Advanced Calculus Midterm Exam

Advanced Calculus Midterm Exam

1. Let f : R² → R² be defined by f (x, y) = (e^xcos y, e^xsin y).

(a) (8 points) For each (x, y) ∈ R², show that there is an open neighborhood U of (x, y) such that f has a (local) C¹ inverse defined on f (U). [You may want to check that a proper theorem is applicable here].

Solution: Since f is smooth, and, for each (x, y) ∈ R², we have det Df = det

µe^xcos y −e^xsin y e^xsin y e^xcos y

¶

= e^2x 6= 0. By the Inverse Function Theorem, there is an open neighborhood U of (x, y) on which f has a (local) C¹ inverse defined on f (U).

(b) (4 points) Does f have a global inverse defined on the R²? Give explanation to your answer.

Solution: Since f (x, y + 2kπ) = f (x, y) for any (x, y) ∈ R² and any k ∈ Z, f is not a one-to-one function on R² and it does not have a global inverse on R².

2. (10 points) Let F (x, y, z) = (xy + 2yz − 3xz, xyz + x − y − 1) for x, y, z ∈ R. Find the differential DF at (x, y, z) = (1, 1, 1), and determine whether it is possible to represent the set S = {(x, y, z) | F (x, y, z) = (0, 0)} as a smooth curve parametrized by z, i.e. whether it is possible to solve for x, y in terms of z, near the point (x, y, z) = (1, 1, 1).

Solution: Direct computation gives that DF |_(1,1,1)=

µy − 3z x + 2y 2y − 3x yz + 1 xz − 1 xy

¶

(1,1,1)

=

µ−2 3 −1

2 0 1

¶ . Since det

µ−2 3

2 0

¶

= −6 6= 0, S can be represented as a smooth curve, parametrized by z, near the point (x, y, z) = (1, 1, 1) by the Implicit Function Theorem.

3. (a) (10 points) Let F : R³ → R³ be defined by F (x, y, z) = (x + y − z, x − y + z, x² + y² + z² − 2yz).

Determine the rank of DF on R³ and determine whether the image set F (R³) is locally a smooth surface or a smooth curve.

Solution: Direct computation gives DF =



1 1 −1

1 −1 1

2x 2y − 2z 2z − 2y



 . Since det

µ1 1 1 −1

¶

= −2 6= 0 and



 −1

1 2z − 2y



 = −



 1

−1 2y − 2z



 , DF has constant rank 2 on R³, and

the set F (R³) is locally a smooth surface.

(b) (10 points) Let F : R³ → R³ be defined by F (x, y, z) = (xy + z, x²y²+ 2xyz + z², 2 − xy − z). Determine the rank of DF on R³ and determine whether the image set F (R³) is locally a smooth surface or a smooth curve.

Solution: Direct computation gives that DF =



 y x 1

2y(xy + z) 2x(xy + z) 2(xy + z)

−y −x −1



 . Since¡

2y(xy + z), 2x(xy + z), 2(xy + z)¢

= 2(xy + z)¡

y, x, 1¢ and¡

− y, −x, −1¢

= −¡

y, x, 1¢

, DF has constant rank 1 on R³, and the set F (R³) is locally a smooth curve.

4. Let f : R³ → R² be defined by f (x, y, z) = (x + y + z, x − y − 2xz), so that f (0, 0, 0) = (0, 0) and Df (0, 0, 0) =

µ1 1 1 1 −1 0

¶ .

Advanced Calculus Midterm Exam (Continued) April 27, 2011 (a) (10 points) Show that we can solve for (x, y) = g(z), i.e. solve for x, y in terms of z,

Solution: In Df (0, 0, 0), since det

µ1 1 1 −1

¶

= −2 6= 0, we can solve for (x, y) = g(z), by the Implicit Function Theorem, i.e. we can solve for x, y in terms of z.

(b) (10 points) Show that Dg(0) = µ−¹₂

−¹₂

¶ .

Solution: By differentiating the components of f with respect to z, we get f_x^∂x_∂z + f_y ^∂y_∂z + f_z|_{at (0,0,0)} =

µ1 1

¶

∂x

∂z + µ 1

−1

¶

∂y

∂z + µ1

0

¶

= µ0

0

¶

which is equivalent to that µ1 1

1 −1

¶ µ_∂x

∂z∂y

∂z

¶

= µ−1

0

¶

⇒ Dg(0) = µ_∂x

∂y∂z

∂z

¶

at z=0

= −¹₂

µ−1 −1

−1 1

¶ µ−1 0

¶

= µ−¹₂

−¹₂

¶

5. (a) (6 points) Let f : R → R be defined by f (x) = x¹³. For each c > 0, prove that f is Lipschitz on [c, ∞).

Solution: For any x, y ∈ [c, ∞), by the Mean Value Theorem, we have

|f (x) − f (y)| = |f⁰(z)(x − y)| = |_3z¹2/3(x − y)| for some point z lying between x, y ∈ [c, ∞)

⇒ |f (x) − f (y)| = |_3z¹2/3(x − y)| ≤ _3c¹2/3|x − y| holds for any x, y ∈ [c, ∞).

This proves that f is Lipschitz on [c, ∞).

(b) (6 points) Let f : R → R be defined by f (x) = x². Prove that f is Not Lipschitz on [1, ∞).

Solution: For any x, y ∈ [1, ∞), by the Mean Value Theorem, we have

|f (x) − f (y)| = |2z(x − y)| for some point z lying between x and y.

By letting x and y go to ∞, we note that z will go to ∞.

Thus,there does not exist a fixed number A such that |f (x)−f (y)| ≤ A|x−y| holds for all x, y ∈ [1, ∞).

Hence, f is not Lipschitz on [1, ∞).

(c) (6 points) Let f, g be Lipschitz maps defined on D ⊂ R^p with ranges in R^q. Prove that f +g is Lipschitz on D.

Solution: Since f, g are Lipschitz on D, there exist constant A, B ≥ 0 such that kf (x) − f (y)k ≤ Akx − yk and kg(x) − g(y)k ≤ Bkx − yk hold for any x, y ∈ D.

This implies that k(f + g)(x) − (f + g)(y)| ≤ kf (x) − f (y)k + kg(x) − g(y)k ≤ (A + B)kx − yk holds for any x, y ∈ D.

Hence, f + g is Lipschitz on D.

(d) (6 points) Give an example of Lipschitz functions f, g : [1, ∞) → R and that the product f g is Not Lipschitz on [1, ∞).

Solution: Let f (x) = g(x) = x for each x ∈ [1, ∞). Then, f and g are Lipschitz (with LIpschitz constant 1), but (f g)(x) = x² is not Lipschitz by part (b).

6. (a) (8 points) Let {f_n} be a sequence of functions defined by f_n(x) = ^x_n for each x ∈ R. Show, without using Arzel`a-Ascoli’s theorem, that f_n converges uniformly on any closed interval [a, b] ⊂ R while the convergence in Not uniform on R.

Solution: For each x ∈ R, we have lim

n→∞f_n(x) = lim

n→∞

x

n = 0 = f (x), and lim

n→∞sup

[a,b]

|f_n(x) − f (x)| = lim

n→∞sup

[a,b]

|x|

n = lim

n→∞max{^|a|_n,^|b|_n} = 0.

Hence, fn converges uniformly on any closed interval [a, b] ⊂ R.

Page 2

Advanced Calculus Midterm Exam (Continued) April 27, 2011

Now, since lim

n→∞sup

R

|f_n(x) − f (x)| = lim

n→∞sup

R

|x|

n ≥ lim

n→∞

|n|

n = 1 6= 0, the convergence is not uniform on R.

(b) (8 points) Let f_n be defined on the interval [0, 1] by the formula f_n(x) =

(0, if x ∈ [0, 1 − _n¹], nx − (n − 1), if x ∈ [1 − ¹_n, 1].

Show that lim

n→∞fn(x) exists on [0, 1], and, without using Arzel`a-Ascoli’s theorem, show that this con- vergence is Not uniform on [0, 1].

Solution: For each x ∈ [0, 1), we have x ∈ [0, 1 − _n¹) if n > _1−x¹ . This implies that f_n(x) = 0 for all n > _1−x¹ and for all x ∈ [0, 1).

Thus, we have lim

n→∞f_n(x) = 0 for each x ∈ [0, 1).

At x = 1, since fn(1) = 1 for all n ∈ N, we have lim

n→∞fn(1) = 1.

Hence, we have lim

n→∞f_n(x) = f (x) = (

0 if x ∈ [0, 1) 1 if x = 1 , and, since lim

n→∞sup

[0,1]

|f_n(x) − f (x)| = lim

n→∞sup

[0,1)

¡nx − (n − 1)¢

= lim

n→∞1 = 1 6= 0, the convergence is not uniform on [0, 1].

7. Let F = {f_n(x) = ^x_nⁿ | x ∈ [0, 1], n = 1, 2, . . .}.

(a) (4 points) Show that F is uniformly bounded on [0, 1].

Solution: Since |f_n(x)| ≤ _n¹ ≤ 1, for each f_n ∈ F and for each x ∈ [0, 1], the set F is uniformly bounded on [0, 1].

(b) (8 points) Show, without using the arzel`a-Ascoli’s theorem, that F is uniformly equicontinuous on [0, 1].

Solution: For each f_n ∈ F and for each x ∈ (0, 1), since |f_n⁰(x)| = |x|ⁿ⁻¹ ≤ 1, by the Mean Value Theorem, we have (∗) · · · |f_n(x) − f_n(y)| ≤ 1 · |x − y| for any x, y ∈ [0, 1] and for all n ∈ N.

This implies that F is uniformly equicontinuous on [0, 1].

(For each ² > 0, we choose δ = ² such that if x, y ∈ [0, 1] and |x − y| < δ, then the inequality (∗) implies that |f_n(x) − f_n(y)| < |x − y| < δ = ² for each f_n∈ F .)

8. (12 points) Let {fn} be a sequence of continuous functions with domain D ⊂ R^p and range in R^q and let this sequence converge uniformly on D to a function f. Prove that f is continuous on D.

Solution: Given ² > 0, since f_n converges uniformly to f on D, there exists an M ∈ N such that kf_n(x) − f (x)k < ₃^² for each x ∈ D.

At any point x ∈ D, since fM is continuous at x, there exists a δ > 0 such that if y ∈ D and kx − yk < δ then kf_M(x) − f_M(y)k < ₃^².

Thus, we have

kf (x) − f (y)k = kf (x) − fM(x) + fM(x) − fM(y) + fM(y) − f (y)k

≤ kf (x) − f_M(x)k + kf_M(x) − f_M(y)k + kf_M(y) − f (y)k < ₃^² + ^²₃ +₃^² = ².

This implies that f is continuous at x.

Since x is an arbitrarily chosen point from D, f is continuous on D.

Page 3