MATH 404 ANALYSIS II - SPRING 2009
SOLUTIONS to HOMEWORK 4
Problem 1. Let f : Rn → Rn be defined by f (x) = |x|2 · x. Show that f is of
class C ∞ , f is one-to-one and f (B1 (0)) = B1 (0). Show that the inverse function of
f : is not differentiable at 0.
2
Solution: We have f (x) = (f1 (x), . . . , fn (x)) where fi (x) = |x| xi . Since x →
n
2
|x| = j=1 x2j is of class C ∞ and x → xi is of class C ∞ , their product is also of
class C ∞ . Hence each fi is of class C ∞ . Next assume that f (x) = f (y). Then
|x|2 x = |y|2 y. This implies that |x|3 = |y|3 . That is, |x| = |y| and |x|2 x =
2
2
|y| x = |y| y which implies that x = y. So, f is one-to-one. If |x| < 1, then
3
2/3
|f (x)| = |x| < 1 so that f (B1 (0)) ⊂ B1 (0). If 0 < |y| < 1, then set x = y/ |y| .
1/3
Then |x| = |y|
< 1 and f (x) = y so that B1 (0) ⊂ f (B1 (0)). The inverse function
2/3
−1
n
n
: R → R of f is given by f −1 (0) = 0 and f −1 (y) = y/ |y| . The ith
f
2/3
component (f −1 )i of f −1 is equal to (f −1 )i (0) = 0 and (f −1 )i (y)) = yi / |y| . So,
t
(f −1 )i (0 + tei ) − (f −1 )i (0)
1
=
= 2/3 → ∞ as t → 0.
t
t · t2/3
t
Hence the partial derivative Di (f −1 )i (0) does not exist and so, f −1 is not differentiable at 0.
Problem 2. Let g : R2 → R2 be given by
g(x, y) = (2ye2x , xey ),
and let f : R2 → R3 be given by
f (x, y) = (3x − y 2 , 2x + y, xy + y 3 ).
(a) Show that there exist a neighborhood U of (0, 1) and V of (2, 0) such that
that g : U → V is a bijection.
(b) Find D(f ◦ g −1 )(2, 0).
Solution: (a) The function g is differentiable with
2x
4ye
2e2x
4 2
Dg(x, y) =
so that Dg(0, 1) =
.
ey
xey
e 0
So, detDF (0, 1) = −2e = 0. By the inverse function theorem there is an open
neighborhood U of the point (1, 0) such that g(U ) is an open neighborhood of the
point g(0, 1) = (2, 0) such that g is one-to one on U .
(b) Let g −1 : g(U ) → U be the inverse of g. Then by the chain rule,
D(f ◦ g −1 )(2, 0) = Df (g −1 (2, 0))Dg −1 (2, 0) = Df (0, 1)[Dg(0, 1)]−1 .
Since

3
Df (x, y) = 2
y

−2y
1 
x + 3y 2


3 −2
1 ,
so that Df (0, 1) = 2
1
3
1
2
one gets

3
D(f ◦ g −1 )(2, 0) = 2
1

−2 4
1 ·
e
3

−1
3
1 
2
2
=
0
2e
1



−2 −2e 18
1 
0
2
1 ·
e
0
=
e −4
2e
3
3e −10
Problem 3. Let Q be a closed rectangle in Rn and let f : Q → R be a bounded
function. Prove:
(a) If f is Riemann
integrable and f vanishes except on a set B having measure
zero, then Q f = 0.
(b) if f vanishes except on
a closed set B having measure zero, then f is Riemann integrable and Q f = 0
(c) If f is Riemann integrable and f (x) > 0 for all x ∈ Q, then Q f > 0.
Solution: (a) Let P be any partition of Q. If R is a subrectangle determined
by P , then whenever R ∩ B = ∅, then there is a point x ∈ R \ B which implies
that inf R f ≤ 0 and supR f ≥ 0. Hence L(f, P ) ≤
0 and U(f, P ) ≥ 0. Since f is
integrable, Q f = L(f ) = supP L(f, P ) ≤ 0 and Q f = U(f ) = inf P U(f, P ) ≥ 0.
So, Q f = 0.
(b) In view of (a), it suffices to show that f is integrable. Write Q = I1 × . . . × In .
Since f is bounded, |f (x)| ≤ M for x ∈ Q. Let ε > 0. Since B has measure zero,
there is a sequence (Qk ) of closed rectangle such that
B⊂
int Qk
and
k=1
∞
v(Qk ) <
k=1
ε
2M
Since B is closed and bounded, it is compact. So, there is N ∈ N such that
N
B ⊂ k=1 Qk . Denote by Pi the partition of the interval Ii by taking the end points
of Ii and end points of the ith components of the rectangles Qk for k = 1, . . . , N .
The let P = (P1 , . . . , Pn ). Every Qk is a union of some subrectangles determined
by P . Denote by Rk for k = 1, . . . , N disjoint collections of subrectangles R such
that R ⊂ Qk . Note that R∈Rk v(R) ≤ v(Qk ). Moreover, if R is the subrectangle
determined by P not contained in any Qk , then inf R f = supR f = 0. So
L(f, P ) =
N k=1 R∈Rk
U(f, P ) =
N
k=1 R∈Rk
(inf f ) · v(R) ≥ −M
R
N v(R) ≥ −M
k=1 R∈Rk
(inf f ) · v(R) ≤
R
M
N
N
v(Qk ) > −
ε
2
v(Qk ) <
ε
2
k=1
v(R) ≤
k=1 R∈Rk
M
N
k=1
which implies that
U(f, P ) − L(f, P ) < ε.
So, f is integrable over Q as claimed.
(c) Since f is integrable, the set of points at which f is discontinuous has measure
0. Let x0 ∈ Q◦ be a point at which f is continuous. Set a := f (x0 ) > 0. Using
the continuity of f , we find a subrectangle Q ⊂ Q containing x0 and such that
f (x) > a/2 for all x ∈ Q . Let P be any partition of Q such that Q is one of the
subrectangles determined by P . Then
f ≥ L(f, P ) ≥ (a/2) · v(Q ) > 0.
Q
3
Problem 4. Show that if A ⊂ Rn is compact and has measure 0 in Rn , then A
has content 0 in Rn .
Solution:
∞ Take ε >
0∞and let (Qj ) be a sequence of closed rectangles such that
A ⊂ j=1 Q◦j and
j=1 v(Qj ) < ε. Since A is compact, there is N such that
N
N
◦
A ⊂ j=1 Qj . Clearly, j=1 v(Qj ) < ε. Hence A has content 0.
Problem 5. Let Q be a closed rectangle in Rn and let f : Q → R. The graph of
f is defined by
G = {(x, y) ∈ Rn+1 |y = f (x)}.
Show that if f is continuous, then G has measure zero in Rn+1 .
Solution: Write Q = [a1 , b1 ] × . . . [an , bn ]. We may assume without lost of
generality that d := b1 − a1 = . . . = bn − an . Take ε > 0. Since Q is compact and f is continuous, f is uniformly continuous. So, there is δ0 such that
n
|f (x) − f (y)| < ε/(2d
√ ) for all x, y ∈ Q satisfying |x − y| < δ. Let N be a positive
integer such that nd/N < δ. Partition each [ai , bi ] into N equal subintervals each
of length d/N . Let P be the partition of Q determined by end points of these
subintervals. The partition P determines the partition of Q into N n subcubes
each having volume equal to dn /N n . If R is such a subcube, then for x, y ∈ R,
n
2
2
≤ n(d/N )2 < δ 2 so that |f (x) − f (y)| <
one has |x − y| =
j=1 |xj − yj |
n
ε/(2d ). Hence the graph of f over the subcube R is contained in the rectangle [f (x0 ) − ε/(2dn ), f (x0 ) + ε/(2dn )] × R where x0 is some point in R. The volume
of this rectangle is equal to (ε/dn ) · (dn /N n ) = ε/N n . Hence the graph of f is
contained in N n rectangles each having volume ε/N n and so the total volume of
these rectangle is equal to ε. Consequently, G(f ) has measure 0 in Rn+1 .