) = lim

k →0

1 k

 ∂f

∂x(a, b+k )−∂f

∂x(a, b)



= ∂2f

∂x ∂y(a, b).

(23)

Prove that

f (x , y ) = (

xy xx22−y+y22

 (x , y ) 6= 0

0 (x , y ) = 0

(24)

Theorem (11.4)

Let H = [a, b] × [c, d ] be a rectangle and suppose that f : H →R is continuous. If

F (y ) = Z b

a

f (x , y )dx , then F is continuous on [c, d ]; i.e.,

y →y0lim

y ∈[c,d ]

Z b a

f (x , y )dx = Z b

a

y →y0lim

y ∈[c,d ]

f (x , y )dx

for all y0∈ [c, d].

(25)

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b].Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0.Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k, it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ. We conclude that F is continuous on [c, d ].

(26)

Proof.

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H.Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k, it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ. We conclude that F is continuous on [c, d ].

(27)

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0.Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such thatk(x, y ) − (z, w )k < δ and (x, y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k, it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ. We conclude that F is continuous on [c, d ].

(28)

Proof.

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H.Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k, it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ. We conclude that F is continuous on [c, d ].

(29)

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such thatk(x, y ) − (z, w )k < δ and (x, y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k,it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ. We conclude that F is continuous on [c, d ].

(30)

Proof.

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k, it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ.We conclude that F is continuous on [c, d ].

(31)

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k,it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ. We conclude that F is continuous on [c, d ].

(32)

Proof.

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k, it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ.We conclude that F is continuous on [c, d ].

(33)

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y0)k, it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ. We conclude that F is continuous on [c, d ].

(34)

Theorem (11.5)

Let H = [a, b] × [c, d ] be a rectangle inR2and let f : H →R. Suppose that f (·, y ) is integrable on [a, b] for each y ∈ [c, d ] and that the partial derivative fy(x , ·) exists on [c, d ] for each x ∈ [a, b]. If the two-variable function fy(x , y ) is continuous on H, then

d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx for all y ∈ [c, d ].

Note: These hypotheses are met if f ∈ C1(H).

(35)

Recall that ”fy(x , ·) exists on [c, d ]” means that fy(x , ·) exists on (c, d ), and

fy(x , c) := lim

h→0+

f (x , c + h) − f (x , c)

h ,

fy(x , d ) := lim

h→0−

f (x , d + h) − f (x , d ) h

both exist (see Definition 4.6). Hence, it suffices to show that

h→0+lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ [c, d ), and

(36)

Proof.

Recall that ”fy(x , ·) exists on [c, d ]” means that fy(x , ·) exists on (c, d ), and

fy(x , c) := lim

h→0+

f (x , c + h) − f (x , c)

h ,

fy(x , d ) := lim

h→0−

f (x , d + h) − f (x , d ) h

both exist (see Definition 4.6).Hence, it suffices to show that

h→0+lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ [c, d ), and

(37)

Recall that ”fy(x , ·) exists on [c, d ]” means that fy(x , ·) exists on (c, d ), and

fy(x , c) := lim

h→0+

f (x , c + h) − f (x , c)

h ,

fy(x , d ) := lim

h→0−

f (x , d + h) − f (x , d ) h

both exist (see Definition 4.6). Hence, it suffices to show that

h→0+lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ [c, d ), and

(38)

Proof.

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ].The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ).Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a). By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(39)

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ]. The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ). Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a). By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(40)

Proof.

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ]. The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ).Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a).By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(41)

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ]. The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ). Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a). By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(42)

Proof.

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ]. The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ). Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a).By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(43)

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ]. The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ). Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a). By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(44)

Proof.

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ]. The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ). Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a). By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(45)

Since |c(x ; h) − y | = c(x ; h) − y ≤ h,it follows that if 0 < h < δ, then

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(46)

Proof.

Since |c(x ; h) − y | = c(x ; h) − y ≤ h, it follows that if 0 < h < δ, then

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(47)

Since |c(x ; h) − y | = c(x ; h) − y ≤ h, it follows that if 0 < h < δ, then

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(48)

Proof.

Since |c(x ; h) − y | = c(x ; h) − y ≤ h, it follows that if 0 < h < δ, then

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(49)

Since |c(x ; h) − y | = c(x ; h) − y ≤ h, it follows that if 0 < h < δ, then

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(50)

Definition (11.6)

Let a < b be extended real numbers, let I be an interval in R, and suppose that f : (a, b) × I → R. The improper integral

Z b a

f (x , y )dx

is said to converge uniformly on I if and only if f (·, y ) is improperly integrable on (a, b) for each y ∈ I and given ε >0 there exist real numbers A, B ∈ (a, b) such that

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

< ε for all a < α < A, B < β < b and all y ∈ I.

(51)

Suppose that a < b are extended real numbers, that I is an interval inR, that f : (a, b) × I → R, and that f (·, y ) is locally integrable on the interval (a, b) for each y ∈ I. If there is a function M : (a, b) →R, absolutely integrable on (a, b), such that

|f (x, y )| ≤ M(x) for all x ∈ (a, b) and y ∈ I, then

Z b a

f (x , y )dx converges uniformly on I.

(52)

Proof.

Let ε > 0.By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I.Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I, we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

(53)

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I. Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I, we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

M(x )dx + Z b

M(x )dx < ε

(54)

Proof.

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I.Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I, we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

(55)

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I. Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I,we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

M(x )dx + Z b

M(x )dx < ε

(56)

Proof.

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I. Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I, we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

(57)

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I. Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I,we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

M(x )dx + Z b

M(x )dx < ε

(58)

Proof.

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I. Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I, we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

(59)

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I. Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I, we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

M(x )dx + Z b

M(x )dx < ε

(60)

Proof.

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I. Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I, we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

(61)

Suppose that a < b are extended real numbers, that c < d are finite real numbers, and that

f : (a, b) × [c, d ] →R is continuous. If F (y ) =

Z b a

f (x , y )dx converges uniformly on [c, d ]; i.e.,

y →y0lim

y ∈[c,d ]

Z b a

f (x , y )dx = Z b

a

y →y0lim

y ∈[c,d ]

f (x , y )dx

for all y0∈ [c, d].

(62)

Proof.

Let ε > 0 and y0∈ [c, d].Choose real numbers A, B such that a < A < B < b and

F (y ) − Z B

A

f (x , y )dx

< ε 3

for all y ∈ [c, d ]. By Theorem 11.4, choose δ > 0 such that

Z B A

(f (x , y ) − f (x , y0))dx

< ε 3 for all y ∈ [c, d ] that satisfy |y − y0| < δ. Then

(63)

Let ε > 0 and y0∈ [c, d]. Choose real numbers A, B such that a < A < B < b and

F (y ) − Z B

A

f (x , y )dx

< ε 3

for all y ∈ [c, d ].By Theorem 11.4, choose δ > 0 such that

Z B A

(f (x , y ) − f (x , y0))dx

< ε 3 for all y ∈ [c, d ] that satisfy |y − y0| < δ. Then

(64)

Proof.

Let ε > 0 and y0∈ [c, d]. Choose real numbers A, B such that a < A < B < b and

F (y ) − Z B

A

f (x , y )dx

< ε 3

for all y ∈ [c, d ]. By Theorem 11.4, choose δ > 0 such that

Z B A

(f (x , y ) − f (x , y0))dx

< ε 3 for all y ∈ [c, d ] that satisfy |y − y0| < δ. Then

(65)

|F (y ) − F (y0)|≤

F (y ) − Z B

A

f (x , y )dx +

Z B A

(f (x , y ) − f (x , y0))dx +

F (y0) − Z B

A

f (x , y0)dx

< ε 3 + ε

3+ ε 3 = ε for all y ∈ [c, d ] that satisfy |y − y0| < δ.

(66)

Proof.

|F (y ) − F (y0)| ≤

F (y ) − Z B

A

f (x , y )dx +

Z B A

(f (x , y ) − f (x , y0))dx +

F (y0) − Z B

A

f (x , y0)dx

< ε 3 + ε

3+ ε 3 = ε for all y ∈ [c, d ] that satisfy |y − y0| < δ.

(67)

Suppose that a < b are extended real numbers, that c < d are finite real numbers, that f : (a, b) × [c, d ] →R is continuous, and that the improper integral

F (y ) = Z b

a

f (x , y )dx

exists for all y ∈ [c, d ]. If fy(x , y ) exists and is continuous on (a, b) × [c, d ] and if

φ(y ) = Z b

a

∂f

∂y(x , y )dx

converges uniformly on [c, d ], then F is differentiable on [c, d ], and F0(y ) = φ(y ); i.e.,

(68)

Theorem (11.9) d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx for all y ∈ [c, d ].

(69)

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung