1
Solutions to selected problems
1. Let
A ⊂ B ⊂ Rn .
Show that int A
⊂ int B
x ∈ int A. Then there
A = [0, 1] and B = [0, 2],
but in general bd A
>0
Solution.
Let
is
x ∈ int B .
If
then bd A
2. Let
A ⊂ Rn
be open and
B
around
u
ball
such that
6⊂ bd B .
B (x) ⊂ A ⊂ B .
= {0, 1} 6⊂ bd B = {0, 2}.
such that
f : A → R continuous with f (u) > 0.
f (x) > f (u)/2 for x ∈ B .
This shows
Show there is an open
f is continuous at u, for = f (u)/2 there is δ > 0 such that f (Bδ (u)) ∈
(f (u) − , f (u) + ). In particular, f (Bδ (u)) > f (u)/2 and we may take B = Bδ (u).
Solution. Since
3.
Suppose
component.
f : Rn → R is continuous and f (u) > 0
n
Prove that f (u) ≥ 0 for all u ∈ R .
u ∈ Rn
if
u
has at least one rational
u = (u1 , . . . , un ) and let {rk } be the rst k
digits in the decimal expansion for u1 , and let uk = (rk , u2 , . . . , un ). Then {uk } → u so
by continuity of f , {f (uk )} → f (u). Also f (uk ) > 0 for all k since the rk 's are rational.
Thus, f (u) ≥ 0.
Solution.
Let
be arbitrary. Write
4. Show that an open ball in
Rn
is bounded.
Br (x) ⊂ Rn be an arbitrary ball. We must show it is contained in a ball
Bs (0) around 0. Let s = r + kxk. If y ∈ Br (x) then kyk ≤ ky − xk + kx|k < r + kxk = s.
Thus, Br (x) ⊂ Bs (0).
Solution. Let
f : A → R be continuous
closed, is f (A) closed?
5. Let
is
with
A ⊂ Rn .
If
A
Solution. The answer to both questions is no. Consider
bounded but
f (A) = (1, ∞)
is not;
A = [1, ∞)
f : Rn → R be continuous
([0, 1]) is sequentially compact.
6. Let
−1
f
Solution.
Since
[0, 1]
is closed and
f
with
f (A)
A
f (x) = 1/x. Then A = (0, 1)
f (A) = (0, 1] is not.
is
is closed but
f (u) ≥ kuk
is continuous,
1
bounded? If
is bounded, is
for every
f −1 ([0, 1])
u ∈ Rn .
Prove that
is closed. It remains to
f −1 ([0, 1]) is also bounded, hence sequentially compact. If x ∈ f −1 ([0, 1]), then
0 ≤ f (x) ≤ 1 and f (x) ≥ kxk, so in particular, kxk ≤ 1. Thus, f −1 ([0, 1]) is bounded.
show
7. Let
A ⊂ Rn
be sequentially compact and
v ∈ Rn \ A.
ku − vk ≤ kx − vk
for all
Prove there is
u∈A
such that
x ∈ A.
(1)
d = inf x∈A kx − vk. Since d + 1/k is not a lower bound for kx − vk over
uk ∈ A such that d ≤ kuk − vk < d + 1/k . Using sequential
compactness, pick {unk } → u. By continuity of vector subraction and the norm k · k,
{kunk − vk} → ku − vk. And by the squeeze theorem in R, ku − vk = d.
Solution. Let
x ∈ A,
we may pick
The point
u
is not unique: if
8. A mapping
F : Rn → Rm
A = {1, −1} ⊂ R
v = 0,
and
then
is Lipschitz continuous if there is
u = ±1
K>0
satisfy (1).
such that
|F (x) − F (y)k ≤ Kkx − yk
for all
x, y ∈ Rn .
Show that a Lipschitz mapping is uniformly continuous.
Solution. Suppose (2) holds for
F.
Let
>0
and pick
kF (x) − F (x)k ≤ Kkx − yk < Kδ = 9. Let
A
(2)
be sequentially compact and
f : A → f (A)
δ = K/.
whenever
Then
kx − yk < δ.
continuous and injective. Show
is continuous. Give an example to show the assumption on
A
f −1
is necessary.
uk = f −1 (vk ),
u = f −1 (v). Suppose {uk } 6→ u. Using sequential compactness of A, pick {unk } → w ∈ A
with w 6= u. By continuity of f , {f (unk )} → f (w). As {f (unk )} = {vnk } → v , we have
f (w) = v . Thus, w = f −1 (v) = u, contradiction.
Solution.
Let
{vk }
be a sequence in
f (A)
such that
{vk } → v ,
and let
A is needed, let
(
x,
0≤x<1
f (x) =
,
4 − x, 2 ≤ x ≤ 3
(
x,
0≤x<1
f −1 (x) =
.
3 − x, 1 ≤ x ≤ 2
To see that sequential compactness of
Note that
f
is continuous on
A = [0, 1) ∪ [2, 3]
2
but
f −1
is not continuous at
1.
10. Let
f : A → Rm
be continuous and
A
f
sequentially compact. Show
is uniformly
continuous.
{uk }, {vk } be sequences in A such that {kuk − vk k} → 0. Suppose
{kf (uk )−f (vk )k} 6→ 0. Then along a subsequence, {kf (unk )−f (vnk )k} → c > 0. (Why?)
Since A is sequentially compact, we can pick sub-subsequences {umn } → u ∈ A and
k
{vmnk } → v ∈ A, and u = v since {kuk −vk k} → 0. By continuity of f , {f (umnk )} → f (u)
and {f (vmn )} → f (v) = f (u). Thus, {kf (umn ) − f (vmn )k} → 0, contradiction.
k
k
k
Solution.
Let
u ∈ Rn is a a limit point of A ⊂ Rn if there is a sequence in A \ {u} that
converges to u. Prove that u is a limit point of A if and only if every open ball around
u contains innitely many points of A \ {u}.
11. We say
u be a limit point of A and δ > 0. Let {uk } in A\{u} be such that {uk } → u,
and pick N such that k ≥ N implies uk ∈ Bδ (A) \ {u}. Notice {uk : k ≥ N } must be
an innite set: if it were nite with elements v1 , . . . , vn , then for = min1≤i≤n kvi − uk
we would have uk ∈
/ B (u) for every k ≥ N .
Solution. Let
A ⊂ Rn
12. Let
at
u
and let
if and only if
f
f
is continuous
Can one make an analogous statement about
limx→u f (x)?
be the characteristic function of
u∈
/ bd A.
A.
Show that
u∈
/ bd A. Then if u ∈ A, there is δ > 0 such that Bδ (u) ⊂ A. Given
> 0, if kx − uk < δ then x ∈ A and so |f (x) − f (u)| = |1 − 1| = 0 < . If u ∈
/ A,
n
there is δ > 0 such that Bδ (u) ⊂ R \ A and an analogous argument holds. Conversely,
n
if u ∈ bd A, then for each k ∈ N there is vk , wk ∈ B1/k (u) such that vk ∈ A, wk ∈ R \ A.
Then {vk } → u and {wk } → u, but {f (vk )} → 1, {f (wk )} → 0.
An analogous statement does not hold for limits. It is true that if u ∈
/ bd(A) then
limx→u f (x) exists in the argument above, just replace kx−uk < δ with 0 < kx−uk < δ .
However, the converse is false: if A = {0} then 0 ∈ bd(A) yet limx→0 f (x) = 0 exists.
(In the above argument, vk and wk could equal u.)
Solution. Suppose
any
f : A → Rm with u ∈ A a limit point
at u, then f is not continuous at u.
13. Let
a limit
of
A ⊂ Rn .
Show that if
f
does not have
f is continuous at u. Then limx→u f (x) =
f (u). To see this, let > 0 and pick δ > 0 such that kf (x) − f (u)k < whenever
kx − uk < δ and x ∈ A; then in particular, kf (x) − f (u)k < whenever 0 < kx − uk < δ
and x ∈ A.
Solution. We prove the contrapositive. Suppose
3
f : A → Rm with u ∈ A a limit
and only if limx→u f (x) = f (u).
14. Let
u
if
point of
Solution. We only need to observe that for any
kf (x) − f (u)k < A ⊂ Rn .
> 0,
x∈A
whenever
Show that
f
is continuous at
the statements
and
0 < kx − uk < δ
and
kf (x) − f (u)k < are equivalent, since
whenever
kf (x) − f (u)k = 0 < x∈A
when
and
kx − uk < δ
kx − uk = 0.
A ⊂ Rn and suppose 0 is a limit point of A. Suppose f : A → R is
f (x) ≥ ckxk2 for all x ∈ A, where c > 0 is constant. Suppose g : A → R is
limx→0 g(x)/kxk2 = 0. Prove there is r > 0 such that f (x) − g(x) ≥ (c/2)kxk2
with 0 < kxk < r .
15. Let
such that
such that
for
x∈A
= c/2 in the denition -δ denition of limx→0 g(x)/kxk2 = 0. Pick
x ∈ A with 0 < kxk < δ , then kg(x)/kxk2 k = kg(x)k/kxk2 < c/2 and so
Solution. We use
δ>0
so that if
kf (x) − g(x)k ≥ kf (x)k − kg(x)k ≥ ckxk2 − (c/2)kxk2 = (c/2)kxk2 .
16. Let
(
f (x, y) =
Show
f
is continuous at
(0, 0) and
(0, 0).
xy 2
,
x2 +y 2
(x, y) 6= (0, 0)
.
(x, y) = (0, 0)
0,
has directional derivatives at
(0, 0)
in every direction,
but is not dierentiable at
(0, 0): for (x, y) 6= (0, 0),
xy 2
= 2|x| ≤ |x| → 0
−
0
|f (x, y) − f (0, 0)| = 2
x
x + y2
+1
Solution. To see
f
is continuous at
as
(x, y) → 0.
y2
(See problem 14 above.) For the directional derivatives
D(a,b) f (0, 0):
f ((0, 0) + t(a, b)) − f (0, 0)
f (ta, tb)
t2 ab2
ab2
= lim
= lim 2 2
=
.
t→0
t→0
t→0 t a + t2 b2
t
t
a2 + b 2
lim
Now note that
D1 f (0, 0) = 0, D2 f (0, 0) = 0
and
f ((0, 0) + (x, y)) − f (0, 0) − 0x − 0y
=
k(x, y)k
4
xy 2
x2 +y 2
− 0x − 0y
xy 2
p
= 2
.
(x + y 2 )3/2
x2 + y 2
(3)
When
x≡y
17. Dene
and
y→0
f : R2 → R
the limit of (3) is
so
f
is not dierentiable at
f (x, y) =
Show the partial derivatives of
(x, y) 6= (0, 0),
resorting to the denition):
1
f
xy
,
x2 +y 2
(x, y) 6= (0, 0)
.
(x, y) = (0, 0)
0,
are not continuous at
(0, 0).
we can calculate partial derivatives as usual (i.e., without
D1 f (x, y) =
y 3 − x2 y
,
(x2 + y 2 )2
D2 f (x, y) =
D1 f (0, y) = y −1 , D1 f (x, 0) = x−1 do not
are not continuous at (0, 0). (See problem 13
x3 − xy 2
.
(x2 + y 2 )2
In particular,
have limits as
Thus, they
above.)
18. Dene
g : R2 → R
g(x, y) =
g
y → 0, x → 0.
by
(
Is
(0, 0).
by
(
Solution. For
2−3/2 6= 0,
x2 y 4
,
x2 +y 2
(x, y) 6= (0, 0)
.
(x, y) = (0, 0)
0,
continuously dierentiable?
Solution. For
(x, y) 6= (0, 0)
we can calculate partial derivatives as usual:
2xy 6
,
D1 g(x, y) = 2
(x + y 2 )2
4x4 y 3 + 2x2 y 5
D2 g(x, y) =
.
(x2 + y 2 )2
Moreover, from the limit denition of partial derivatives,
g(t, 0) − g(0, 0)
0−0
= lim
= 0,
t→0
t
t
g(0, 0) − g(0, t)
0−0
D2 g(0, 0) = lim
= lim
= 0.
t→0
t→0
t
t
D1 g(0, 0) = lim
t→0
Since
D1 g(x, y) and D2 g(x, y) are rational functions, they are continuous except at their
(x, y) = (0, 0). Thus, to establish continuity of D1 g and D2 g we need only
asymptotes
1D
if
≡ Dei f ≡
∂f
∂xi
5
lim(x,y)→(0,0) Di g(x, y) = 0
check that
for
i = 1, 2.
For
(x, y) 6= (0, 0),
6
2
2xy
2xy
= 4
|D1 g(x, y) − 0| = 4
≤ 2|x|y 2 → 0 as (x, y) → (0, 0)
2
2
2
4
2x
x
y4 + y2 + 1 x + 2x y + y
4x4 y 3 + 2x2 y 5 4x2 y + 2y 3 =
|D2 g(x, y) − 0| = 4
≤ 2x2 |y| + |y|3 → 0 as (x, y) → (0, 0).
x + 2x2 y 2 + y 4 x22 + 2 + y22 y
This shows
x
D1 g(x, y) and D2 g(x, y) are continuous; thus, g
19. Suppose
g : R2 → R
is continuously dierentiable.
|g(x, y)| ≤ x2 + y 2 for
both x and y at (0, 0).
has the property
has partial derivatives with respect to
(x, y) ∈ R2 .
Show
g
g(0, 0) = 0 and thus
g(0, t) − g(0, 0) t2 g(t, 0) − g(0, 0) t2 ≤ = |t|,
≤ = |t|.
t
t
t
t
Solution. Our assumption on
Taking limits as
20. Suppose
t→0
g
all
forces
in the above expressions, we nd that
f : R2 → R
has rst-order partial derivatives and
D1 f (x, y) = D2 f (x, y) = 0
Show that
Solution.
f ≡c
Fix
c ∈ R;
for some
y0 ∈ R
D1 g(0, 0) = D2 g(0, 0) = 0.
that is,
and consider
f
for all
(x, y) ∈ R2 .
is a constant function.
g : R → R
dened by
g(x) = f (x, y0 ).
Suppose
g
is nonconstant. Then g(a) 6= g(b) for some a < b, and by MVT there is r ∈ (a, b) such
0
that D1 f (r, y0 ) = g (r) = [g(b) − g(a)]/(b − a) 6= 0, contradiction. Thus, g is constant,
2
say g ≡ c. Let (x, y) ∈ R . By repeating the argument above, we nd that the function
h : R → R dened by h(z) = f (x, z) is constant. Since h(y0 ) = g(x) we must have h ≡ c,
2
and in particular f (x, y) = c. Since (x, y) ∈ R was arbitrary, we can conclude f ≡ c.
21. Given
φ, ψ : R2 → R,
a function
D1 f (x, y) = φ(x, y)
and
f : R2 → R
is called a potential function for
D2 f (x, y) = ψ(x, y)
Show that when a potential function exists for
φ, ψ ,
for all
φ, ψ
if
(x, y) ∈ R2 .
it is unique up to an additive con-
stant. Then show that if there is a potential function for
φ, ψ
and
φ, ψ
are continuously
dierentiable, then
D1 ψ(x, y) = D2 φ(x, y)
for all
6
(x, y) ∈ R2 .
(4)
Solution. Suppose
f
and
g
are two potential functions for
Di h(x, y) = Di f (x, y) − Di g(x, y) = 0
so by Problem 20,
h
is constant.
Thus,
in (4) follows from the assumption
D1 f
f
for all
φ, ψ
and let
h = f − g.
Then
(x, y) ∈ R2 , i = 1, 2,
g dier by a constant. The statement
D2 f are continuously dierentiable and
and
and
Theorem 13.10.
22. Dene
f : R2 → R
by
(
f (x, y) =
x3 y−xy 3
,
x2 +y 2
0,
(x, y) 6= (0, 0)
.
(x, y) = (0, 0)
Show that D1 f (0, y) = −y for all y ∈ R and D2 f (x, 0) = x
D2 D1 f (0, 0) = −1 but D1 D2 f (0, 0) = 1.
Solution.
Away from
(0, 0)
for all
x ∈ R.
we can calculate partial derivatives of
without resorting to the limit denition. Thus, for
f
Conclude that
as usual, i.e.,
(x, y) 6= (0, 0),
(x2 + y 2 )(3x2 y − y 3 ) − (x3 y − xy 3 )(2x)
,
(x2 + y 2 )2
(x2 + y 2 )(x3 − 3xy 2 ) − (x3 y − xy 3 )2y
D2 f (x, y) =
.
(x2 + y 2 )2
D1 f (x, y) =
Thus, for
y 6= 0
and
x 6= 0,
D1 f (0, y) =
y 2 (−y 3 )
= −y,
(y 2 )2
D2 f (x, 0) =
x 2 x3
= x,
(x2 )2
D1 f (0, 0) = 0 = D2 f (0, 0). (Check
this! See Problem 18 for a similar calculation.) Thus, D2 D1 f (0, y) = −1 and D1 D2 f (x, 0) =
1 for all x, y ∈ R. In particular, D2 D1 f (0, 0) = −1 but D1 D2 f (0, 0) = 1.
and, from the limit denition of partial derivatives,
A ⊂ R2 be an open set containing (x0 , y0 ). Prove
(x, y) ∈ A whenever |x − x0 | < 2r and |y − y0 | < 2r.
23. Let
that there is
r>0
such that
√
0 < r < /(2 2), then
√
p
p
√
k(x, y) − (x0 , y0 )k = (x − x0 )2 + (y − y0 )2 < (2r)2 + (2r)2 = 8r2 = 2 2r < Solution. Pick
whenever
>0
such that
|x − x0 | < 2r
and
B (x0 , y0 ) ⊂ A.
|y − y0 | < 2r.
7
If
f : Rn → R and g : R → R are continuously
∇(g ◦ f )(x) in terms of ∇f (x) and g 0 (f (x)).
24. Suppose
for
dierentiable. Find a formula
Solution. By the denition of partial derivative and the mean value theorem,
f (x + tei ) − f (x)
g(f (x + tei )) − g(f (x))
= lim g 0 (zt )
,
t→0
t→0
t
t
Di (g ◦ f )(x) = lim
zt is on the line segment between f (x) and f (x + tei ). Notice limt→0 g 0 (zt ) =
g (f (x)) due to continuity of f and g 0 , and Di f (x) := limt→0 [f (x + tei ) − f (x)]/t. Thus,
where
0
Di (g ◦ f )(x) = g 0 (f (x))Di f (x),
that is,
i = 1, . . . , n,
∇(g ◦ f )(x) = ∇f (x) g 0 (f (x)).
25. Suppose
any nonzero
f : Rn → R
c ∈ R.
is such that
Dv f (x)
exists. Prove that
Dcv f (x) = cDv f (x)
for
Solution. This follows from the computation
f (x + tcv) − f (x)
t→0
t
f (x + tcv) − f (x)
f (x + sv) − f (x)
= lim c
= c lim
= cDv f (x).
t→0
s→0
tc
s
Dcv f (x) = lim
f : Rn → R has rst-order partial
for f , that is, there is > 0 such that
26. Suppose
minimizer
f (x + h) ≥ f (x)
Prove that
for
derivatives and that
x ∈ Rn
h ∈ B (0).
is a local
(5)
∇f (x) = 0.
Solution. Due to the assumption (5), for
i = 1, . . . , n,
f (x + tei ) − f (x)
≥ 0,
t→0
t
f (x − tei ) − f (x)
D−ei f (x) = lim
≥ 0,
t→0
t
Dei f (x) = lim
and by problem 25,
Dei f (x) = −D−ei f (x) ≤ 0.
Thus
Dei f (x) = 0,
27. Consider
f (x, y, z) = xyz + x2 + y 2 .
8
and so
∇f (x) = 0.
Find
θ ∈ (0, 1)
such that
f (1, 1, 1) − f (0, 0, 0) = D1 f (θ, θ, θ) + D2 f (θ, θ, θ) + D3 f (θ, θ, θ).
Solution. Note that
D1 f (x, y, z) = yz + 2x,
and
f (1, 1, 1) − f (0, 0, 0) = 3.
28. Dene
f : R2 → R
D2 f (x, y, z) = xz + 2y,
Thus, we solve
3θ2 + 4θ = 3
x2 +y 2
,
|y|
x
f (x, y) =
0,
b) Prove
f
f
is not continuous at
if
y 6= 0,
if
y = 0.
c∈R
θ = − 32 +
√
13
.
3
(0, 0).
has directional derivatives in all directions at
c) Prove for any
to get
by
( √
a) Prove
D3 f (x, y, z) = xy.
there is
p
(0, 0).
such that
kpk = 1
and
Dp f (0, 0) = c.
Does this contradict Corollary 13.18?
this curve shows
b) When
f
2
√x
and x
1−x2
is not continuous at zero, since f (0, 0)
Solution. a) Note that
f (x, y) ≡ 1
when
y=
6= 0. Approaching (0, 0)
= 0 6= 1.
along
b 6= 0,
r
√
f (ta, tb) − f (0, 0)
ta t2 a2 + t2 b2
a2
= lim
=a
+ 1.
D(a,b) f (0, 0) = lim
t→0
t→0
t
t|tb|
b2
Also,
f (ta, 0) − f (0, 0)
0−0
= lim
= 0.
t→0
t
t
in every direction at (0, 0).
D(a,0) f (0, 0) = lim
t→0
Thus,
f
has directional derivatives
c) We show such
r
a
p
exists by solving the equations
a2
+ 1 = c,
b2
a2 + b2 = 1
...
This does not contradict the corollary since
f
Problem 33 below.)
9
a= √
c
,
1 + c2
b= √
1
.
1 + c2
is not continuously dierentiable. (See also
f : Rn → R is continuously dierentiable and let K = {x ∈ Rn : kxk = 1}.
Show there is a point x ∈ K at which f |K attains its smallest value. Now suppose
n
whenever p ∈ R is a unit vector, h∇f (p), pi > 0. Show that then kxk < 1.
29. Suppose
Solution.
K
is sequentially compact and
f
h∇f (p), pi > 0
f |K attains a smallest value
p ∈ Rn with kpk = 1. Then
is continuous, so
by the extreme value theorem (Theorem 11.22).
Let
and
f (p − tp) − f (p)
f (p − tp) − f (p) − h∇f (p), −tpi
= lim
+ h∇f (p), pi.
t→0
t
t
This shows that for t > 0 suciently close to zero, f (p − tp) − f (p) < 0. Thus, the
minimum of f |K cannot be attained at p. For an -δ proof of this, let = h∇f (p), pi and
−1
pick δ > 0 such that |t [f (p − tp) − f (p)] + h∇f (p), pi| < whenever 0 < |t| < δ . Then
f (p − tp) − f (p) < 0 whenever 0 < t < δ ; why?
0 = lim
t→0
30. Prove that
sin(2x + 2y) − 2x − 2y
p
= 0.
(x,y)→(0,0)
x2 + y 2
lim
Solution. Let
f (x, y) = sin(2x + 2y). Then D1 f (x, y) = 2 cos(2x + 2y) = D2 f (x, y) are
f is continuously dierentiable, so since D1 f (0, 0) = D2 f (0, 0) = 2,
continuous. Thus,
by the rst order approximation theorem,
sin(2x + 2y) − 2x − 2y
= 0.
(x,y)→(0,0)
k(x, y)k
lim
31. Suppose
f : R2 → R
is continuous and
lim
a, b ∈ R.
Prove
[f (x, y) − (f (0, 0) + ax + by)] = 0.
(x,y)→(0,0)
Is it true that
f (x, y) − [f (0, 0) + ax + by]
p
= 0?
(x,y)→(0,0)
x2 + y 2
lim
(6)
(7)
f is continuous, f (x, y) → f (0, 0) = 0 as (x, y) → (0, 0). Moreover, ax +
by → 0 as (x, y) → (0, 0). This proves (6). Notice (7) is false unless f is also dierentiable
at (0, 0) with D1 f (0, 0) = a, D2 f (0, 0) = b. (See the rst order approximation theorem.)
Solution. Since
32. Dene
f : R2 → R
by
(p
2
x2 + y 2 sin yx ,
f (x, y) =
0,
10
if
if
x 6= 0,
.
x=0
f
Show
f is not
(0, 0).
but
at
(0, 0) and has directional derivatives in every direction at (0, 0),
dierentiable at (0, 0) that is, there is no tangent plane to the graph of f
is continuous at
p
p
2
|f (x, y) − f (0, 0)| = x2 + y 2 | sin xy 2 | ≤ x2 + y 2 → 0 as (x, y) → 0.
continuous at (0, 0). Since t 7→ sin(tc) is continuous for c ∈ R, when a 6= 0,
Solution. Notice
Thus,
f
is
f (ta, tb) − f (0, 0)
t→0
t
√
2 2
2
2
√
tb2 t a + t2 b2 sin t tab
= 0.
= lim a2 + b2 sin
= lim
t→0
t→0
t
a D(a,b) f (0, 0) = lim
Also,
tives
(0,0)
= limt→0 0−0
= 0. Thus, f has directional derivaD(0,b) f (0, 0) = limt→0 f (0,tb)−f
t
t
in every direction at (0, 0). Note that D1 f (0, 0) = D2 f (0, 0) = 0 but the limit
f (x, y) − f (0, 0) − 0x − 0y
y2
= lim sin
(x,y)→(0,0)
(x,y)→(0,0)
k(x, y)k
x
lim
does not exist (e.g., take
x ≡ y3
and let
y → 0).
Thus,
f
is not dierentiable at
(0, 0).
f : Rn → R is homogeneous, that is, f (0) = 0 and f (tx) = tf (x) for all
t ∈ R and x ∈ Rn . Prove that if f is dierentiable, then f (x) = h∇f (0), xi, in particular,
f is linear2 . Thus, if f is homogeneous and not linear, it cannot be dierentiable.
33. Suppose
Solution. Let
0 = lim+
t→0
f
be dierentiable and homogeneous, and
u ∈ Rn
a unit vector. Then
f (tu) − f (0) − h∇f (0), tui
tf (u) − th∇f (0), ui
= lim+
= f (u) − h∇f (0), ui.
t→0
ktuk
tkuk
This shows that
f (u) = h∇f (0), ui
and thus for
t ∈ R,
f (tu) = tf (u) = th∇f (0), ui = h∇f (0), tui.
Any nonzero
x ∈ Rn
is a scalar times a unit vector,
x
x = kxk kxk
.
f : Rn → R be k times continuously dierentiable,
φ(t) = f (x + th) for t ∈ R. Prove that for all s ∈ R,
34.
Let
φ(k) (s) = Dhk f (x + sh),
Thus, we are done.
let
h ∈ Rn
Dhk f = Dh Dh . . . Dh f.
|
{z
}
k times
Write formulas for
2f
φ(k) (s) = Dhk f (x + sh)
when
k=1
and
k = 2.
is linear i it is homogeneous and additive: f (x + y) = f (x) + f (y) for all x, y ∈ Rn .
11
and dene
Solution. For
k=1
this follows from the computation
f (x + (s + t)h) − f (x + sh)
φ(s + t) − φ(s)
= lim
t→0
t→0
t
t
f ((x + sh) + th) − f (x + sh)
= lim
= Dh f (x + sh).
t→0
t
(k−1)
general case follows from induction: if φ
(s) = Dhk−1 f (x + sh) for all s ∈ R, then
φ0 (s) = lim
The
Dhk−1 f (x + (s + t)h) − Dhk−1 f (x + sh)
φ(k−1) (s + t) − φ(k−1) (s)
= lim
φ (s) = lim
t→0
t→0
t
t
Dhk−1 f ((x + sh) + th) − Dhk−1 f (x + sh)
= lim
t→0
t
k−1
k
= Dh Dh f (x + sh) = Dh f (x + sh).
(k)
When
k = 1,
φ0 (s) = Dh f (x + sh) =
n
X
hi Di f (x + sh) = h∇f (x + sh), hi.
i=1
When
k = 2,
n
X
φ00 (s) = Dh2 f (x + sh) = Dh
!
hj Dj f (x + sh)
j=1
=
=
n
X
hi Di
i=1
n X
n
X
n
X
!
hj Dj f (x + sh)
j=1
hi hj Di Dj f (x + sh)
i=1 j=1
2
where the last step uses the fact that
= h∇ f (x + sh)h, hi,
P P
hAh, hi = ni=1 nj=1 hi hj aij
f : R2 → R by f (x, y) = exy + x2 + 2xy .
φ (0) in the following two ways:
35. Dene
compute
Dene
when
is
n × n.
for
t∈R
A
φ(t) = f (2t, 3t)
00
i) By using single-variable theory, i.e., dierentiating the single-variable function
ii) By thinking of
φ00 (0)
as a directional derivative of
3 or
3
Problem 34,
in the direction of
formula (14.11) in the text
12
φ;
h = (2, 3).
φ(t) = e6t + 16t2 , so direct computation gives φ00 (0) = 44.
φ00 (0) = h∇2 f (0, 0)h, hi where h = (2, 3). We compute
2 3
2
∇ f (0, 0) =
3 0
Solution. i) Note that
ii) By
2
f
and
and
h∇2 f (0, 0)h, hi = h(13, 6), (2, 3)i = 44.
f : R2 → R be twice continuously dierentiable. Suppose that ∇f (0, 0) = (0, 0)
2
2
and that for some h ∈ R , h∇ f (0, 0)h, hi > 0. Using single variable theory, prove there
is r > 0 such that
f (th) > f (0, 0),
0 < |t| < r.
36. Let
Solution. Dene
φ(t) = f (th).
By Taylor's theorem in 1D,
1
φ(t) = φ(0) + φ0 (0)t + φ00 (0)t2 + o(t2 )
2
where
limt→0 o(t2 )/t2 = 0.
By Problem 34, this can be rewritten
1
f (th) = f (0, 0) + h∇f (0, 0), hit + h∇2 f (0, 0)h, hit2 + o(t2 ).
2
Since
∇f (0, 0) = (0, 0), h∇f (0, 0), hi = 0
and we can rewrite again to get
1 2
o(t2 )
f (th) − f (0, 0)
=
h∇
f
(0,
0)h,
hi
+
.
t2
2
t2
limt→0 o(t2 )/t2 = 0 and h∇2 f (0, 0)h, hi > 0, the RHS of this equation is positive
for suciently small t. Thus, f (th) > f (0, 0) for suciently small t. More precisely, let
= 12 h∇2 f (0, 0)h, hi and choose δ > 0 such that |o(t2 )/t2 | < whenever 0 < |t| < δ .
Then f (th) > f (0, 0) for 0 < |t| < δ , so we can take r = δ .
Since
37. In Problem 36, suppose additionally that
h∇2 f (0, 0)h, hi > 0
for all nonzero
Is this enough to directly conclude the origin is a local minimum of
Solution. We would have to show that
not depend on
h.
we might
for all
f?
0 < |t| < r,
r
r = rh
r = inf h rh , but
where
The trouble with using Problem 36 here is that we had
h. To get
get r = 0.
depends on
f (th) > f (0, 0)
h ∈ Rn .
an
r
which is independent of
h,
we could try
does
that
then
Note that in this situation the origin is indeed a local minimizer see for instance
Theorem 14.22. But this conclusion cannot be reached directly from Problem 36.
a 6= 0, and dene p(t) = at2 + 2bt + c. Show that p(t) > 0 for
2
all t if and only if a > 0 and ac − b > 0. Show also that p(t) < 0 for all t if and only if
2
a < 0 and ac − b > 0.
38. Let
a, b, c ∈ R
with
13
p(t) = 0 when
√
−b ± b2 − ac
.
t=
a
Solution. The quadratic equation shows that
ac − b2 > 0. Then p(t) = 0 has no solutions. So by the intermediate value
theorem, either p(t) > 0 for all t or p(t) < 0 for all t. To nish the proof, notice that
limt→∞ p(t)/t2 = a. If a > 0 then p(t) > 0 for suciently large t, hence for all t; while if
a < 0 then p(t) < 0 for suciently large t, hence for all t.
Suppose
2 × 2 matrices
g(x, y) = x2 + 8xy + y 2 .
39. Find
associated with the quadratic functions
h(x, y) = x2 − y 2
and
Solution. They are
1 0
A=
,
0 −1
With
z = (x, y)
we have
h(z) = hAz, zi
1 4
B=
.
4 1
and
g(z) = hBz, zi.
Q : R → R by Q(x) = x4 . Show that there is no c > 0 such
x 6= 0. Explain why this does not contradict Proposition 14.16.
40. Dene
for all
that
Q(x) ≥ cx2
c > 0, if x2 < c then Q(x) = x4 = x2 x2 < cx2 . This does not contradict
Proposition 14.16 because Q is not a quadratic function: it cannot be written in the form
Q(x) = hAx, xi for a matrix A. (Such a matrix A would have to 1 × 1, and so hAx, xi
2
would be a scalar constant times x .)
Solution. For any
41. Show that the point
(−1, 1)
is the minimizer of the function
f : R2 → R
dened by
f (x, y) = (2x + 3y)2 + (x + y − 1)2 + (x + 2y − 2)2 .
Solution. Notice
∇f (x, y) = (12x + 18y − 6, 18x + 28y − 10)
12 18
2
∇ f (x, y) ≡
.
18 28
In particular, ∇f (x, y) = 0 ⇔ (x, y)
∇2 f (x, y) is positive denite for all
and
= (−1, 1), and since 12 > 0, 12·28−18·18 = 12 > 0,
(x, y). Thus, f has a local minimum at (−1, 1). We
argue this must be a global minimum. Suppose to the contrary that f (r, s) < f (−1, 1).
2
Let L(t) = (−1, 1) + t(r, s) and dene g = f ◦ L. Since ∇f (−1, 1) = (0, 0) and ∇ f (x, y)
0
00
is always positive denite, we have g (0) = 0 and g (t) > 0 for all t ∈ R. (To see this,
14
apply Theorem 14.12 or Problem 34 above.) Thus,
g(1) = f (r, s) < f (−1, 1) = g(0),
42. Suppose
∇f (x) =
0
is a global minimum for
f : Rn → R
h∇2 f (x)u, ui > 0,
x
But
contradiction.
is twice continuously dierentiable. Let
0. Suppose there are u, v ∈ Rn such that
Prove that
g.
x ∈ Rn
be such that
h∇2 f (x)v, vi < 0.
is neither a local maximum nor local minimum of
Solution. By the second order approximation theorem
f.
4
t2
1
f (x + tu) − f (x) = h∇f (x), tui + h∇2 f (x)tu, tui + o((tu)2 ) = h∇2 f (x)u, ui + o((tu)2 ).
2
2
where
limt→0 o((tu)2 )/ktuk2 = 0.
Rearranging, we nd that
f (x + tu) − f (x)
1 2
o((tu)2 )
=
h∇
f
(x)u,
ui
+
t2
2
t2
limt→0 o((tu)2 )/t2 = 0.
ciently small t. An analogous
where
We can conclude that the LHS above is positive for suargument shows the LHS of
f (x + tv) − f (x)
1 2
o((tv)2 )
=
h∇
f
(x)v,
vi
+
t2
2
t2
is negative for suciently small
t.
similar proofs in Problems 29, 36 and 43 below.)
43. Suppose
f : Rn → R
-δ delta proofs here; see however
Thus, f has a saddle point at x.
(We will not give
is twice continuously dierentiable,
positive denite. Prove that there exist
c>0
and
f (x + h) − f (x) ≥ ckhk2
δ>0
if
∇f (x) = 0,
and
∇2 f (x)
is
such that
khk < δ.
Solution. By the second order approximation theorem,
1
1
f (x + h) − f (x) = h∇f (x), hi + h∇2 f (x)h, hi + o(h2 ) = h∇2 f (x)h, hi + o(h2 ).
2
2
Thus,
f (x + h) − f (x)
1
=
2
khk
2
h
h
∇ f (x)
,
khk khk
2
4 Remark
+
o(h2 )
.
khk2
on notation: we write o(hp ) to represent a function of h which, when divided by khkp ,
approaches 0 as h → 0. (This is a slight abuse of notation, since if h is a vector hp is not dened.)
15
Let
c=
1
2
∇ f (x)u, u .
: kuk=1 4
min
n
u∈R
u on the unit sphere {u ∈ Rn : kuk = 1},
2
since the unit sphere is sequentially compact. Since ∇ f (x) is positive denite, this
2
2
means that c > 0. Now choose δ > 0 such that |o(h )/khk | < c whenever 0 < khk < δ .
Then if 0 < khk < δ , we have
f (x + h) − f (x)
h
h
o(h2 )
1
2
∇
f
(x)
,
+
=
≥ 2c − c = c.
khk2
2
khk khk
khk2
Notice the minimum above is attained at some
n
44. Suppose f : R → R is twice continuously dierentiable. Suppose that ∇f (x) = 0
2
and ∇ f (x) is the matrix of all zeros. Show that x could be a local maximum, a local
minimum, or neither.
n = 2 and consider the functions f (x, y) = −x4 − y 4 , f (x, y) = x4 − y 4
4
4
and f (x, y) = x + y ; these functions all satisfy the assumptions above and they have,
respectively, a local maximum, saddle point, and local minimum at (0, 0).
Solution.
Let
45. Which of the following functions
f : R2 → R2
are linear?
x
f (x, y) = (−y, e )
2
ii) f (x, y) = (x − y , 2y)
iii) f (x, y) = 17(x, y)
i)
2f (1, 0) = (0, 2e) 6= (0, e2 ) = f (2, 0), and in ii), notice 2f (0, 1) =
(−2, 4) 6= (−4, 4) = f (0, 2). So the functions in i) and ii) are not linear. The function in
Solution. In i), note that
iii) is linear with matrix
17 0
.
0 17
46. Show there is no linear mapping
and
T : R2 → R2 having the property that T (1, 1) = (4, 0)
T (−2, −2) = (0, 1).
Solution. Suppose
T
were linear with
T (1, 1) = (4, 0)
and
T (−2, −2) = (0, 1).
Then
(0, 0) = T (0, 0) = T (2(1, 1) + (−2, −2)) = 2T (1, 1) + T (−2, −2) = (8, 0) + (0, 1) = (8, 1),
contradiction.
16
(x, y) in the plane R2 , dene T (x, y) to be the point on the line ` =
{(x, y) ∈ R : y = 2x} that is closest to (x, y). Show that T : R2 → R2 is linear and nd
47.
For a point
2
its associated matrix.
Solution. Observe that
T (x, y) = (r, 2r)
where
r
is the minimizer of
f (r) = (x − r)2 + (y − 2r)2 .
Notice
all
r,
f 0 (r) = 10r − 2x − 4y = 0
this is the unique global
48. Suppose
A
is an
m×n
r = x/5 + 2y/5. Since f 00 (r) = 10 > 0
minimizer. Thus, T is linear with matrix
1/5 2/5
.
2/5 4/5
if and only if
matrix. Dene
f : Rn → Rm
for
by
f (x) = Ax.
Prove that
Df (x) = A
for all
x ∈ Rn .
Solution. This follows from uniqueness of the derivative: since for every
x ∈ Rn ,
A(x + h) − Ax − Ah
0
f (x + h) − f (x) − Ah
= lim
= lim
= 0,
h→0
h→0 khk
h→0
khk
khk
lim
we must have
49.
Df (x) = A
for every
x ∈ Rn .
Give a proof of the rst order approximation theorem based on the mean value
theorem.
Solution. Suppose
f1 , . . . , f m .
f : Rn → Rm
is continuously dierentiable with component functions
By the mean value theorem,
fi (x + h) − fi (x) = h∇fi (z (i) ), hi,
z (i)
is on the line segment between x and x + h. Let x be xed and let B(h) be
(i)
the matrix whose ith row is ∇fi (z ). Combining the component equations above,
where
f (x + h) − f (x) = B(h)h.
Thus,
f (x + h) − f (x) − Df (x)h
f (x + h) − f (x) − B(h)h B(h)h − Df (x)h
= lim
+
lim
h→0
h→0
khk
khk
khk
B(h)h − Df (x)h
h
= lim
= lim [B(h) − Df (x)]
.
h→0
h→0
khk
khk
17
limh→0 B(h) − Df (x) = 0 by continuity
h/khk is bounded. Thus, the limit above is 0.
Note
of partial derivatives of
f.
Moreover,
To see why, note that the generalized
Cauchy-Schwartz inequality (Theorem 14.13) yields
k[B(h) − Df (x)]hk
≤ lim kB(h) − Df (x)k = 0.
h→0
h→0
khk
lim
f : R2 → R2 is continuously dierentiable with f (x, y) = (ψ(x, y), φ(x, y)).
2
Dene g : R
→ R2 by g(x, y) = 21 [ψ(x, y)2 + φ(x, y)2 ]. Show that Dg(x0 , y0 ) =
f (x0 , y0 )Df (x0 , y0 ). Use this to show that if (x0 , y0 ) is a minimizer of g : R2 → R
and Df (x0 , y0 ) is invertible, then f (x0 , y0 ) = 0.
50. Suppose
h(x, y) = 21 [x2 + y 2 ].
g(x, y) = (h ◦ f )(x, y). By the chain rule,
D1 ψ(x, y) D2 ψ(x, y)
Dg(x, y) = Dh(f (x, y))Df (x, y) = ψ(x, y) φ(x, y)
D1 φ(x, y) D2 φ(x, y)
Solution. Let
Then
= f (x, y)Df (x, y).
The result follows when x = x0 , y =
Df (x0 , y0 )−1 and using the fact that
51. Suppose
ψ : R2 → R
y0 by multiplying both sides above on the right by
Dg(x0 , y0 ) = 0 since (x0 , y0 ) is a minimizer of g .
is continuously dierentiable and dene
g : R2 → R
g(s, t) = ψ(s2 t, s).
Find
∂g/∂s(s, t)
and
∂g/∂t(s, t).
Solution. By the chain rule,
∂g
∂g ∂x ∂g ∂y
=
+
,
∂s
∂x ∂s ∂y ∂s
∂g
∂g ∂x ∂g ∂y
=
+
.
∂t
∂x ∂t
∂y ∂t
That is,
∂g
(s, t) = D1 ψ(s2 t, s)2st + D2 ψ(s2 t, s),
∂s
∂g
(s, t) = D1 ψ(s2 t, s)s2 .
∂t
52. Suppose
g, h : R2
have continuous second derivatives and dene
u(s, t) = g(s − t) + h(s + t).
18
by
Prove that
∂ 2u
∂ 2u
(s, t) − 2 (s, t) = 0
∂t2
∂s
(s, t) ∈ R2 .
for all
Solution. By the chain rule,
∂u(s, t)
= −g 0 (s − t) + h0 (s + t)
∂t
∂u(s, t)
= g 0 (s − t) + h0 (s + t),
∂s
and
∂ 2u
(s, t) = g 00 (s − t) + h00 (s + t),
∂s
∂ 2u
(s, t) = −(−g 00 (s − t)) + h00 (s + t).
∂s
The result follows.
53. Let
a∈U
U, V ⊂ R
be such that
be open, suppose f : U → V is dierentiable and bijective, and let
f 0 (a) = 0. Prove that f −1 : V → U is not dierentiable at b = f (a).
g := f −1
Solution. Suppose
is dierentiable at
b.
Then by the chain rule,
(g ◦ f )0 (a) = g 0 (b)f 0 (a) = 0.
But
(g ◦ f )(x) = x
for all
x ∈ U,
54. For the following mappings
the point
(0, 0)
f (0, 0)
f (x, y) = (x + x2 + e
x2 y 2
(1, 0) =
g(0, 0).
f (0, 0) = (1, 0)
Df
contradiction.
apply the inverse function theorem at
, −x + y + sin(xy)),
Df (0, 0) =
−1
f, g : R2 → R2 ,
and
Solution. Since
and
(g ◦ f )0 (a) = 1,
and calculate the partial derivatives of the components of the inverse
mapping at the points
a)
so that
and
1 0
,
−1 1
g(0, 0) = (1, 1),
1 0
−1 1
−1
b)
g(x, y) = (ex+y , ex−y ).
1 1
Dg(0, 0) =
1 −1
by the inverse function theorem
−1 1 1 1 1
Dg (1, 1) =
= 21 2 1 .
−2
1 −1
2
1 0
=
,
1 1
−1
−1
second) row of Df
(1, 0) contains the partial derivatives of the rst
−1
−1
(resp. second) component of f
at the point (1, 0), and similarly for g
.
The rst (resp.
19
55. Dene
f : R2 → R2
theorem applies at every
f (x, y) = (ex cos y, ex sin y).
2
point (x, y) ∈ R , but that f is
by
Show that the inverse function
not one-to-one. Does the latter
statement contradict the former?
Solution. Observe that
x
e cos y −ex sin y
Df (x, y) =
ex sin y ex cos y
det Df (x, y) = ex 6= 0 for all (x, y) ∈ R2 . Thus, the inverse function applies at
(x, y) ∈ R2 . However, f is not one-to-one since f (x, y) = f (x, y + 2π). There is no
and so
all
contradiction here: a function can be locally invertible at every point without having a
global inverse.
56. For a pair of real numbers
a, b,
consider the system of nonlinear equations
3 2
x + x2 cos y + xyex y = a,
y + x5 + y 3 − x2 cos(xy) = b.
(8)
Use the inverse function theorem to show that there is some positive number
2
2
2
if a + b < r , then this system of equations has at least one solution.
r
such that
3 2
f (x, y) = (x + x2 cos y + xyex y , y + x5 + y 3 − x2 cos(xy)). Then
3 2
3 2
3 2
3 2
1 + 2x cos y + yex y + 3x3 yex y −x2 sin y + xex y + 2xy 2 ex y
Df (x, y) =
5x4 − 2x cos(xy) + x2 y sin(xy)
1 + 3y 2 + x3 sin(xy)
Solution. Let
and so
1 0
Df (x, y) =
0 1
which is nonsingular. So by the inverse function theorem, there are neighborhoods
U
of
f : U → V is invertible. Thus, the system (8)
of equations can be solved when (a, b) ∈ V (uniquely, if one considers only (x, y) ∈ U ).
Since V is open we may pick r > 0 such that Br (0, 0) ⊂ V , which completes the proof.
(0, 0)
and
V
57. Suppose
of
f (0, 0) = (0, 0)
ψ : R2 → R
such that
is continuously dierentiable and dene
f : R2 → R2
by
f (x, y) = (ψ(x, y), −ψ(x, y)).
Explain both geometrically and analytically why the hypotheses of the inverse function
2
theorem must fail for f at every point (x, y) ∈ R .
Solution. The analytic explanation is that
D1 ψ(x, y)
D
ψ(x,
y)
2
=0
det Df (x, y) = −D1 ψ(x, y) −D2 ψ(x, y)
20
for every
(x, y).
The geometric explanation is that Im f is the line
y = x,
which is
one-dimensional, and an injective continuous function dened on the plane cannot have
a
1-dimensional
58. Dene
that
f
image.
f (x) = xn
is not stable if
x ∈ R.
n > 1.
for
Prove that
f
is bijective if
n≥1
is odd. Then show
n ≥ 1 be odd. To see that f is onto, let y ∈ R. By the intermediate value
√
theorem there is x between 0 and n y such that f (x) = y . To see that f is one-to-one,
suppose f (a) = f (b) > 0 for a 6= b. Then a, b > 0 since n is odd, and by the mean value
0
n−1
theorem there is c between a and b such that f (c) = nc
= 0, a contradiction since
c > min{a, b} > 0. Similarly, f (a) = f (b) < 0 only if a = b. The other possibility is that
f (a) = f (b) = 0, but f (x) = xn = 0 only when x = 0. To see that f is not stable when
n > 1, we use the dierence of powers formula
Solution. Let
n
n
f (x) − f (y) = x − y = (x − y)
n−1
X
xn−1−i y i .
i=0
Suppose
f
p
c. If y = 2x where 0 < x < n−1 c/(2n),
n−1
f (x) − f (y) X
n−1−i i =
x
y = 2nxn−1 < c,
x−y were stable with constant
then
i=0
contradiction.
f : (a, b) → R be dierentiable. Prove
0
only if |f (x)| ≥ c for all x ∈ (a, b).
59. Let
if and
Solution. Suppose that
|f 0 (x)| ≥ c
for all
that
x ∈ (a, b).
f
is stable with stability constant
If there were
x, y ∈ (a, b)
c
such that
|f (x) − f (y)| < c|x − y|
then by the mean value theorem there would be a point r between x and y such that
|f 0 (r)| = |f (x) − f (y)|/|x − y| < c, contradiction. Thus, f must be stable. Conversely,
suppose
and so
f
|f (x) − f (y)| ≥ c|x − y|
f (x) − f (y) 0
|f (y)| = lim
≥ c.
x→y
x−y is stable with constant
2
60. Dene g(x, y) = (x , y) for
2
2
that g : R → R is stable.
c.
Then
(x, y) ∈ R2 .
for all
x, y ∈ (a, b)
Show there is no neighborhood of
21
(0, 0)
such
Suppose
g
were stable with stability constant
c.
Then for
x∈R
such that
0 < |x| < c,
kg(x, 0) − g(0, 0)k
x2
=
= |x| < c,
k(x, 0) − (0, 0)k
|x|
contradiction.
61. Is the sum of stable mappings also a stable mapping?
Solution. No. For example,
but
f +g ≡0
62.
Let
U ⊂ Rn
and
be open and suppose
stable. Prove for each
Solution. Let
f (x) = x
g(x) = −x,
dened for
x ∈ R,
are both stable,
is not.
x ∈ U.
x ∈ U , Df (x)
f : U → Rn
is continuously dierentiable and
is nonsingular.
By the rst order approximation theorem,
f (x + h) − f (x) = Df (x)h + o(h).
Suppose
f
has stability constant
c.
Then
kDf (x)hk ≥ kDf (x)h + o(h)k − ko(h)k
= kf (x + h) − f (x)k − ko(h)k ≥ ckhk − ko(h)k.
Thus,
This shows that
63. For a point
Df (x) h ≥ c − ko(h)k → c
khk khk
Df (x)u 6= 0 for all unit vectors u. It
(ρ, θ, φ) ∈ R3 ,
as
h → 0.
follows that
Df (x)
is nonsingular.
dene
f (ρ, θ, φ) = (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ).
At what points does the inverse function apply to
f?
Solution. Since
sin φ cos θ −ρ sin φ sin θ ρ cos φ cos θ
det Df (ρ, θ, φ) = sin φ sin θ ρ sin φ cos θ ρ cos φ sin θ cos φ
0
−ρ sin φ −ρ sin φ sin θ ρ cos φ cos θ
sin φ cos θ −ρ sin φ sin θ
− ρ sin φ = cos φ sin φ sin θ ρ sin φ cos θ ρ sin φ cos θ ρ cos φ sin θ = −ρ2 sin φ cos2 φ − ρ2 sin3 φ = −ρ2 sin φ,
22
the inverse function theorem applies when
64. Dene
f (x) = x2
for
ρ 6= 0
and
φ
is not an integer multiple of
π.
x ∈ R.
U ⊂ R is open and 0 ∈
/ U , then f (U ) is open.
if U ⊂ R is open and 0 ∈ U , then f (U ) is not open.
a) Show that if
b) Show that
U ⊂ R is a union of open intervals. Thus, we only have to show
that if U = (a, b) ⊂ R and 0 ∈
/ (a, b), then f (U ) is open. If 0 < a < b and U = (a, b),
then f (U ) = (f (a), f (b)), while if a < b < 0 and U = (b, a), then f (U ) = (f (b), f (a)).
Solution. a) Any open set
b) Let
image
U ⊂ R with 0 ∈ U . Then 0 ∈ f (U ) but for every > 0, (−, ) 6⊂ f (U )
of f contains no negative numbers. Thus, f (U ) is not open.
Give an example of a continuously dierentiable mapping f : R
n
property that there is no open subset U ⊂ R such that f (U ) is open.
65.
f (x) ≡ 0. f is continuously
U ⊂ R , f (U ) = {0} is not open.
Solution. Let
open
n
→ Rn
since the
with the
dierentiable (since it is linear), but for every
n
(x, y) ∈ R2 . Show that the assumptions of
Dini's Theorem do not hold at the point (0, 0). Explain geometrically why the conclusion
of Dini's theorem does not hold at (0, 0).
66. Consider the equation
x2 + y 2 = 0
for
f (x, y) = x2 + y 2 . Since D1 f (0, 0) = D2 f (0, 0) = 0, the assumptions of
2
2
Dini's theorem do not hold at (0, 0). Note that the solution set for x + y = 0 consists
only of the point (0, 0). Thus, y cannot be written as a function of x (or vice-versa) on
an open set containing 0.
Solution.
Let
67. Consider the equation
e2x−y + cos(x2 + xy) − 2 − 2y = 0,
(x, y) ∈ R2 .
Does the set of solutions of this equation in a neighborhood of the solution
x as a function
the point 0.
dene
at
Solution. Let
of
y
(0, 0) implicitly
or vice-versa? If so, compute the derivative of this function(s)
f (x, y) = e2x−y + cos(x2 + xy) − 2 − 2y
and note that
∇f (x, y) = (2e2x−y − (2x + y) sin(x2 + xy), −e2x−y − x sin(x2 + xy) − 2).
23
D1 f (0, 0) = 2 6= 0 and D2 f (0, 0) = −3 6= 0. So the set of solutions denes x as
function of y , and vice-versa, in a neighborhood of (0, 0). Writing ψ and φ for these
Thus,
a
functions, respectively, we have
D1 f (ψ(x), x)ψ 0 (x) + D2 f (ψ(x), x) = 0,
D1 f (x, φ(x)) + D2 f (x, φ(x))φ0 (x) = 0,
for
x
in a neighborhood of
ψ 0 (0) = −
0.
Thus,
D2 f (0, 0)
3
= ,
D1 f (0, 0)
2
φ0 (0) = −
D1 f (0, 0)
2
= .
D2 f (0, 0)
3
O ⊂ R2 be open and f : O → R continuously dierentiable. Suppose that
f (a, b) = 0 and ∇f (a, b) 6= (0, 0). Show that ∇f (a, b) is orthogonal to the tangent line
at (a, b) of the implicitly dened function.
68.
Let
∇f (a, b) 6= (0, 0), we can assume without loss of generality that D2 f (a, b) 6=
0. So by Dini's theorem, in a neighborhood of (a, b) the equation f (x, y) = 0 implicitly
denes y = φ(x) and moreover
Solution. Since
D1 f (x, φ(x)) + D2 f (x, φ(x))φ0 (x) = 0.
Setting
x = a,
this rewrites as
h∇f (a, b), (1, φ0 (a))i = 0.
Observe that
(1, φ0 (a))
x = a,
x 7→ (x, φ(x)).
is the tangent line to the implicitly dened function at
since in a neighborhood of
(a, b)
the graph of that function is the curve
φ : R → R is continuously dierentiable and that φ0 (a) 6= 0.
2
2
Set b = φ(a) and dene f : R → R by f (x, y) = y − φ(x) for (x, y) ∈ R . Apply Dini's
2
theorem to the function f : R → R at the point (a, b) and compare the result with the
conclusion of the Inverse Function Theorem applied to φ at a.
69. Suppose the function
D1 f (a, b) = −φ0 (a) 6= 0, by Dini's theorem, there is a continuously
dierentiable function ψ dened on a neighborhood of b such that x = ψ(y) uniquely
solves the equation f (x, y) = y − φ(x) = 0 in a neighborhood of (a, b). Thus, for y in a
−1
neighborhood of b, ψ(y) = φ (y) and
Solution.
Since
0 = D1 f (ψ(y), y)ψ 0 (y) + D2 f (ψ(y), y) = −φ0 (φ−1 (y))(φ−1 )0 (y) + 1.
24
(9)
Since
φ0 (a) 6= 0,
the Inverse Function Theorem shows that in a neighborhood of
a continuously dierentiable inverse and for
y
b, φ
has
in this neighborhood,
(φ−1 )0 (y) = φ0 (φ−1 (y))−1 .
(10)
Observe that equations (9) and (10) are simply rearrangements of one another.
70. Suppose
f : R2 → R
is continuously dierentiable and
D2 f (x, y) ≥ c
for all
x∈R
and if
Solution. Fix
0
φ (y) ≥ c > 0.
x
f (x, y) = 0,
and dene
Thus,
φ
then
is such that
(x, y) ∈ R2 .
Prove there is a continuously dierentiable function
for all
c>0
g:R→R
(11)
such that
f (x, g(x)) = 0
y = g(x).
φ:R→R
φ(y) = f (x, y). The assumption
(why?), call it g(x). That is,
by
has a unique root
(11) implies
φ(g(x)) = f (x, g(x)) = 0.
f (x, y) = 0 then y = g(x). To see
that g is continuously dierentiable at an arbitrary point a ∈ R, note that by (11), Dini's
theorem holds at the point (a, g(a)). Since g is unique, it must agree with the implicit
function furnished by Dini's theorem in a neighborhood of a. Since the latter function is
continuously dierentiable at a, so is g .
Uniqueness of the root (for arbitrary
x)
shows that if
71. Consider the linear system of equations
a11 x + a12 y + a13 z = 0,
a21 x + a22 y + a23 z = 0,
Dene
ν = (a11 , a12 , a13 )
and
β = (a21 , a22 , a23 ).
(12)
(x, y, z) ∈ R3 .
Show that if
ν × β 6= (0, 0, 0),
then
the equations (12) denes two of the variables as a function of the remaining variable.
Interpret this geometrically.
Solution. Dene
f (x, y, z) = (a11 x + a12 y + a13 z, a21 x + a22 y + a23 z)
and observe that
a12 a13 a11 a13 a12 a12 (1, 0, 0) − ν × β = a12 a23 (0, 1, 0) + a21 a22 (0, 0, 1)
a22 a23 = |D(y,z) f (x, y, z)|(1, 0, 0) − |D(x,z) f (x, y, z)|(0, 1, 0) + |D(x,y) f (x, y, z)|(0, 0, 1).
25
Since
ν ×β 6= (0, 0, 0), one of the determinants above must be nonzero.
Thus, the Implicit
Function Theorem shows that (12) implicitly denes two of the variables as a function
3
of the third. Geometrically, each equation of (12) represents a plane in R , and the
assumption
ν × β 6= (0, 0, 0)
guarantees that these two planes are not parallel. Thus, the
planes intersect in a line, which is why we can solve for two of the variables as a function
of the third.
72.
y 3 − x2 = 0 for (x, y) ∈ R2 . Does the
point (0, 0)? Does this equation dene one of
Consider the solutions of the equation
Implicit Function Theorem apply at the
the components of a solution
(x, y)
as a function of the other component?
(0, 0): letting
f (x, y) = 0
that y is not a
Solution. The hypotheses of the Implicit Function Theorem do not hold at
f (x, y) = y 3 − x2 , we have D1 f (0, 0) = D2 f (0, 0) = 0.
2/3
does dene y as a function of x, namely y = x
.
continuously dierentiable function of x.)
26
However, the equation
(Note, however,