Differentiation Written by

Differentiation

Written by Men-Gen Tsai email: b89902089@ntu.edu.tw 1. Let f be defined for all real x, and suppose that

|f (x) − f (y)| ≤ (x − y)² for all real x and y. Prove that f is constant.

Proof: |f (x) − f (y)| ≤ (x − y)² for all real x and y. Fix y, |f (x)−f (y) x−y | ≤

|x − y|. Let x → y, therefore, 0 ≤ lim

x→y

f (x) − f (y) x − y ≤ lim

x→y|x − y| = 0

It implies that (f (x) − f (y))/(x − y) → 0 as x → y. Hence f⁰(y) = 0, f = const.

2. Suppose f⁰(x) > 0 in (a, b). Prove that f is strictly increasing in (a, b), and let g be its inverse function. Prove that g is differentible, and that

g⁰(f (x)) = 1

f⁰(x) (a < x < b).

Proof: For every pair x > y in (a, b), f (x) − f (y) = f⁰(c)(x − y) where y < c < x by Mean-Value Theorem. Note that c ∈ (a, b) and f⁰(x) > 0 in (a, b), hence f⁰(c) > 0. f (x) − f (y) > 0, f (x) > f (y) if x > y, f is strictly increasing in (a, b).

Let ∆g = g(x₀+ h) − g(x₀). Note that x₀ = f (g(x₀)), and thus, (x₀+ h) − x₀ = f (g(x₀+ h)) − f (g(x₀)),

h = f (g(x₀) + ∆g) − f (g(x₀)) = f (g + ∆g) − f (g).

Thus we apply the fundamental lemma of differentiation, h = [f⁰(g) + η(∆g)]∆g,

1

f⁰(g) + η(∆g) = ∆g h

Note that f⁰(g(x)) > 0 for all x ∈ (a, b) and η(∆g) → 0 as h → 0, thus,

h→0lim∆g/h = lim

h→0

1

f⁰(g) + η(∆g) = 1 f⁰(g(x)). Thus g⁰(x) = _f0(g(x))¹ , g⁰(f (x)) = _f0¹(x).

3. Suppose g is a real function on R¹, with bounded derivative (say

|g⁰| ≤ M ). Fix  > 0, and define f (x) = x + g(x). Prove that f is one-to-one if  is small enough. (A set of admissible values of  can be determined which depends only on M .)

Proof: For every x < y, and x, y ∈ R, we will show that f (x) 6= f (y).

By using Mean-Value Theorem:

g(x) − g(y) = g⁰(c)(x − y) where x < c < y, (x − y) + ((x) − g(y)) = (g⁰(c) + 1)(x − y), that is,

f (x) − f (y) = (g⁰(c) + 1)(x − y). (∗)

Since |g⁰(x)| ≤ M , −M ≤ g⁰(x) ≤ M for all x ∈ R. Thus 1 − M ≤

g⁰(c) + 1 ≤ 1 + M , where x < c < y. Take c = _2M¹ , and g⁰(c) + 1 > 0 where x < c < y for all x, y. Take into equation (*), and f (x)−f (y) < 0 since x − y < 0, that is, f (x) 6= f (y), that is, f is one-to-one (injective).

4. If

C₀+C₁

2 + ... + C_n−1

n + C_n n + 1 = 0,

where C₀, ..., C_n are real constants, prove that the equation C0+ C1x + ... + Cn−1xⁿ⁻¹+ Cnxⁿ= 0 has at least one real root between 0 and 1.

Proof: Let f (x) = C₀x + ... + _n+1^Cn xⁿ⁺¹. f is differentiable in R¹ and f (0) = f (1) = 0. Thus, f (1) − f (0) = f⁰(c) where c ∈ (0, 1) by Mean-Value Theorem. Note that

f⁰(x) = C₀+ C₁x + ... + C_n−1xⁿ⁻¹+ C_nxⁿ.

Thus, c ∈ (0, 1) is one real root between 0 and 1 of that equation.

5. Suppose f is defined and differentiable for every x > 0, and f⁰(x) → 0 as x → +∞. Put g(x) = f (x + 1) − f (x). Prove that g(x) → 0 as x → +∞.

Proof: f (x + 1) − f (x) = f⁰(c)(x + 1 − x) where x < c < x + 1 by Mean-Value Theorem. Thus, g(x) = f⁰(c) where x < c < x + 1, that is,

x→+∞lim g(x) = lim

x→+∞f⁰(c) = lim

c→+∞f⁰(c) = 0.

6. Suppose

(a) f is continuous for x ≥ 0, (b) f⁰(x) exists for x > 0, (c) f (0) = 0,

(d) f⁰ is monotonically increasing.

Put

g(x) = f (x)

x (x > 0) and prove that g is monotonically increasing.

Proof: Our goal is to show g⁰(x) > 0 for all x > 0

⇔ g⁰(x) = ^{xf0(x)−f (x)}_x2 > 0 ⇔ f⁰(x) > ^{f (x)}_x .

Since f⁰(x) exists, f (x) − f (0) = f⁰(c)(x − 0) where 0 < c < x by Mean-Value Theorem. ⇒ f⁰(c) = ^{f (x)}_x where 0 < c < x. Since f⁰ is monotonically increasing, f⁰(x) > f⁰(c), that is, f⁰(x) > ^{f (x)}_x for all x > 0.

7. Suppose f⁰(x), g⁰(x) exist, g⁰(x) 6= 0, and f (x) = g(x) = 0. Prove that limt→x

f (t)

g(t) = f⁰(x) g⁰(x). (This holds also for complex functions.)

Proof:

f⁰(t)

g⁰(t) = lim_t→x f (t)−f (x) t−x

lim_t→x ^g(t)−g(x)_t−x = lim_t→xf (t) limt→xg(t) = lim

t→x

f (t) g(t) Surely, this holds also for complex functions.

8. Suppose f⁰(x) is continuous on [a, b] and  > 0. Prove that there exists δ > 0 such that

|f (t) − f (x)

t − x − f⁰(x)| < 

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. (This could be expressed by saying f is uniformly differentiable on [a, b] if f⁰ is contin- uous on [a, b].) Does this hold for vector-valued functions too?

Proof: Since f⁰(x) is continuous on a compact set [a, b], f⁰(x) is uni- formly continuous on [a, b]. Hence, for any  > 0 there exists δ > 0 such that

|f⁰(t) − f⁰(x)| < 

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. Thus, f⁰(c) =

f (t)−f (x)

t−x where c between t and x by Mean-Value Theorem. Note that 0 < |c − x| < δ and thus |f⁰(c) − f⁰(x)| < , thus,

|f (t) − f (x)

t − x − f⁰(x)| <  whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b.

Note: It does not hold for vector-valued functions. If not, take f (x) = (cos x, sin x),

[a, b] = [0, 2π], and x = 0. Hence f⁰(x) = (− sin x, cos x). Take any 1 >  > 0, there exists δ > 0 such that

|f (t) − f (0)

t − 0 − f⁰(0)| < 

whenever 0 < |t| < δ by our hypothesis. With calculating,

|(cos t − 1 t ,sin t

t ) − (0, 1)| < 

|(cos t − 1 t ,sin t

t − 1)| <  (cos t − 1

t )² + (sin t

t − 1)² < ² <  2

t² + 1 − 2(cos t + sin t) t <  since 1 >  > 0. Note that

2

t² + 1 − 4 t < 2

t² + 1 − 2(cos t + sin t) t

But _t²2 + 1 − ⁴_t → +∞ as t → 0. It contradicts.

9. Let f be a continuous real function on R¹, of which it is known that f⁰(x) exists for all x 6= 0 and that f⁰(x) → 0 as x → 0. Dose it follow that f⁰(0) exists?

Note: We prove a more general exercise as following.

Suppose that f is continuous on an open interval I containing x₀, sup- pose that f⁰ is defined on I except possibly at x0, and suppose that f⁰(x) → L as x → x₀. Prove that f⁰(x₀) = L.

Proof of the Note: Using L’Hospital’s rule:

h→0lim

f (x₀ + h) − f (x₀)

h = lim

h→0f⁰(x₀+ h) By our hypothesis: f⁰(x) → L as x → x₀. Thus,

h→0lim

f (x₀+ h) − f (x₀)

h = L,

Thus f⁰(x₀) exists and

f⁰(x₀) = L.

10. Suppose f and g are complex differentiable functions on (0, 1), f (x) → 0, g(x) → 0, f⁰(x) → A, g⁰(x) → B as x → 0, where A and B are complex numbers, B 6= 0. Prove that

x→0lim f (x) g(x) = A

B. Compare with Example 5.18. Hint:

f (x)

g(x) = (f (x)

x − A) x

g(x) + A x g(x).

Apply Theorem 5.13 to the real and imaginary parts of ^{f (x)}_x and ^g(x)_x .

Proof: Write f (x) = f₁(x) + if₂(x), where f₁(x), f₂(x) are real-valued functions. Thus,

df (x)

dx = df1(x)

dx + idf2(x) dx , Apply L’Hospital’s rule to ^f1_x^(x) and ^f2_x^(x), we have

x→0lim f1(x)

x = lim

x→0f₁⁰(x)

x→0lim f₂(x)

x = lim

x→0f₂⁰(x) Combine f1(x) and f2(x), we have

x→0lim f₁(x)

x + i lim

x→0

f₂(x)

x = lim

x→0

f₁(x)

x + if₂(x) x = lim

x→0

f (x) x or

x→0lim f1(x)

x + i lim

x→0

f2(x)

x = lim

x→0f₁⁰(x) + i lim

x→0f₂⁰(x) = lim

x→0f⁰(x) Thus, lim_x→0^{f (x)}_x = lim_x→0f⁰(x). Similarly, lim_x→0 ^g(x)_x = lim_x→0g⁰(x).

Note that B 6= 0. Thus,

x→0lim f (x)

g(x) = lim

x→0(f (x)

x − A) x

g(x)+ A x g(x)

= (A − A)1 B + A

B = A B.

Note: In Theorem 5.13, we know g(x) → +∞ as x → 0. (f (x) = x, and g(x) = x + x²e^x2i ).

11. Suppose f is defined in a neighborhood of x, and suppose f ”(x) exists.

Show that

h→0lim

f (x + h) + f (x − h) − 2f (x)

h² = f⁰⁰(x).

Show by an example that the limit may exist even if f ”(x) dose not.

Hint: Use Theorem 5.13.

Proof: By using L’Hospital’s rule: (respect to h.)

h→0lim

f (x + h) + f (x − h) − 2f (x)

h² = lim

h→0

f⁰(x + h) − f⁰(x − h) 2h

Note that

f⁰⁰(x) = 1

2(f⁰⁰(x) + f⁰⁰(x))

= 1

2(lim

h→0

f⁰(x + h) − f⁰(x)

h + lim

h→0

f⁰(x − h) − f⁰(x)

−h )

= 1

2lim

h→0

f⁰(x + h) − f⁰(x − h) h

= lim

h→0

f⁰(x + h) − f⁰(x − h) 2h

Thus,

f (x + h) + f (x − h) − 2f (x)

h² → f⁰⁰(x)

as h → 0. Counter-example: f (x) = x|x| for all real x.

12. If f (x) = |x|³, compute f⁰(x), f⁰⁰(x) for all real x, and show that f⁽³⁾(0) does not exist.

Proof: f⁰(x) = 3|x|² if x 6= 0. Consider f (h) − f (0)

h = |h|³ h

Note that |h|/h is bounded and |h|² → 0 as h → 0. Thus, f⁰(0) = lim

h→0

f (h) − f (0)

h = 0.

Hence, f⁰(x) = 3|x|² for all x. Similarly, f⁰⁰(x) = 6|x|.

Thus,

f⁰⁰(h) − f (0)

h = 6|h|

h

Since ^|h|_h = 1 if h > 0 and = −1 if h < 0, f⁰⁰⁰(0) does not exist.

13. Suppose a and c are real numbers, c > 0, and f is defined on [−1, 1] by f (x) =







x^asin (x^−c) (if x 6= 0),

0 (if x = 0).

Prove the following statements:

(a) f is continuous if and only if a > 0.

(b) f⁰(0) exists if and only if a > 1.

(c) f⁰ is bounded if and only if a ≥ 1 + c.

(d) f⁰ is continuous if and only if a > 1 + c.

(e) f⁰⁰(0) exists if and only if a > 2 + c.

(f) f⁰⁰ is bounded if and only if a ≥ 2 + 2c.

(g) f⁰⁰ is continuous if and only if a > 2 + 2c.

Proof: For (a): (⇒) f is continuous iff for any sequence {x_n} → 0 with xn 6= 0, x^a_nsin x^−c_n → 0 as n → ∞. In particular, take

x_n= ( 1

2nπ + π/2)^1c > 0

and thus x^a_n → 0 as n → ∞. Hence a > 0. (If not, then a = 0 or a < 0.

When a = 0, x^a_n = 1. When a < 0, x^a_n = 1/x^−a_n → ∞ as n → ∞. It contradicts.)

(⇐) f is continuous on [−1, 1] − {0} clearly. Note that

−|x^a| ≤ x^asin (x^−c) ≤ |x^a|,

and |x^a| → 0 as x → 0 since a > 0. Thus f is continuous at x = 0.

Hence f is continuous.

For (b): f⁰(0) exists iff x^a−1sin (x^−c) → 0 as x → 0. In the previous proof we know that f⁰(0) exists if and only if a − 1 > 0. Also, f⁰(0) = 0.

14. Let f be a differentiable real function defined in (a, b). Prove that f is convex if and only if f⁰ is monotonically increasing. Assume next that f⁰⁰(x) exist for every x ∈ (a, b), and prove that f is convex if and only if f⁰⁰(x) ≥ 0 for all x ∈ (a, b).

15. Suppose a ∈ R¹, f is a twice-differentiable real function on (a, ∞), and M₀, M₁, M₂ are the least upper bounds of |f (x)|, |f⁰(x)|, |f⁰⁰(x)|, respectively, on (a, ∞). Prove that

M₁² ≤ 4M₀M₂. Hint: If h > 0, Taylor’s theorem shows that

f⁰(x) = 1

2h[f (x + 2h) − f (x)] − hf⁰⁰(ξ) for some ξ ∈ (x, x + 2h). Hence

|f⁰(x)| ≤ hM2+M₀ h .

To show that M₁² = 4M₀M₂ can actually happen, take a = −1, define

f (x) =







2x²− 1 (−1 < x < 0),

x²−1

x²+1 (0 ≤ x < ∞),

and show that M₀ = 1, M₁ = 4, M₂ = 4. Does M₁² ≤ 4M₀M₂ hold for vector-valued functions too?

Proof: Suppose h > 0. By using Taylor’s theorem:

f (x + h) = f (x) + hf⁰(x) + h² 2 f⁰⁰(ξ)

for some x < ξ < x + 2h. Thus

h|f⁰(x)| ≤ |f (x + h)| + |f (x)| + h²

2 |f⁰⁰(ξ)|

h|f⁰(x)| ≤ 2M0 +h² 2 M2. h²M₂− 2h|f⁰(x)| + 4M₀ ≥ 0 (∗)

Since equation (*) holds for all h > 0, its determinant must be non- positive.

4|f⁰(x)|²− 4M2(4M0) ≤ 0

|f⁰(x)|² ≤ 4M₀M₂ (M₁)² ≤ 2M₀M₂ Note: There is a similar exercise:

Suppose f (x)(−∞ < x < +∞) is a twice-differentiable real function, and

Mk= sup

−∞<x<+∞|f^(k)(x)| < +∞ (k = 0, 1, 2).

Prove that M₁² ≤ 2M₀M₂.

Proof of Note:

f (x + h) = f (x) + f⁰(x)h + f⁰⁰(ξ₁) 2 h²

(x < ξ₁ < x + h or x > ξ₁ > x + h) . . . (*) f (x − h) = f (x) − f⁰(x)h + f⁰⁰(ξ₂)

2 h²

(x − h < ξ₂ < x or x − h > ξ₂ > x) . . . (**) (*) minus (**):

f (x + h) − f (x − h) = 2f⁰(x)h + h²

2(f⁰⁰(ξ₁) − f⁰⁰(ξ₂)).

2h|f⁰(x)| ≤ |2hf⁰(x)|

2h|f⁰(x)| ≤ |f (x + h)| + |f (x − h)| + h²

2 (|f⁰⁰(ξ₁)| + |f⁰⁰(ξ₂)|) 2h|f⁰(x)| ≤ 2M₀+ h²M₂

M₂h²− 2|f⁰(x)|h + 2M₀ ≥ 0

Since this equation holds for all h, its determinant must be non-positive:

4|f⁰(x)|²− 4M₂(2M₀) ≤ 0,

|f⁰(x)|² ≤ 2M₀M₂ Thus

M₁² ≤ 2M₀M₂

16. Suppose f is twice-differentiable on (0, ∞), f⁰⁰ is bounded on (0, ∞), and f (x) → 0 as x → ∞. Prove that f⁰(x) → 0 as x → ∞. Hint: Let a → ∞ in Exercise 15.

Proof: Suppose a ∈ (0, ∞), and M₀, M₁, M₂ are the least upper bounds of |f (x)|, |f⁰(x)|, |f⁰⁰(x)| on (a, ∞). Hence, M₁² ≤ 4M₀M₂. Let a → ∞, M0 = sup |f (x)| → 0. Since M2 is bounded, therefore M₁² → 0 as a → ∞. It implies that sup |f⁰(x)| → 0 as x → ∞.

17. Suppose f is a real, three times differentiable function on [−1, 1], such that

f (−1) = 0, f (0) = 0, f (1) = 1, f⁰(0) = 0.

Prove that f⁽³⁾(x) ≥ 3 for some x ∈ (−1, 1).

Note that equality holds for ¹₂(x³+ x²).

Hint: Use Theorem 5.15, with α = 0 and β = 1, −1, to show that there exist s ∈ (0, 1) and t ∈ (−1, 0) such that

f⁽³⁾(s) + f⁽³⁾(t) = 6.

Proof: By Theorem 5.15, we take α = 0, β = 1, f (1) = f (0) + f⁰(0) + f⁰⁰(0)

2 + f⁽³⁾(s) 6 where s ∈ (0, 1). Take α = 0, andβ = −1,

f (−1) = f (0) − f⁰(0) + f⁰⁰(0)

2 − f⁽³⁾(t) 6 where t ∈ (−1, 0). Thus

1 = f⁰⁰(0)

2 +f⁽³⁾(s)

6 , s ∈ (0, 1) (∗) 0 = f⁰⁰(0)

2 − f⁽³⁾(s)

6 , s ∈ (−1, 0) (∗∗) Equation (*) - equation (**):

f⁽³⁾(s)

6 + f⁽³⁾(t)

6 , s ∈ (0, 1), t ∈ (−1, 0).

f⁽³⁾(s) + f⁽³⁾(t) = 6, s, t ∈ (−1, 1).

f⁽³⁾(x) ≥ 3 f or some x ∈ (−1, 1).

Theorem 5.15: Suppose f is a real function on [a, b], n is a positive integer, f⁽ⁿ⁻¹⁾ is continuous on [a, b], f⁽ⁿ⁾(t) exists for every t ∈ (a, b).

Let α, β be distinct points of [a, b], and define P (t) =

n−1

X

k=0

f^(k)(α)

k! (t − α)^k.

Then there exists a point x between α and β such that f (β) = P (β) +f⁽ⁿ⁾(x)

n! (β − α)ⁿ.

18. Suppose f is a real function on [a, b], n is a positive integer, and f⁽ⁿ⁻¹⁾ exists for every t ∈ [a, b]. Let α, β, and P be as in Taylor’s theorem (5.15). Define

Q(t) = f (t) − f (β) t − β for t ∈ [a, b], t 6= β, differentiate

f (t) − f (β) = (t − β)Q(t)

n − 1 times at t = α, and derive the following version of Taylor’s theorem:

f (β) = P (β) + Q⁽ⁿ⁻¹⁾(α)

(n − 1)! (β − α)ⁿ.

19. Suppose f is defined in (−1, 1) and f⁰(0) exists. Suppose −1 < α_n <

β_n < 1, α_n→ 0, and β_n→ 0 as n → ∞. Define the difference quotients D_n= f (βn) − f (αn)

β_n− α_n Prove the following statements:

(a) If α_n < 0 < β_n, then lim D_n= f⁰(0).

(b) If 0 < α_n< β_nand {β_n/(β_n−α_n)} is bounded, then lim D_n= f⁰(0).

(c) If f⁰ is continuous in (−1, 1), then lim D_n = f⁰(0).

Give an example in which f is differentiable in (−1, 1) (but f⁰ is not continuous at 0) and in which α_n, β_n tend to 0 in such a way that

lim D_n exists but is different from f⁰(0).

Proof: For (a):

D_n= f (β_n) − f (0) β_n

β_n

β_n− α_n +f (α_n) − f (0) α_n

−α_n β_n− α_n Note that

f⁰(0) = lim

n→∞

f (α_n) − f (0)

α_n = lim

n→∞

f (β_n) − f (0) β_n Thus for any  > 0, there exists N such that

L −  < f (α_n) − f (0)

α_n < L + , L −  < f (β_n) − f (0)

β_n < L + ,

whenever n > N where L = f⁰(0) respectively. Note that β_n/(β_n− α_n) and −α_n/(β_n− α_n) are positive. Hence,

β_n

β_n− α_n(L − ) < f (β_n) − f (0) β_n

β_n

β_n− α_n < β_n

β_n− α_n(L + )

−α_n βn− αn

(L − ) < f (α_n) − f (0) αn

−α_n βn− αn

< −α_n βn− αn

(L + ) Combine two inequations,

L −  < Dn< L +  Hence, lim D_n = L = f⁰(0).

For (b): We process as above prove, but note that −α_n/(β_n− α_n) < 0.

Thus we only have the following inequations:

β_n

β_n− α_n(L − ) < f (β_n) − f (0) β_n

β_n

β_n− α_n < β_n

β_n− α_n(L + )

−α_n

β_n− α_n(L + ) < f (α_n) − f (0) α_n

−α_n

β_n− α_n < −α_n

β_n− α_n(L − ) Combine them:

L − β_n+ α_n

β_n− α_n < D_n< L + β_n+ α_n β_n− α_n Note that {β_n/(β_n− α_n)} is bounded, ie,

| β_n βn− αn

| ≤ M

for some constant M . Thus

|βn+ αn

β_n− α_n| = | 2βn

β_n− α_n − 1| ≤ 2M + 1 Hence,

L − (2M + 1) < D_n< L + (2M + 1)

Hence, lim D_n = L = f⁰(0).

For (c): By using Mean-Value Theorem, D_n = f⁰(t_n) where t_n is between α_n and β_n. Note that

min{αn, βn} < tn< max{αn, βn} and

max{α_n, β_n} = 1

2(α_n+ β_n+ |α_n− β_n|) min{αn, βn} = 1

2(αn+ βn− |αn− βn|)

Thus, max{α_n, β_n} → 0 and min{α_n, β_n} → 0 as α_n→ 0 and β_n→ 0.

By squeezing principle for limits, t_n → 0. With the continuity of f⁰, we have

lim D_n = lim f⁰(t_n) = f⁰(lim t_n) = f⁰(0).

Example: Let f be defined by

f (x) =







x²sin(1/x) (x 6= 0),

0 (x = 0).

Thus f⁰(x) is not continuous at x = 0, and f⁰(0) = 0. Take α_n =

1

π/2+2nπ and β_n= _2nπ¹ . It is clear that α_n→ 0, and β_n→ 0 as n → ∞.

Also,

D_n = −4nπ

π(π/2 + 2nπ) → −2 π

as n → ∞. Thus, lim D_n= −2/π exists and is different from 0 = f⁰(0).

20.

21.

22. Suppose f is a real function on (−∞, ∞). Call x a fixed point of f if f (x) = x.

(a) If f is differentiable and f⁰(t) 6= 1 for every real t, prove that f has at most one fixed point.

(b) Show that the function f defined by f (t) = t + (1 + e^t)⁻¹

has no fixed point, although 0 < f⁰(t) < 1 for all real t.

(c) However, if there is a constant A < 1 such that |f⁰(t)| ≤ A for all real t, prove that a fixed point x of f exists, and that x = lim xn, where x₁ is an arbitrary real number and

x_n+1 = f (x_n) for n = 1, 2, 3, ...

(d) Show that the process describe in (c) can be visualized by the zig- zag path

(x₁, x₂) → (x₂, x₂) → (x₂, x₃) → (x₃, x₃) → (x₃, x₄) → ....

Proof: For (a): If not, then there exists two distinct fixed points, say x and y, of f . Thus f (x) = x and f (y) = y. Since f is differeniable, by applying Mean-Value Theorem we know that

f (x) − f (y) = f⁰(t)(x − y)

where t is between x and y. Since x 6= y, f⁰(t) = 1. It contradicts.

For (b): We show that 0 < f⁰(t) < 1 for all real t first:

f⁰(t) = 1 + (−1)(1 + e^t)⁻²e^t= 1 − e^t (1 + e^t)² Since

e^t> 0

(1 + e^t)² = (1 + e^t)(1 + e^t) > 1(1 + e^t) = 1 + e^t> e^t> 0 for all real t, thus

(1 + e^t)⁻²e^t> 0 (1 + e^t)⁻²e^t< 1 for all real t. Hence 0 < f⁰(t) < 1 for all real t.

Next, since f (t) − t = (1 − e^t)⁻¹ > 0 for all real t, f (t) has no fixed point.

For (c): Suppose x_n+16= x_nfor all n. (If x_n+1 = x_n, then x_n= x_n+1= ... and x_n is a fixed point of f ).

By Mean-Value Theorem,

f (x_n+1) − f (x_n) = f⁰(t_n)(x_n+1− x_n)

where t_n is betweem x_n and x_n+1. Thus,

|f (x_n+1) − f (x_n)| = |f⁰(t_n)||(x_n+1− x_n)|

Note that |f⁰(t_n)| is bounded by A < 1, f (x_n) = x_n+1, and f (x_n+1) = x_n+2. Thus

|xn+2− xn+1| ≤ A|xn+1− xn|

|x_n+1− x_n| ≤ CAⁿ⁻¹

where C = |x₂− x₁|. For two positive integers p > q,

|x_p− x_q| ≤ |x_p − x_p−1| + ... + |x_q+1− x_q|

= C(A^q−1+ A^q+ ... + A^p−2)

≤ CA^q−1 1 − A. Hence

|x_p − x_q| ≤ CA^q−1 1 − A.

Hence, for any  > 0, there exists N = [log_A^(1−A)_C ] + 2 such that

|x_p− x_q| <  whenever p > q ≥ N . By Cauchy criterion we know that {x_n} converges to x. Thus,

n→∞lim x_n+1 = f ( lim

n→∞x_n) since f is continuous. Thus,

x = f (x).

x is a fixed point of f .

For (d): Since x_n+1 = f (x_n), it is trivial.

23.

24.

25. Suppose f is twice differentiable on [a, b], f (a) < 0, f (b) > 0, f⁰(x) ≥ δ > 0, and 0 ≤ f⁰⁰(x) ≤ M for all x ∈ [a, b]. Let ξ be the unique point in (a, b) at which f (ξ) = 0.

Complete the details in the following outline of Newton’s method for computing ξ.

(a) Choose x₁ ∈ (ξ, b), and define {x_n} by x_n+1= x_n− f (x_n)

f⁰(x_n).

Interpret this geometrically, in terms of a tangent to the graph of f . (b) Prove that x_n+1 < x_n and that

n−>oolim x_n = ξ.

(c) Use Taylor’s theorem to show that x_n+1− ξ = f⁰⁰(t_n)

2f⁰(xn)(x_n− ξ)² for some t_n ∈ (ξ, x_n).

(d) If A = ^M_2δ, deduce that

0 ≤ x_n+1− ξ ≤ 1

A[A(x₁− ξ)]²ⁿ. (Compare with Exercise 16 and 18, Chap. 3)

(e) Show that Newton’s method amounts to finding a fixed point of the function g defined by

g(x) = x − f (x) f⁰(x). How does g⁰(x) behave for x near ξ?

(f) Put f (x) = x^1/3 on (−∞, ∞) and try Newton’s method. What happens?

Proof: For (a): You can see the picture in the following URL:

http://archives.math.utk.edu/visual.calculus/3/newton.5/1.html.

For (b): We show that x_n ≥ x_n+1 ≥ ξ. (induction). By Mean-Value Theorem, f (x_n) − f (ξ) = f⁰(c_n)(x_n − ξ) where c_n ∈ (ξ, x_n). Since f⁰⁰≥ 0, f⁰ is increasing and thus

f (x_n)

x_n− ξ = f⁰(c_n) ≤ f⁰(x_n) = f (x_n) x_n− x_n+1 f (x_n)(x_n− ξ) ≤ f (x_n)(x_n− x_n+1)

Note that f (x_n) > f (ξ) = 0 since f⁰ ≥ δ > 0 and f is strictly increasing.

Thus,

x_n− ξ ≤ x_n− x_n+1 ξ ≤ x_n+1

Note that f (x_n) > 0 and f⁰(x_n) > 0. Thus x_n+1 < x_n. Hence, x_n> x_n+1≥ ξ.

Thus, {xn} converges to a real number ζ. Suppose ζ 6= ξ, then x_n+1 = x_n− f (x_n)

f⁰(x_n)

Note that _f^{f (x}0(xⁿn⁾) > ^{f (ζ)}_δ . Let α = ^{f (ζ)}_δ > 0, be a constant. Thus, x_n+1< x_n− α

for all n. Thus, x_n < x₁− (n − 1)α, that is, x_n → −∞ as n → ∞. It contradicts. Thus, {x_n} converges to ξ.

For (c): By using Taylor’s theorem,

f (ξ) = f (xn) + f⁰(xn)(ξ − xn) + f⁰⁰(t_n) 2(x_n− ξ)² 0 = f (x_n) + f⁰(x_n)(ξ − x_n) + f⁰⁰(t_n)

2(x_n− ξ)²

0 = f (x_n)

f⁰(x_n) − x_n+ ξ + f⁰⁰(t_n)

2f⁰(x_n)(x_n− ξ)² x_n+1− ξ = f⁰⁰(t_n)

2f⁰(x_n)(x_n− ξ)² where t_n ∈ (ξ, x_n).

For (d): By (b) we know that 0 ≤ x_n+1− ξ for all n. Next by (c) we know that

x_n+1− ξ = f⁰⁰(t_n)

2f⁰(x_n)(x_n− ξ)² Note that f⁰⁰ ≤ M and f⁰ ≥ δ > 0. Thus

x_n+1− ξ ≤ A(x_n− ξ)² ≤ 1

A(A(x₁− ξ))²ⁿ by the induction. Thus,

0 ≤ x_n+1− ξ ≤ 1

A[A(x₁− ξ)]²ⁿ.

For (e): If x0 is a fixed point of g(x), then g(x0) = x0, that is, x₀− f (x₀)

f⁰(x₀) = x₀ f (x0) = 0.

It implies that x₀ = ξ and x₀ is unique since f is strictly increasing.

Thus, we choose x₁ ∈ (ξ, b) and apply Newton’s method, we can find out ξ. Hence we can find out x₀.

Next, by calculating

g⁰(x) = f (x)f⁰⁰(x) f⁰(x)² 0 ≤ g⁰(x) ≤ f (x)M

δ².

As x near ξ from right hand side, g⁰(x) near f (ξ) = 0.

For (f ): x_n+1 = x_n− _f^{f (x}0(xⁿn⁾) = −2x_n by calculating. Thus, x_n= (−2)ⁿ⁻¹x₁

for all n, thus {xn} does not converges for any choice of x1, and we cannot find ξ such that f (ξ) = 0 in this case.

26. Suppose f is differentiable on [a, b], f (a) = 0, and there is a real number A such that |f⁰(x)| ≤ A|f (x)| on [a, b]. Prove that f (x) = 0 for all x ∈ [a, b]. Hint: Fix x₀ ∈ [a, b], let

M0 = sup |f (x)|, M1 = sup |f⁰(x)|

for a ≤ x ≤ x₀. For any such x,

|f (x)| ≤ M₁(x₀− a) ≤ A(x₀− a)M₀.

Hence M₀ = 0 if A(x₀− a) < 1. That is, f = 0 on [a, x₀]. Proceed.

Proof: Suppose A > 0. (If not, then f = 0 on [a, b] clearly.) Fix x₀ ∈ [a, b], let

M₀ = sup |f (x)|, M₁ = sup |f⁰(x)|

for a ≤ x ≤ x₀. For any such x,

f (x) − f (a) = f⁰(c)(x − a)

where c is between x and a by using Mean-Value Theorem. Thus

|f (x)| ≤ M₁(x − a) ≤ M₁(x₀ − a) ≤ A(x₀− a)M₀

Hence M₀ = 0 if A(x₀ − a) < 1. That is, f = 0 on [a, x₀] by taking x₀ = a + _2A¹ . Repeat the above argument by replacing a with x₀, and note that _2A¹ is a constant. Hence, f = 0 on [a, b].

27. Let φ be a real function defined on a rectangle R in the plane, given by a ≤ x ≤ b, α ≤ y ≤ β. A solution of the initial-value problem

y⁰ = φ(x, y), y(a) = c (α ≤ c ≤ β)

is, by definition, a differentiable function f on [a, b] such that f (a) = c, α ≤ f (x) ≤ β, and

f⁰(x) = φ(x, f (x)) (a ≤ x ≤ b)

Prove that such a problem has at most one solution if there is a constant A such that

|φ(x, y₂) − φ(x, y₁)| ≤ A|y₂− y₁| whenever (x, y₁) ∈ R and (x, y₂) ∈ R.

Hint: Apply Exercise 26 to the difference of two solutions. Note that this uniqueness theorem does not hold for the initial-value problem

y⁰ = y^1/2, y(0) = 0,

which has two solutions: f (x) = 0 and f (x) = x²/4. Find all other solutions.

Proof: Suppose y₁ and y₂ are solutions of that problem. Since

|φ(x, y2) − φ(x, y1)| ≤ A|y2− y1|,

y(a) = c, y₁⁰ = φ(x, y₁), and y⁰₂ = φ(x, y₂), by Exercise 26 we know that y₁− y₂ = 0, y₁ = y₂. Hence, such a problem has at most one solution.

Note: Suppose there is initial-value problem y⁰ = y^1/2, y(0) = 0.

If y^1/2 6= 0, then y^1/2dy = dx. By integrating each side and noting that y(0) = 0, we know that f (x) = x²/4. With y^1/2 = 0, or y = 0. All solutions of that problem are

f (x) = 0 and f (x) = x²/4

Why the uniqueness theorem does not hold for this problem? One reason is that there does not exist a constant A satisfying

|y₁⁰ − y₂⁰| ≤ A|y₁− y₂|

if y₁ and y₂ are solutions of that problem. (since 2/x → ∞ as x → 0 and thus A does not exist).

28. Formulate and prove an analogous uniqueness theorem for systems of differential equations of the form

y_j⁰ = φ_j(x, y₁, ..., y_k), y_j(a) = c_j (j = 1, ..., k) Note that this can be rewritten in the form

y⁰ = φ(x, y), y(a) = c

where y = (y₁, ..., y_k) ranges over a k-cell, φ is the mapping of a (k + 1)- cell into the Euclidean k-space whose components are the function φ₁, ..., φ_k, and c is the vector (c₁, ..., c_k). Use Exercise 26, for vector- valued functions.

Theorem: Let φ_j(j = 1, ..., k) be real functions defined on a rectangle R_j in the plane given by a ≤ x ≤ b, α_j ≤ y_j ≤ β_j.

A solution of the initial-value problem

y⁰_j = φ(x, y_j), y_j(a) = c_j (α_j ≤ c_j ≤ β_j)

is, by definition, a differentiable function f_j on [a, b] such that f_j(a) = c_j, α_j ≤ f_j(x) ≤ β_j, and

f_j⁰(x) = φ_j(x, f_j(x)) (a ≤ x ≤ b)

Then this problem has at most one solution if there is a constant A such that

|φ_j(x, y_j2) − φ_j(x, y_j1)| ≤ A|y_j2 − y_j1| whenever (x, y_j1) ∈ R_j and (x, y_j2) ∈ R_j.

Proof: Suppose y₁ and y₂ are solutions of that problem. For each components of y₁ and y₂, say y_1j and y_2j respectively, y_1j = y_2j by using Exercise 26. Thus, y₁ = y₂

29. Specialize Exercise 28 by considering the system y_j⁰ = yj+1 (j = 1, ..., k − 1),

y⁰_k= f (x) −

k

X

j=1

g_j(x)y_j

where f, g₁, ..., g_k are continuous real functions on [a, b], and derive a uniqueness theorem for solutions of the equation

y^(k)+ g_k(x)y^(k−1)+ ... + g₂(x)y⁰+ g₁(x)y = f (x), subject to initial conditions

y(a) = c1, y⁰(a) = c1, ..., y^(k−1)(a) = ck.

Theorem: Let R_j be a rectangle in the plain, given by a ≤ x ≤ b, min y_j ≤ y_j ≤ max y_j. (since y_j is continuous on the compact set, say

[a, b], we know that y_j attains minimal and maximal.) If there is a constant A such that







|y_j+1,1− y_j+1,2| ≤ A|y_j,1− y_j,2| (j < k)

|^Pk_j=1g_j(x)(y_j,1− y_j,2)| ≤ A|y_k,1− y_k,2| whenever (x, y_j,1) ∈ R_j and (x, y_j,2) ∈ R_j.

Proof: Since the system y₁⁰, ..., y_k⁰ with initial conditions satisfies a fact that there is a constant A such that |y₁⁰ −y₂⁰| ≤ A|y₁−y₂|, that system has at most one solution. Hence,

y^(k)+ g_k(x)y^(k−1)+ ... + g₂(x)y⁰+ g₁(x)y = f (x), with initial conditions has at most one solution.