(b) (4 points) Define what it means to say that f is a uniformly continuous function on D

Advanced Calculus Midterm Exam II May 18, 2011

There are 8 questions with total 126 points in this exam.

1. Let {fn} be a sequence of functions with domain D ⊂ R^p and range in R^q.

(a) (4 points) Define what it means to say that {f_n} is pointwise convergent to f on D.

(b) (4 points) Define what it means to say that {f_n} is uniformly convergent to f on D.

(c) (4 points) Define what it means to say that {f_n} is Not uniformly convergent to f on D.

2. Let f be a function with domain D ⊂ R^p and range in R^q.

(a) (4 points) Define what it means to say that f is a continuous function on D.

(b) (4 points) Define what it means to say that f is a uniformly continuous function on D.

(c) (4 points) Define what it means to say that f is a Lipschitz function on D..

3. (10 points) Let f : R² → R² be defined by f (x, y) = (x²− y², 2xy). For each (x, y) 6= (0, 0), show that there is an open neighborhood U of (x, y) such that f has a (local) C¹ inverse defined on f (U).

Solution: Since f is smooth and det Df =

¯¯

¯¯2x −2y

2y 2x

¯¯

¯¯ = 4(x²+ y²) 6= 0 for each (x, y) 6= (0, 0), there is an open neighborhood U of (x, y) on which f has a (local) C¹ inverse defined on f (U) by the Inverse Function Theorem.

4. (10 points) In the system

3x + 2y + z²+ u + v² = 0 4x + 3y + z + u²+ v + w + 2 = 0 x + z + w + u²+ 2 = 0,

discuss the solvability for u, v, w in terms of x, y, z near the point (x, y, z, u, v, w) = (0, 0, 0, 0, 0, −2).

Solution: Let F (x, y, z, u, v, w) = (3x + 2y + z²+ u + v² , 4x + 3y + z + u²+ v + w + 2 , x + z + w + u²+ 2).

Direct computation gives that DF |(0,0,0,0,0,−2)=



3 2 2z 1 2v 0

4 3 1 2u 1 1

1 0 1 2u 0 1





(0,0,0,0,0,−2)

=



3 2 0 1 0 0 4 3 1 0 1 1 1 0 1 0 0 1



 .

Since det DF |(u,v,w)=(0,0,−2) =

¯¯

¯¯

¯¯

1 0 0 0 1 1 0 0 1

¯¯

¯¯

¯¯= 1 6= 0, one can solve for u, v, w in terms of x, y, z near the point (x, y, z, u, v, w) = (0, 0, 0, 0, 0, −2) by the Implicit Function Theorem.

5. (10 points) Let f : R² → R³ be defined by f (x, y) = (x + y³, xy, y + y²). Is the range of f a two-dimensional surface or a one-dimensional curve near (0, 0)?

Solution: SinceDF |_(0,0) =



1 3y²

y x

0 1 + 2y





(0,0)

=



1 0 0 0 0 1



 has rank 2, the range a smooth surface near (0, 0).

6. Let f : R⁵ → R² be defined by f (x₁, x₂, y₁, y₂, y₃) = (2e^x1 + x₂y₁ − 4y₂ + 3 , x₂cos x₁− 6x₁+ 2y₁ − y₃) so that f (0, 1, 3, 2, 7) = (0, 0) and Df (0, 1, 3, 2, 7) =

µ 2 3 1 −4 0

−6 1 2 0 −1

¶ .

Advanced Calculus Midterm Exam II (Continued) May 18, 2011 (a) (8 points) Show that we can solve for (x₁, x₂) = g(y₁, y₂, y₃) i.e. solve for x₁, x₂ in terms of y₁, y₂, y₃,

near (y1, y2, y3) = (3, 2, 7).

Solution: Since det Dxf |_{(x1,x2)=(0,1)} =

¯¯

¯¯ 2 3

−6 1

¯¯

¯¯ = 20 6= 0, one can solve for x¹, x2 in terms of y1, y2, y3, i.e. there exists a C¹ function g such that (x1, x2) = g(y) = g(y1, y2, y3), for those (x1, x2, y1, y2, y3) satisfying that f (x₁, x₂, y₁, y₂, y₃) = (0, 0) near (y₁, y₂, y₃) = (3, 2, 7) by the Implicit Function Theorem.

(b) (10 points) Show that Dg(3, 2, 7) = −1 20

µ1 −3

6 2

¶ µ1 −4 0

2 0 −1

¶ .

Solution: Part (a) implies that f (x₁, x₂, y₁, y₂, y₃) = f (g(y), y) near (y₁, y₂, y₃) = (3, 2, 7). Using chain rule and differentiating f (x, y) = f (g(y), y) with respect to y_i, for i = 1, 2, 3, we obtain that

fi,x1

∂g1

∂y_j + fi,x2

∂g2

∂y_j + ∂fi

∂y_j = 0 for i = 1, 2 and j = 1, 2, 3.

⇔ £

f_i,x1 f_i,x2¤





∂g₁

∂y₁

∂g₁

∂y₂

∂g₁

∂y₃

∂g₂

∂y1

∂g₂

∂y2

∂g₂

∂y3



 +

·∂f_i

∂y1

∂f_i

∂y2

∂f_i

∂y3

¸

=£

0 0 0¤

for i = 1, 2.

⇔

·f1,x1 f1,x2

f_2,x1 f_2,x2

¸ 



∂g₁

∂y1

∂g₁

∂y2

∂g₁

∂y3

∂g2

∂y₁

∂g2

∂y₂

∂g2

∂y₃



 +





∂f₁

∂y1

∂f₁

∂y2

∂f₁

∂y3

∂f2

∂y₁

∂f2

∂y₂

∂f2

∂y₃



 = 0

⇔ Dg =





∂g₁

∂y₁

∂g₁

∂y₂

∂g₁

∂y₃

∂g₂

∂y₁

∂g₂

∂y₂

∂g₂

∂y₃



 = −

·f_1,x1 f_1,x2 f_2,x1 f_2,x2

¸₋₁ 



∂f₁

∂y₁

∂f₁

∂y₂

∂f₁

∂y₃

∂f₂

∂y₁

∂f₂

∂y₂

∂f₂

∂y₃



 .

Evaluating at (0, 1, 3, 2, 7), we obtain Dg(3, 2, 7) = −

· 2 3

−6 1

¸₋₁ ·

1 −4 0

2 0 −1

¸

= −1 20

·1 −3

6 2

¸ ·1 −4 0

2 0 −1

¸ .

7. (a) (8 points) Let f : R → R be defined by f (x) = 1

1 + x². Prove that f is uniformly continuous on R.

[Hint: You may use the Mean Value Theorem and the inequality 2ab

a² + b² ≤ 1 when a²+ b² 6= 0.]

Solution: For each x, y ∈ R, by the Mean Value Theorem, |f (x) − f (y)| = |f⁰(z)(x − y)| =

| 2z

(1 + z²)²||x − y| holds for some z lying between x and y. Using the inequality 2ab

a²+ b² ≤ 1 when a²+ b² 6= 0, we have |f (x) − f (y)| ≤ 1

1 + z²|x − y| ≤ |x − y| for each x, y ∈ R. Hence, f is uniformly continuous on R since it is Lipschitz there.

(b) (8 points) Let g(x) = tan x for x ∈ [0,π

2). Prove that g is Not Lipschitz on [0,π 2).

Solution: For each x, y ∈ [0,π

2), by the Mean Value Theorem, | tan x − tan y| = sec²z|x − y| holds for some z lying between x and y. Since lim

x,y→(π/2)⁻sec²z = lim

z→(π/2)⁻sec²z = ∞, g is not Lipschitz on [0,π

2).

(c) (8 points) Let f (x) = 1 2

µ x + 2

x

¶

for x ∈ S = [1, ∞). Prove that f is a contraction mapping of S, and find the fixed point of f .

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Advanced Calculus Midterm Exam II (Continued) May 18, 2011

Solution: The Mean Value Theorem implies that |f (x) − f (y)| = |f⁰(z)(x − y)| holds for some z lying between x, y ∈ [1, ∞). Since |f⁰(z)| = |1

2− 1 x²| ≤ 1

2 < 1, we obtain that |f (x) − f (y)| ≤ 1 2|x − y|

holds for all x, y ∈ [1, ∞) which implies that f is a contraction mapping of S.

A point x ∈ S is a fixed point of f if f (x) = x ⇔ x² = 2, x ∈ S ⇔ x =√ 2.

8. Let {f_n} be a sequence of functions defined by f_n(x) = nx

1 + nx² for each x ∈ [0, 1].

(a) (6 points) Find the limit f (x) = lim

n→∞f_n(x) for each x ∈ [0, 1]. [Hint: x ∈ [0, 1] = {0} ∪ (0, 1].]

Solution: f (x) = lim

n→∞f_n(x) =





0 if x = 0, 1

x if x ∈ (0, 1].

(b) (6 points) Show that the convergence in Not uniform on [0, 1].

Solution: Since f is not continuous at x = 0 the convergence is not uniform on [0, 1].

9. Let {fn} be a sequence of functions defined by fn(x) =√

nxⁿ(1 − x) for each x ∈ [0, 1].

(a) (6 points)Find max

x∈[0,1]f_n(x).

Solution: Since f_n⁰(x) = n√

n xⁿ⁻¹(1 − x) −√

n xⁿ = √

nxⁿ⁻¹£

n − (n + 1)x¤

= 0 when x = n n + 1, we obtain that max

x∈[0,1]f_n(x) = f_n( n

n + 1) = √ n¡ n

n + 1

¢_n¡

1 − n n + 1

¢=

√n n + 1

¡1 − 1 n + 1

¢_n . (b) (6 points) Find the limit f (x) = lim

n→∞f_n(x) for each x ∈ [0, 1]. [Hint: x ∈ [0, 1] = (0, 1) ∪ {0, 1}. ] Solution: f (x) = lim

n→∞fn(x) =

(0 if x ∈ {0, 1}

0 if x ∈ (0, 1).

= 0 for each x ∈ [0, 1].

(c) (6 points) Show that the convergence is uniform on [0, 1].

Solution: For each x ∈ [0, 1], since |fn(x) − f (x)| = |fn(x)| ≤ fn( n n + 1) =

√n n + 1

¡1 − 1 n + 1

¢_n

√ = n n + 1

¡1 − 1 n + 1

¢_n+1¡

1 − 1 n + 1

¢₋₁

→ 0, the convergence is uniform on [0, 1].

10. Let f, g be uniformly continuous maps defined on D ⊂ R^p with ranges in R^q. (a) Prove that f + g is uniformly continuous on D.

Solution: For each ² > 0 since f, g are uniformly continuous on D, there exists a δ > 0 such that if x, y ∈ D and kx − yk < δ then kf (x) − f (y)k < ² and kg(x) − g(y)k < ²

⇒ k(f + g)(x) − (f + g)(y)k = kf (x) − f (y) + g(x) − g(y)k ≤ kf (x) − f (y)k + kg(x) − g(y)k < 2².

Hence, that f + g is uniformly continuous on D.

(b) If f and g are bounded on D (by M). Prove that the product f g is uniformly continuous on D.

Solution: Assume that kf (x)k, kg(x)k ≤ M for each x ∈ D.

Given ² > 0 since f, g are uniformly continuous on D, there exists a δ > 0 such that if x, y ∈ D and kx − yk < δ then kf (x) − f (y)k < ² and kg(x) − g(y)k < ²

⇒ k(f g)(x) − (f g)(y)k = kf (x)g(x) − f (y)g(x) + f (y)g(x) − f (y)g(y)k ≤ kf (x) − f (y)kkg(x)k + kf (y)kkg(x) − g(y)k < 2M².

Hence, that f g is uniformly continuous on D.

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