Advanced Calculus (II)

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Advanced Calculus (II)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

WEN-CHINGLIEN Advanced Calculus (II)

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Ch11: Differentiability on R

11.2: Definition of Differentiability

Definition (11.12)

Let f be a vector function from n variables to m variables.

(i) f is said to be differentiable at a pointa ∈ Rⁿif and only if there is an open set V containinga such that f : V → R^m and there is a T ∈ L(Rⁿ;R^m)such that the function

ε(h) := f (a + h) − f (a) − T (h)

(defined forh sufficiently small) satisfies ^ε(h)_khk → 0 as h → 0.

(ii) f is said to be differentiable on a set E if and only if E is nonempty, and f is differentiable at every point in E .

W -C L Advanced Calculus (II)

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Theorem (11.13)

Let f be a vector function. If f is differentiable ata, then f is continuous ata.

Proof.

Suppose that f is differentiable ata. Then by limh→0 f (a+h)−f (a)−Bh

khk =0,there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number, we

conclude from the Squeeze Theorem that f (a + h) → f (a) ash → 0; i.e., f is continuous at a.

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Proof.

Suppose that f is differentiable ata.Then by limh→0 f (a+h)−f (a)−Bh

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ.By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

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Theorem (11.13)

Proof.

khk =0,there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm,it follows that

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Proof.

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ.By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ.Since kBk is a finite real number, we

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Theorem (11.13)

Proof.

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm,it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number,we

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Proof.

khk =0, there is an m × n matrix B and a δ >0 such that kf (a + h) − f (a) − Bhk ≤ khk for all khk < δ. By the triangle inequality (see Theorem 8.6iii) and the definition of the operator norm, it follows that

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ.Since kBk is a finite real number, we

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Theorem (11.13)

Proof.

kf (a + h) − f (a)k ≤ kBkkhk + khk for khk < δ. Since kBk is a finite real number,we

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Proof.

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Theorem (11.14)

Let f be a vector function. If f is differentiable ata, then all first-order partial derivatives of f exists ata.

Proof.

Let B = [bij]be an m × n matrix that satisfies limh→0 f (a+h)−f (a)−Bh

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0.Since khk = t, we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Be_j.

Take the limit of this identity as t → 0+, using limh→0 f (a+h)−f (a)−Bh

khk =0 and the definition of matrix multiplication. We obtain

t→0+lim

f (a + tej) −f (a)

t =Be_j = (b_1j, . . . ,b_mj).

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Proof.

khk =0.Fix 1 ≤ j ≤ n and seth = tej for some t > 0. Since khk = t,we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Be_j.

t→0+lim

f (a + tej) −f (a)

t =Be_j = (b_1j, . . . ,b_mj).

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Theorem (11.14)

Proof.

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0.Since khk = t, we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Be_j.

t→0+lim

f (a + tej) −f (a)

t =Be_j = (b_1j, . . . ,b_mj).

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Proof.

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0. Since khk = t,we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Be_j.

khk =0 and the definition of matrix multiplication.We obtain

t→0+lim

f (a + tej) −f (a)

t =Be_j = (b_1j, . . . ,b_mj).

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Theorem (11.14)

Proof.

khk =0. Fix 1 ≤ j ≤ n and set h = tej for some t > 0. Since khk = t, we have

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Be_j.

t→0+lim

f (a + tej) −f (a)

t =Be_j = (b_1j, . . . ,b_mj).

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Proof.

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Be_j.

khk =0 and the definition of matrix multiplication.We obtain

t→0+lim

f (a + tej) −f (a)

t =Be_j = (b_1j, . . . ,b_mj).

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Theorem (11.14)

Proof.

f (a + h) − f (a) − Bh

khk := f (a + tej) −f (a)

t − Be_j.

t→0+lim

f (a + tej) −f (a)

t =Be_j = (b_1j, . . . ,b_mj).

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A similar argument shows that the limit of this quotient as t → 0− also exists and equals (b1j, . . .bmj).Since a vector function converges if and only if each of its components converges (see Theorem 9.15), it follows that the

first-order partial derivative of each component fi with respect to xj exists ata and satisfies

∂fi

∂xj

(a) = bij

for i = 1, 2, . . . , m.In particular,

(4) B = ∂fi

∂xj

(a)

m×n

:=







∂f1

∂x1(a) · · · _∂x^∂f¹

n(a) ... . .. ...

∂fm

∂x₁(a) · · · ^∂f_∂x^m

n(a)





.

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Proof.

A similar argument shows that the limit of this quotient as t → 0− also exists and equals (b1j, . . .bmj). Since a vector function converges if and only if each of its components converges (see Theorem 9.15),it follows that the

∂fi

∂xj

(a) = bij

for i = 1, 2, . . . , m. In particular,

(4) B = ∂fi

∂xj

(a)

m×n

:=







∂f1

∂x1(a) · · · _∂x^∂f¹

n(a) ... . .. ...

∂fm

∂x₁(a) · · · ^∂f_∂x^m

n(a)





.

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A similar argument shows that the limit of this quotient as t → 0− also exists and equals (b1j, . . .bmj). Since a vector function converges if and only if each of its components converges (see Theorem 9.15), it follows that the

∂fi

∂xj

(a) = bij

for i = 1, 2, . . . , m.In particular,

(4) B = ∂fi

∂xj

(a)

m×n

:=







∂f1

∂x1(a) · · · _∂x^∂f¹

n(a) ... . .. ...

∂fm

∂x₁(a) · · · ^∂f_∂x^m

n(a)





.

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Proof.

A similar argument shows that the limit of this quotient as t → 0− also exists and equals (b1j, . . .bmj). Since a vector function converges if and only if each of its components converges (see Theorem 9.15), it follows that the

∂fi

∂xj

(a) = bij

for i = 1, 2, . . . , m. In particular,

(4) B = ∂fi

∂xj

(a)

m×n

:=







∂f1

∂x1(a) · · · _∂x^∂f¹

n(a) ... . .. ...

∂fm

∂x₁(a) · · · ^∂f_∂x^m

n(a)





.

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Df (a) := [∂fi

∂xj

(a)]m×n, i = 1, 2, . . . , m j = 1, 2, . . . , n

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Theorem (11.15)

Let V be open inRⁿ, leta ∈ V , and suppose that

f : V →R^m. If all first-order partial derivatives of f exist in V and are continuous ata, then f is differentiable at a.

Note: These hypotheses are met if f is C¹on V . Proof.

Since a function converges if and only if its components converge (see Theorem 9.15), we may suppose that m = 1. By definition,then, it suffices to show that

h→0lim

f (a + h) − f (a) − ∇f (a) · h

khk =0.

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Since a function converges if and only if its components converge (see Theorem 9.15),we may suppose that m = 1. By definition, then, it suffices to show that

h→0lim

f (a + h) − f (a) − ∇f (a) · h

khk =0.

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Theorem (11.15)

Since a function converges if and only if its components converge (see Theorem 9.15), we may suppose that m = 1. By definition,then, it suffices to show that

h→0lim

f (a + h) − f (a) − ∇f (a) · h

khk =0.

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Since a function converges if and only if its components converge (see Theorem 9.15), we may suppose that m = 1. By definition, then, it suffices to show that

h→0lim

f (a + h) − f (a) − ∇f (a) · h

khk =0.

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Proof.

Leta = (a1, . . . ,an).Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h₁, . . . ,hn) 6=0 in B_r(0).By

telescoping and using the one-dimensional Mean Value Theorem, we can choose numbers cj between aj and aj +hj such that

f (a + h)−f (a) = f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

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Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V .Fixh = (h₁, . . . ,hn) 6=0 in B_r(0). By

telescoping and using the one-dimensional Mean Value Theorem,we can choose numbers cj between aj and aj +hj such that

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

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Proof.

Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h₁, . . . ,hn) 6=0 in B_r(0).By

f (a + h)−f (a)=f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

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Leta = (a1, . . . ,an). Choose r > 0 so small that Br(a) ⊂ V . Fix h = (h₁, . . . ,hn) 6=0 in B_r(0). By

telescoping and using the one-dimensional Mean Value Theorem,we can choose numbers cj between aj and aj +hj such that

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

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Proof.

f (a + h)−f (a)=f (a1+h1, . . . ,an+hn)−f (a1,a2+h2, . . . ,an+hn)

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

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+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

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Proof.

+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

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+ . . . +f (a1, . . . ,an−1,an+hn) −f (a1, . . . ,an)

=

n

X

j=1

hj

∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn).

Therefore,

(5) f (a + h)−f (a)−∇f (a)·h = h · δ,

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Proof.

where δ ∈Rⁿ is the vector with components δj = ∂f

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n; i.e., kδk → 0 as h → 0.

Moreover, by the Cauchy-Schwarz Inequality and (5),

(6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0. Thus f is differentiable at a by definition.

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∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Since the first-order partial derivatives of f are continuous ata, δj → 0 for each 1 ≤ j ≤ n;i.e., kδk → 0 as h → 0.

(6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

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Proof.

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

(6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

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∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

Moreover, by the Cauchy-Schwarz Inequality and (5), (6) 0 ≤ |f (a + h) − f (a) − ∇f (a) · h|

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0.Thus f is differentiable ata by definition.

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Proof.

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

khk = h · δ

khk ≤ kδk.

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∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

khk = h · δ

khk ≤ kδk.

It follows from the Squeeze Theorem that first quotient in (6) converges to 0 ash → 0.Thus f is differentiable ata by definition.

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Proof.

∂xj

(a1, . . . ,aj−1,cj,aj+1+hj+1, . . . ,an+hn)−∂f

∂xj

(a1, . . . ,an).

khk = h · δ

khk ≤ kδk.

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Prove that

f (x , y ) =( (x²+y²)sin√ ¹

x²+y² (x , y ) 6= (0, 0)

0 (x , y ) = (0, 0)

is differentiable onR²but not continuously differentiable at (0, 0).

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Advanced Calculus (II)