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(ii) Give an example of a function f on [−π, π] such that the Fourier series of f converges uniformly

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FINAL FOR ADVANCED CALCULUS

Date: Monday, Jun 17, 2002 Instructor: Shu-Yen Pan

No credit will be given for an answer without reasoning.

1. [10%]

(i) Give an example of a set A ⊂ Rn such that A is of measure zero but not of volume zero.

(ii) Give an example of a function f on [−π, π] such that the Fourier series of f converges uniformly.

(iii) Give an example of a function f on [−π, π] such that the Fourier series of f converges to f at every point x in [−π, π] but not uniformly on [−π, π].

2. [10%] Integrate the function f (x, y, z) = x2+ y2+ z2over the set B = { (x, y, z) | x2+ y2+ z2≤ 1 }.

3. [10%] Suppose that A ⊂ Rn and A has zero volume. Suppose f : A → R is a bounded function. Prove that f is integrable andR

Af = 0.

4. [10%] Let fn(x) =Pn

k=1 1

2ksin kx for x ∈ R.

(i) Show that limn→∞fn(x) exists for all x.

(ii) Show that the sequence converges uniformly on the whole real line.

(iii) Show that R

0 (limn→∞fn(t)) dt = 0

5. [10%] Let ϕn(x) = 1πsin nx. Check that { ϕn(x) | n ∈ N } is an orthonormal family of functions on [0, 2π].

6. [10%] Compute the Fourier series a20 +P

n=1(ancos nx + bnsin nx) of the function f (x) = x on the interval [−π, π].

7. [10%] Let V be a Hilbert space and let ϕ0, ϕ1, ϕ2, . . . be a complete orthonormal set. Let c0, c1, c2, . . . be complex numbers and suppose thatP

k=0|ck|2< ∞. Let fn=Pn

k=0ckϕk. (i) Show that kfn− fmk2=Pn

k=m+1|ck|2. (ii) Show that fn is a Cauchy sequence in V.

(iii) Show that P

k=0ckϕk is the Fourier series of some f . 8. [10%] Let f : R2→ R be defined by

f (x, y) =



xy(x2−y2)

x2+y2 , if (x, y) 6= (0, 0);

0, if (x, y) = (0, 0).

Show that ∂x∂y2f and ∂y∂x2f exist at (0, 0) but are not equal.

9. [10%] Let f (x) = cos xx . Show thatR

1 f (x) dx converges butR

1 |f (x)| dx does not converge.

10. [10%] Let f : R → R be of class C1 and u = f (x), v = −y + xf (x). If f0(x0) 6= 0, show that this transformation is invertible near (x0, y0) and the inverse has the form x = f−1(u), y = −v + uf−1(u).

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