1 Solutions to selected problems

1. Let A ⊂ B ⊂ Rn. Show that intA ⊂ intB but in general bdA 6⊂ bdB.

Solution. Let x ∈ intA. Then there is ε > 0 such that Bε(x) ⊂ A ⊂ B. This showsx ∈ intB. If A = [0, 1] and B = [0, 2], then bdA = {0, 1} 6⊂ bdB = {0, 2}.

2. Let A ⊂ Rn be open and f : A→ R continuous with f(u) > 0. Show there is an openball B around u such that f(x) > f(u)/2 for x ∈ B.

Solution. Since f is continuous at u, for ε = f(u)/2 there is δ > 0 such that f(Bδ(u)) ∈(f(u)− ε, f(u) + ε). In particular, f(Bδ(u)) > f(u)/2 and we may take B = Bδ(u).

3. Suppose f : Rn → R is continuous and f(u) > 0 if u has at least one rationalcomponent. Prove that f(u) ≥ 0 for all u ∈ Rn.

Solution. Let u ∈ Rn be arbitrary. Write u = (u1, . . . , un) and let {rk} be the �rst kdigits in the decimal expansion for u1, and let uk = (rk, u2, . . . , un). Then {uk} → u soby continuity of f , {f(uk)} → f(u). Also f(uk) > 0 for all k since the rk's are rational.Thus, f(u) ≥ 0.

4. Show that an open ball in Rn is bounded.

Solution. Let Br(x) ⊂ Rn be an arbitrary ball. We must show it is contained in a ballBs(0) around 0. Let s = r+ ‖x‖. If y ∈ Br(x) then ‖y‖ ≤ ‖y− x‖+ ‖x|‖ < r+ ‖x‖ = s.Thus, Br(x) ⊂ Bs(0).

5. Let f : A → R be continuous with A ⊂ Rn. If A is bounded, is f(A) bounded? If Ais closed, is f(A) closed?

Solution. The answer to both questions is no. Consider f(x) = 1/x. Then A = (0, 1) isbounded but f(A) = (1,∞) is not; A = [1,∞) is closed but f(A) = (0, 1] is not.

6. Let f : Rn → R be continuous with f(u) ≥ ‖u‖ for every u ∈ Rn. Prove thatf−1([0, 1]) is sequentially compact.

Solution. Since [0, 1] is closed and f is continuous, f−1([0, 1]) is closed. It remains to

1

show f−1([0, 1]) is also bounded, hence sequentially compact. If x ∈ f−1([0, 1]), then0 ≤ f(x) ≤ 1 and f(x) ≥ ‖x‖, so in particular, ‖x‖ ≤ 1. Thus, f−1([0, 1]) is bounded.

7. Let A ⊂ Rn be sequentially compact and v ∈ Rn \ A. Prove there is u ∈ A such that

‖u− v‖ ≤ ‖x− v‖ for all x ∈ A. (1)

Solution. Let d = infx∈A ‖x − v‖. Since d + 1/k is not a lower bound for ‖x − v‖ overx ∈ A, we may pick uk ∈ A such that d ≤ ‖uk − v‖ < d + 1/k. Using sequentialcompactness, pick {unk

} → u. By continuity of vector subraction and the norm ‖ · ‖,{‖unk

− v‖} → ‖u− v‖. And by the squeeze theorem in R, ‖u− v‖ = d.

The point u is not unique: if A = {1,−1} ⊂ R and v = 0, then u = ±1 satisfy (1).

8. A mapping F : Rn → Rm is Lipschitz continuous if there is K > 0 such that

|F (x)− F (y)‖ ≤ K‖x− y‖ (2)

for all x, y ∈ Rn. Show that a Lipschitz mapping is uniformly continuous.

Solution. Suppose (2) holds for F . Let ε > 0 and pick δ = K/ε. Then

‖F (x)− F (x)‖ ≤ K‖x− y‖ < Kδ = ε whenever ‖x− y‖ < δ.

9. Let A be sequentially compact and f : A→ f(A) continuous and injective. Show f−1

is continuous. Give an example to show the assumption on A is necessary.

Solution. Let {vk} be a sequence in f(A) such that {vk} → v, and let uk = f−1(vk),u = f−1(v). Suppose {uk} 6→ u. Using sequential compactness of A, pick {unk

} → w ∈ Awith w 6= u. By continuity of f , {f(unk

)} → f(w). As {f(unk)} = {vnk

} → v, we havef(w) = v. Thus, w = f−1(v) = u, contradiction.

To see that sequential compactness of A is needed, let

f(x) =

{x, 0 ≤ x < 1

4− x, 2 ≤ x ≤ 3,

f−1(x) =

{x, 0 ≤ x < 1

3− x, 1 ≤ x ≤ 2.

Note that f is continuous on A = [0, 1) ∪ [2, 3] but f−1 is not continuous at 1.

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10. Let f : A → Rm be continuous and A sequentially compact. Show f is uniformlycontinuous.

Solution. Let {uk}, {vk} be sequences in A such that {‖uk − vk‖} → 0. Suppose{‖f(uk)−f(vk)‖} 6→ 0. Then along a subsequence, {‖f(unk

)−f(vnk)‖} → c > 0. (Why?)

Since A is sequentially compact, we can pick sub-subsequences {umnk} → u ∈ A and

{vmnk} → v ∈ A, and u = v since {‖uk−vk‖} → 0. By continuity of f , {f(umnk

)} → f(u)and {f(vmnk

)} → f(v) = f(u). Thus, {‖f(umnk)− f(vmnk

)‖} → 0, contradiction.

11. We say u ∈ Rn is a a limit point of A ⊂ Rn if there is a sequence in A \ {u} thatconverges to u. Prove that u is a limit point of A if and only if every open ball aroundu contains in�nitely many points of A \ {u}.

Solution. Let 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 bean in�nite set: if it were �nite with elements v1, . . . , vn, then for ε = min1≤i≤n ‖vi − u‖we would have uk /∈ Bε(u) for every k ≥ N .

12. Let A ⊂ Rn and let f be the characteristic function of A. Show that f is continuousat u if and only if u /∈ bdA. Can one make an analogous statement about limx→u f(x)?

Solution. Suppose u /∈ bdA. Then if u ∈ A, there is δ > 0 such that Bδ(u) ⊂ A. Givenany ε > 0, if ‖x − u‖ < δ then x ∈ A and so |f(x) − f(u)| = |1 − 1| = 0 < ε. If u /∈ A,there is δ > 0 such that Bδ(u) ⊂ Rn \ A and an analogous argument holds. Conversely,if u ∈ bdA, then for each k ∈ N there is vk, wk ∈ B1/k(u) such that vk ∈ A, wk ∈ Rn \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) thenlimx→u f(x) exists � in the argument above, just replace ‖x−u‖ < δ with 0 < ‖x−u‖ < 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.)

13. Let f : A → Rm with u ∈ A a limit point of A ⊂ Rn. Show that if f does not havea limit at u, then f is not continuous at u.

Solution. We prove the contrapositive. Suppose f is continuous at u. Then limx→u f(x) =f(u). To see this, let ε > 0 and pick δ > 0 such that ‖f(x) − f(u)‖ < ε whenever‖x− u‖ < δ and x ∈ A; then in particular, ‖f(x)− f(u)‖ < ε whenever 0 < ‖x− u‖ < δand x ∈ A.

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14. Let f : A → Rm with u ∈ A a limit point of A ⊂ Rn. Show that f is continuous atu if and only if limx→u f(x) = f(u).

Solution. We only need to observe that for any ε > 0, the statements

‖f(x)− f(u)‖ < ε whenever x ∈ A and 0 < ‖x− u‖ < δ

and‖f(x)− f(u)‖ < ε whenever x ∈ A and ‖x− u‖ < δ

are equivalent, since ‖f(x)− f(u)‖ = 0 < ε when ‖x− u‖ = 0.

15. Let A ⊂ Rn and suppose 0 is a limit point of A. Suppose f : A → R is such thatf(x) ≥ c‖x‖2 for all x ∈ A, where c > 0 is constant. Suppose g : A → R is such thatlimx→0 g(x)/‖x‖2 = 0. Prove there is r > 0 such that f(x)− g(x) ≥ (c/2)‖x‖2 for x ∈ Awith 0 < ‖x‖ < r.

Solution. We use ε = c/2 in the de�nition ε-δ de�nition of limx→0 g(x)/‖x‖2 = 0. Pickδ > 0 so that if x ∈ A with 0 < ‖x‖ < δ, then ‖g(x)/‖x‖2‖ = ‖g(x)‖/‖x‖2 < c/2 and so

‖f(x)− g(x)‖ ≥ ‖f(x)‖ − ‖g(x)‖ ≥ c‖x‖2 − (c/2)‖x‖2 = (c/2)‖x‖2.

16. Let

f(x, y) =

{xy2

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

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

Show f is continuous at (0, 0) and has directional derivatives at (0, 0) in every direction,but is not di�erentiable at (0, 0).

Solution. To see f is continuous at (0, 0): for (x, y) 6= (0, 0),

|f(x, y)− f(0, 0)| =∣∣∣∣ xy2

x2 + y2− 0

∣∣∣∣ = |x|x2

y2+ 1≤ |x| → 0 as (x, y)→ 0.

(See problem 14 above.) For the directional derivatives D(a,b)f(0, 0):

limt→0

f((0, 0) + t(a, b))− f(0, 0)t

= limt→0

f(ta, tb)

t= lim

t→0

t2ab2

t2a2 + t2b2=

ab2

a2 + b2.

Now note that D1f(0, 0) = 0, D2f(0, 0) = 0 and

f((0, 0) + (x, y))− f(0, 0)− 0x− 0y

‖(x, y)‖=

xy2

x2+y2− 0x− 0y√x2 + y2

=xy2

(x2 + y2)3/2. (3)

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When x ≡ y and y → 0 the limit of (3) is 2−3/2 6= 0, so f is not di�erentiable at (0, 0).

17. De�ne f : R2 → R by

f(x, y) =

{xy

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

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

Show the partial derivatives of f are not continuous at (0, 0).

Solution. For (x, y) 6= (0, 0), we can calculate partial derivatives �as usual� (i.e., withoutresorting to the de�nition):1

D1f(x, y) =y3 − x2y(x2 + y2)2

, D2f(x, y) =x3 − xy2

(x2 + y2)2.

In particular, D1f(0, y) = y−1, D1f(x, 0) = x−1 do not have limits as y → 0, x → 0.Thus, they are not continuous at (0, 0). (See problem 13 above.)

18. De�ne g : R2 → R by

g(x, y) =

{x2y4

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

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

Is g continuously di�erentiable?

Solution. For (x, y) 6= (0, 0) we can calculate partial derivatives �as usual�:

D1g(x, y) =2xy6

(x2 + y2)2, D2g(x, y) =

4x4y3 + 2x2y5

(x2 + y2)2.

Moreover, from the limit de�nition of partial derivatives,

D1g(0, 0) = limt→0

g(t, 0)− g(0, 0)t

= limt→0

0− 0

t= 0,

D2g(0, 0) = limt→0

g(0, 0)− g(0, t)t

= limt→0

0− 0

t= 0.

Since D1g(x, y) and D2g(x, y) are rational functions, they are continuous except at theirasymptotes (x, y) = (0, 0). Thus, to establish continuity of D1g and D2g we need only

1Dif ≡ Deif ≡∂f∂xi

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check that lim(x,y)→(0,0)Dig(x, y) = 0 for i = 1, 2. For (x, y) 6= (0, 0),

|D1g(x, y)− 0| =∣∣∣∣ 2xy6

x4 + 2x2y2 + y4

∣∣∣∣ =∣∣∣∣∣ 2xy2

x4

y4+ 2x2

y2+ 1

∣∣∣∣∣ ≤ 2|x|y2 → 0 as (x, y)→ (0, 0)

|D2g(x, y)− 0| =∣∣∣∣ 4x4y3 + 2x2y5

x4 + 2x2y2 + y4

∣∣∣∣ =∣∣∣∣∣ 4x2y + 2y3

x2

y2+ 2 + y2

x2

∣∣∣∣∣ ≤ 2x2|y|+ |y|3 → 0 as (x, y)→ (0, 0).

This shows D1g(x, y) and D2g(x, y) are continuous; thus, g is continuously di�erentiable.

19. Suppose g : R2 → R has the property |g(x, y)| ≤ x2 + y2 for all (x, y) ∈ R2. Show ghas partial derivatives with respect to both x and y at (0, 0).

Solution. Our assumption on g forces g(0, 0) = 0 and thus∣∣∣∣g(t, 0)− g(0, 0)t

∣∣∣∣ ≤ ∣∣∣∣t2t∣∣∣∣ = |t|, ∣∣∣∣g(0, t)− g(0, 0)t

∣∣∣∣ ≤ ∣∣∣∣t2t∣∣∣∣ = |t|.

Taking limits as t→ 0 in the above expressions, we �nd that D1g(0, 0) = D2g(0, 0) = 0.

20. Suppose f : R2 → R has �rst-order partial derivatives and

D1f(x, y) = D2f(x, y) = 0 for all (x, y) ∈ R2.

Show that f ≡ c for some c ∈ R; that is, f is a constant function.

Solution. Fix y0 ∈ R and consider g : R → R de�ned by g(x) = f(x, y0). Suppose gis nonconstant. Then g(a) 6= g(b) for some a < b, and by MVT there is r ∈ (a, b) suchthat D1f(r, y0) = g′(r) = [g(b) − g(a)]/(b − a) 6= 0, contradiction. Thus, g is constant,say g ≡ c. Let (x, y) ∈ R2. By repeating the argument above, we �nd that the functionh : R→ R de�ned by h(z) = f(x, z) is constant. Since h(y0) = g(x) we must have h ≡ c,and in particular f(x, y) = c. Since (x, y) ∈ R2 was arbitrary, we can conclude f ≡ c.

21. Given φ, ψ : R2 → R, a function f : R2 → R is called a potential function for φ, ψ if

D1f(x, y) = φ(x, y) and D2f(x, y) = ψ(x, y) for all (x, y) ∈ R2.

Show that when a potential function exists for φ, ψ, 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 continuouslydi�erentiable, then

D1ψ(x, y) = D2φ(x, y) for all (x, y) ∈ R2. (4)

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Solution. Suppose f and g are two potential functions for φ, ψ and let h = f − g. Then

Dih(x, y) = Dif(x, y)−Dig(x, y) = 0 for all (x, y) ∈ R2, i = 1, 2,

so by Problem 20, h is constant. Thus, f and g di�er by a constant. The statementin (4) follows from the assumption D1f and D2f are continuously di�erentiable andTheorem 13.10.

22. De�ne f : R2 → R by

f(x, y) =

{x3y−xy3x2+y2

, (x, y) 6= (0, 0)

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

Show that D1f(0, y) = −y for all y ∈ R and D2f(x, 0) = x for all x ∈ R. Conclude thatD2D1f(0, 0) = −1 but D1D2f(0, 0) = 1.

Solution. Away from (0, 0) we can calculate partial derivatives of f �as usual,� i.e.,without resorting to the limit de�nition. Thus, for (x, y) 6= (0, 0),

D1f(x, y) =(x2 + y2)(3x2y − y3)− (x3y − xy3)(2x)

(x2 + y2)2,

D2f(x, y) =(x2 + y2)(x3 − 3xy2)− (x3y − xy3)2y

(x2 + y2)2.

Thus, for y 6= 0 and x 6= 0,

D1f(0, y) =y2(−y3)(y2)2

= −y, D2f(x, 0) =x2x3

(x2)2= x,

and, from the limit de�nition of partial derivatives, D1f(0, 0) = 0 = D2f(0, 0). (Checkthis! See Problem 18 for a similar calculation.) Thus,D2D1f(0, y) = −1 andD1D2f(x, 0) =1 for all x, y ∈ R. In particular, D2D1f(0, 0) = −1 but D1D2f(0, 0) = 1.

23. Let A ⊂ R2 be an open set containing (x0, y0). Prove that there is r > 0 such that(x, y) ∈ A whenever |x− x0| < 2r and |y − y0| < 2r.

Solution. Pick ε > 0 such that Bε(x0, y0) ⊂ A. If 0 < r < ε/(2√2), then

‖(x, y)− (x0, y0)‖ =√

(x− x0)2 + (y − y0)2 <√

(2r)2 + (2r)2 =√8r2 = 2

√2r < ε

whenever |x− x0| < 2r and |y − y0| < 2r.

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24. Suppose f : Rn → R and g : R→ R are continuously di�erentiable. Find a formulafor ∇(g ◦ f)(x) in terms of ∇f(x) and g′(f(x)).

Solution. By the de�nition of partial derivative and the mean value theorem,

Di(g ◦ f)(x) = limt→0

g(f(x+ tei))− g(f(x))t

= limt→0

g′(zt)f(x+ tei)− f(x)

t,

where zt is on the line segment between f(x) and f(x + tei). Notice limt→0 g′(zt) =

g′(f(x)) due to continuity of f and g′, and Dif(x) := limt→0[f(x+ tei)− f(x)]/t. Thus,

Di(g ◦ f)(x) = g′(f(x))Dif(x), i = 1, . . . , n,

that is, ∇(g ◦ f)(x) = ∇f(x) g′(f(x)).

25. Suppose f : Rn → R is such that Dvf(x) exists. Prove that Dcvf(x) = cDvf(x) forany nonzero c ∈ R.

Solution. This follows from the computation

Dcvf(x) = limt→0

f(x+ tcv)− f(x)t

= limt→0

cf(x+ tcv)− f(x)

tc= c lim

s→0

f(x+ sv)− f(x)s

= cDvf(x).

26. Suppose f : Rn → R has �rst-order partial derivatives and that x ∈ Rn is a localminimizer for f , that is, there is ε > 0 such that

f(x+ h) ≥ f(x) for h ∈ Bε(0). (5)

Prove that ∇f(x) = 0.

Solution. Due to the assumption (5), for i = 1, . . . , n,

Deif(x) = limt→0

f(x+ tei)− f(x)t

≥ 0,

D−eif(x) = limt→0

f(x− tei)− f(x)t

≥ 0,

and by problem 25, Deif(x) = −D−eif(x) ≤ 0. Thus Deif(x) = 0, and so ∇f(x) = 0.

27. Considerf(x, y, z) = xyz + x2 + y2.

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Find θ ∈ (0, 1) such that

f(1, 1, 1)− f(0, 0, 0) = D1f(θ, θ, θ) +D2f(θ, θ, θ) +D3f(θ, θ, θ).

Solution. Note that

D1f(x, y, z) = yz + 2x, D2f(x, y, z) = xz + 2y, D3f(x, y, z) = xy.

and f(1, 1, 1)− f(0, 0, 0) = 3. Thus, we solve 3θ2 + 4θ = 3 to get θ = −23+√133.

28. De�ne f : R2 → R by

f(x, y) =

{x√x2+y2

|y| , if y 6= 0,

0, if y = 0.

a) Prove f is not continuous at (0, 0).b) Prove f has directional derivatives in all directions at (0, 0).c) Prove for any c ∈ R there is p such that

‖p‖ = 1 and Dpf(0, 0) = c.

Does this contradict Corollary 13.18?

Solution. a) Note that f(x, y) ≡ 1 when y = x2√1−x2 and x 6= 0. Approaching (0, 0) along

this curve shows f is not continuous at zero, since f(0, 0) = 0 6= 1.

b) When b 6= 0,

D(a,b)f(0, 0) = limt→0

f(ta, tb)− f(0, 0)t

= limt→0

ta√t2a2 + t2b2

t|tb|= a

√a2

b2+ 1.

Also,

D(a,0)f(0, 0) = limt→0

f(ta, 0)− f(0, 0)t

= limt→0

0− 0

t= 0.

Thus, f has directional derivatives in every direction at (0, 0).

c) We show such p exists by solving the equations

a

√a2

b2+ 1 = c, a2 + b2 = 1 . . . a =

c√1 + c2

, b =1√

1 + c2.

This does not contradict the corollary since f is not continuously di�erentiable. (See alsoProblem 33 below.)

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29. Suppose f : Rn → R is continuously di�erentiable and let K = {x ∈ Rn : ‖x‖ = 1}.Show there is a point x ∈ K at which f |K attains its smallest value. Now supposewhenever p ∈ Rn is a unit vector, 〈∇f(p), p〉 > 0. Show that then ‖x‖ < 1.

Solution. K is sequentially compact and f is continuous, so f |K attains a smallest valueby the extreme value theorem (Theorem 11.22). Let p ∈ Rn with ‖p‖ = 1. Then〈∇f(p), p〉 > 0 and

0 = limt→0

f(p− tp)− f(p)− 〈∇f(p),−tp〉t

= limt→0

f(p− tp)− f(p)t

+ 〈∇f(p), p〉.

This shows that for t > 0 su�ciently close to zero, f(p − tp) − f(p) < 0. Thus, theminimum of f |K cannot be attained at p. For an ε-δ proof of this, let ε = 〈∇f(p), p〉 andpick δ > 0 such that |t−1[f(p− tp)− f(p)] + 〈∇f(p), p〉| < ε whenever 0 < |t| < δ. Thenf(p− tp)− f(p) < 0 whenever 0 < t < δ; why?

30. Prove that

lim(x,y)→(0,0)

sin(2x+ 2y)− 2x− 2y√x2 + y2

= 0.

Solution. Let f(x, y) = sin(2x + 2y). Then D1f(x, y) = 2 cos(2x + 2y) = D2f(x, y) arecontinuous. Thus, f is continuously di�erentiable, so since D1f(0, 0) = D2f(0, 0) = 2,by the �rst order approximation theorem,

lim(x,y)→(0,0)

sin(2x+ 2y)− 2x− 2y

‖(x, y)‖= 0.

31. Suppose f : R2 → R is continuous and a, b ∈ R. Prove

lim(x,y)→(0,0)

[f(x, y)− (f(0, 0) + ax+ by)] = 0. (6)

Is it true that

lim(x,y)→(0,0)

f(x, y)− [f(0, 0) + ax+ by]√x2 + y2

= 0? (7)

Solution. Since 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 di�erentiableat (0, 0) with D1f(0, 0) = a, D2f(0, 0) = b. (See the �rst order approximation theorem.)

32. De�ne f : R2 → R by

f(x, y) =

{√x2 + y2 sin y2

x, if x 6= 0,

0, if x = 0.

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Show f is continuous at (0, 0) and has directional derivatives in every direction at (0, 0),but f is not di�erentiable at (0, 0) � that is, there is no tangent plane to the graph of fat (0, 0).

Solution. Notice |f(x, y) − f(0, 0)| =√x2 + y2| sin y2

x2| ≤

√x2 + y2 → 0 as (x, y) → 0.

Thus, f is continuous at (0, 0). Since t 7→ sin(tc) is continuous for c ∈ R, when a 6= 0,

D(a,b)f(0, 0) = limt→0

f(ta, tb)− f(0, 0)t

= limt→0

√t2a2 + t2b2 sin t2b2

ta

t= lim

t→0

√a2 + b2

∣∣∣∣sin tb2a∣∣∣∣ = 0.

Also, D(0,b)f(0, 0) = limt→0f(0,tb)−f(0,0)

t= limt→0

0−0t

= 0. Thus, f has directional deriva-tives in every direction at (0, 0). Note that D1f(0, 0) = D2f(0, 0) = 0 but the limit

lim(x,y)→(0,0)

f(x, y)− f(0, 0)− 0x− 0y

‖(x, y)‖= lim

(x,y)→(0,0)sin

y2

x

does not exist (e.g., take x ≡ y3 and let y → 0). Thus, f is not di�erentiable at (0, 0).

33. Suppose f : Rn → R is homogeneous, that is, f(0) = 0 and f(tx) = tf(x) for allt ∈ R and x ∈ Rn. Prove that if f is di�erentiable, then f(x) = 〈∇f(0), x〉, in particular,f is linear2. Thus, if f is homogeneous and not linear, it cannot be di�erentiable.

Solution. Let f be di�erentiable and homogeneous, and u ∈ Rn a unit vector. Then

0 = limt→0+

f(tu)− f(0)− 〈∇f(0), tu〉‖tu‖

= limt→0+

tf(u)− t〈∇f(0), u〉t‖u‖

= f(u)− 〈∇f(0), u〉.

This shows that f(u) = 〈∇f(0), u〉 and thus for t ∈ R,

f(tu) = tf(u) = t〈∇f(0), u〉 = 〈∇f(0), tu〉.

Any nonzero x ∈ Rn is a scalar times a unit vector, x = ‖x‖ x‖x‖ . Thus, we are done.

34. Let f : Rn → R be k times continuously di�erentiable, let h ∈ Rn and de�neφ(t) = f(x+ th) for t ∈ R. Prove that for all s ∈ R,

φ(k)(s) = Dkhf(x+ sh), Dk

hf = DhDh . . . Dh︸ ︷︷ ︸k times

f.

Write formulas for φ(k)(s) = Dkhf(x+ sh) when k = 1 and k = 2.

2f is linear i� it is homogeneous and additive: f(x+ y) = f(x) + f(y) for all x, y ∈ Rn.

11

Solution. For k = 1 this follows from the computation

φ′(s) = limt→0

φ(s+ t)− φ(s)t

= limt→0

f(x+ (s+ t)h)− f(x+ sh)

t

= limt→0

f((x+ sh) + th)− f(x+ sh)

t= Dhf(x+ sh).

The general case follows from induction: if φ(k−1)(s) = Dk−1h f(x+ sh) for all s ∈ R, then

φ(k)(s) = limt→0

φ(k−1)(s+ t)− φ(k−1)(s)

t= lim

t→0

Dk−1h f(x+ (s+ t)h)−Dk−1

h f(x+ sh)

t

= limt→0

Dk−1h f((x+ sh) + th)−Dk−1

h f(x+ sh)

t= DhD

k−1h f(x+ sh) = Dk

hf(x+ sh).

When k = 1,

φ′(s) = Dhf(x+ sh) =n∑i=1

hiDif(x+ sh) = 〈∇f(x+ sh), h〉.

When k = 2,

φ′′(s) = D2hf(x+ sh) = Dh

(n∑j=1

hjDjf(x+ sh)

)

=n∑i=1

hiDi

(n∑j=1

hjDjf(x+ sh)

)

=n∑i=1

n∑j=1

hihjDiDjf(x+ sh)

= 〈∇2f(x+ sh)h, h〉,where the last step uses the fact that 〈Ah, h〉 =

∑ni=1

∑nj=1 hihjaij when A is n× n.

35. De�ne f : R2 → R by f(x, y) = exy + x2 + 2xy. De�ne φ(t) = f(2t, 3t) for t ∈ R andcompute φ′′(0) in the following two ways:

i) By using single-variable theory, i.e., di�erentiating the single-variable function φ;

ii) By thinking of φ′′(0) as a directional derivative of f in the direction of h = (2, 3).

Solution. i) Note that φ(t) = e6t2+ 16t2, so direct computation gives φ′′(0) = 44.

ii) By3 Problem 34, φ′′(0) = 〈∇2f(0, 0)h, h〉 where h = (2, 3). We compute

∇2f(0, 0) =

(2 33 0

)3or formula (14.11) in the text

12

and 〈∇2f(0, 0)h, h〉 = 〈(13, 6), (2, 3)〉 = 44.

36. Let f : R2 → R be twice continuously di�erentiable. Suppose that ∇f(0, 0) = (0, 0)and that for some h ∈ R2, 〈∇2f(0, 0)h, h〉 > 0. Using single variable theory, prove thereis r > 0 such that

f(th) > f(0, 0), 0 < |t| < r.

Solution. De�ne φ(t) = f(th). By Taylor's theorem in 1D,

φ(t) = φ(0) + φ′(0)t+1

2φ′′(0)t2 + o(t2)

where limt→0 o(t2)/t2 = 0. By Problem 34, this can be rewritten

f(th) = f(0, 0) + 〈∇f(0, 0), h〉t+ 1

2〈∇2f(0, 0)h, h〉t2 + o(t2).

Since ∇f(0, 0) = (0, 0), 〈∇f(0, 0), h〉 = 0 and we can rewrite again to get

f(th)− f(0, 0)t2

=1

2〈∇2f(0, 0)h, h〉+ o(t2)

t2.

Since limt→0 o(t2)/t2 = 0 and 〈∇2f(0, 0)h, h〉 > 0, the RHS of this equation is positive

for su�ciently small t. Thus, f(th) > f(0, 0) for su�ciently small t. More precisely, letε = 1

2〈∇2f(0, 0)h, h〉 and choose δ > 0 such that |o(t2)/t2| < ε whenever 0 < |t| < δ.

Then f(th) > f(0, 0) for 0 < |t| < δ, so we can take r = δ.

37. In Problem 36, suppose additionally that 〈∇2f(0, 0)h, h〉 > 0 for all nonzero h ∈ Rn.Is this enough to directly conclude the origin is a local minimum of f?

Solution. We would have to show that f(th) > f(0, 0) for all 0 < |t| < r, where r doesnot depend on h. The trouble with using Problem 36 here is that we had r = rh thatdepends on h. To get an r which is independent of h, we could try r = infh rh, but thenwe might get r = 0.

Note that in this situation the origin is indeed a local minimizer � see for instanceTheorem 14.22. But this conclusion cannot be reached directly from Problem 36.

38. Let a, b, c ∈ R with a 6= 0, and de�ne p(t) = at2 + 2bt + c. Show that p(t) > 0 forall t if and only if a > 0 and ac− b2 > 0. Show also that p(t) < 0 for all t if and only ifa < 0 and ac− b2 > 0.

13

Solution. The quadratic equation shows that p(t) = 0 when

t =−b±

√b2 − aca

.

Suppose ac − b2 > 0. Then p(t) = 0 has no solutions. So by the intermediate valuetheorem, either p(t) > 0 for all t or p(t) < 0 for all t. To �nish the proof, notice thatlimt→∞ p(t)/t

2 = a. If a > 0 then p(t) > 0 for su�ciently large t, hence for all t; while ifa < 0 then p(t) < 0 for su�ciently large t, hence for all t.

39. Find 2 × 2 matrices associated with the quadratic functions h(x, y) = x2 − y2 andg(x, y) = x2 + 8xy + y2.

Solution. They are

A =

(1 00 −1

), B =

(1 44 1

).

With z = (x, y) we have h(z) = 〈Az, z〉 and g(z) = 〈Bz, z〉.

40. De�ne Q : R→ R by Q(x) = x4. Show that there is no c > 0 such that Q(x) ≥ cx2

for all x 6= 0. Explain why this does not contradict Proposition 14.16.

Solution. For any c > 0, if x2 < c then Q(x) = x4 = x2x2 < cx2. This does not contradictProposition 14.16 because Q is not a quadratic function: it cannot be written in the formQ(x) = 〈Ax, x〉 for a matrix A. (Such a matrix A would have to 1 × 1, and so 〈Ax, x〉would be a scalar constant times x2.)

41. Show that the point (−1, 1) is the minimizer of the function f : R2 → R de�ned 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) and

∇2f(x, y) ≡(12 1818 28

).

In particular, ∇f(x, y) = 0⇔ (x, y) = (−1, 1), and since 12 > 0, 12·28−18·18 = 12 > 0,∇2f(x, y) is positive de�nite for all (x, y). Thus, f has a local minimum at (−1, 1). Weargue this must be a global minimum. Suppose to the contrary that f(r, s) < f(−1, 1).Let L(t) = (−1, 1)+ t(r, s) and de�ne g = f ◦L. Since ∇f(−1, 1) = (0, 0) and ∇2f(x, y)is always positive de�nite, 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, 0 is a global minimum for g. Butg(1) = f(r, s) < f(−1, 1) = g(0), contradiction.

42. Suppose f : Rn → R is twice continuously di�erentiable. Let x ∈ Rn be such that∇f(x) = 0. Suppose there are u, v ∈ Rn such that

〈∇2f(x)u, u〉 > 0, 〈∇2f(x)v, v〉 < 0.

Prove that x is neither a local maximum nor local minimum of f .

Solution. By the second order approximation theorem4

f(x+ tu)−f(x) = 〈∇f(x), tu〉+ 1

2〈∇2f(x)tu, tu〉+ o((tu)2) = t2

2〈∇2f(x)u, u〉+ o((tu)2).

where limt→0 o((tu)2)/‖tu‖2 = 0. Rearranging, we �nd that

f(x+ tu)− f(x)t2

=1

2〈∇2f(x)u, u〉+ o((tu)2)

t2

where limt→0 o((tu)2)/t2 = 0. We can conclude that the LHS above is positive for su�-

ciently small t. An analogous argument shows the LHS of

f(x+ tv)− f(x)t2

=1

2〈∇2f(x)v, v〉+ o((tv)2)

t2

is negative for su�ciently small t. (We will not give ε-δ delta proofs here; see howeversimilar proofs in Problems 29, 36 and 43 below.) Thus, f has a �saddle point� at x.

43. Suppose f : Rn → R is twice continuously di�erentiable, ∇f(x) = 0, and ∇2f(x) ispositive de�nite. Prove that there exist c > 0 and δ > 0 such that

f(x+ h)− f(x) ≥ c‖h‖2 if ‖h‖ < δ.

Solution. By the second order approximation theorem,

f(x+ h)− f(x) = 〈∇f(x), h〉+ 1

2〈∇2f(x)h, h〉+ o(h2) =

1

2〈∇2f(x)h, h〉+ o(h2).

Thus,f(x+ h)− f(x)

‖h‖2=

1

2

⟨∇2f(x)

h

‖h‖,h

‖h‖

⟩+o(h2)

‖h‖2.

4Remark on notation: we write o(hp) to represent a function of h which, when divided by ‖h‖p,approaches 0 as h→ 0. (This is a slight abuse of notation, since if h is a vector hp is not de�ned.)

15

Let

c = minu∈Rn : ‖u‖=1

1

4

⟨∇2f(x)u, u

⟩.

Notice the minimum above is attained at some u on the unit sphere {u ∈ Rn : ‖u‖ = 1},since the unit sphere is sequentially compact. Since ∇2f(x) is positive de�nite, thismeans that c > 0. Now choose δ > 0 such that |o(h2)/‖h‖2| < c whenever 0 < ‖h‖ < δ.Then if 0 < ‖h‖ < δ, we have

f(x+ h)− f(x)‖h‖2

=1

2

⟨∇2f(x)

h

‖h‖,h

‖h‖

⟩+o(h2)

‖h‖2≥ 2c− c = c.

44. Suppose f : Rn → R is twice continuously di�erentiable. Suppose that ∇f(x) = 0and ∇2f(x) is the matrix of all zeros. Show that x could be a local maximum, a localminimum, or neither.

Solution. Let n = 2 and consider the functions f(x, y) = −x4 − y4, f(x, y) = x4 − y4and f(x, y) = x4 + y4; these functions all satisfy the assumptions above and they have,respectively, a local maximum, saddle point, and local minimum at (0, 0).

45. Which of the following functions f : R2 → R2 are linear?

i) f(x, y) = (−y, ex)ii) f(x, y) = (x− y2, 2y)iii) f(x, y) = 17(x, y)

Solution. In i), note that 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 iniii) is linear with matrix (

17 00 17

).

46. Show there is no linear mapping T : R2 → R2 having the property that T (1, 1) = (4, 0)and 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

47. For a point (x, y) in the plane R2, de�ne T (x, y) to be the point on the line ` ={(x, y) ∈ R2 : y = 2x} that is closest to (x, y). Show that T : R2 → R2 is linear and �ndits 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 f ′(r) = 10r− 2x− 4y = 0 if and only if r = x/5 + 2y/5. Since f ′′(r) = 10 > 0 forall r, this is the unique global minimizer. Thus, T is linear with matrix(

1/5 2/52/5 4/5

).

48. Suppose A is an m× n matrix. De�ne f : Rn → Rm 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,

limh→0

f(x+ h)− f(x)− Ah‖h‖

= limh→0

A(x+ h)− Ax− Ah‖h‖

= limh→0

0

‖h‖= 0,

we must have Df(x) = A for every x ∈ Rn.

49. Give a proof of the �rst order approximation theorem based on the mean valuetheorem.

Solution. Suppose f : Rn → Rm is continuously di�erentiable with component functionsf1, . . . , fm. By the mean value theorem,

fi(x+ h)− fi(x) = 〈∇fi(z(i)), h〉,

where z(i) is on the line segment between x and x + h. Let x be �xed and let B(h) bethe matrix whose ith row is ∇fi(z(i)). Combining the component equations above,

f(x+ h)− f(x) = B(h)h.

Thus,

limh→0

f(x+ h)− f(x)−Df(x)h‖h‖

= limh→0

[f(x+ h)− f(x)−B(h)h

‖h‖+B(h)h−Df(x)h

‖h‖

]= lim

h→0

B(h)h−Df(x)h‖h‖

= limh→0

[B(h)−Df(x)] h‖h‖

.

17

Note limh→0B(h) − Df(x) = 0 by continuity of partial derivatives of f . Moreover,h/‖h‖ is bounded. Thus, the limit above is 0. To see why, note that the generalizedCauchy-Schwartz inequality (Theorem 14.13) yields

limh→0

‖[B(h)−Df(x)]h‖‖h‖

≤ limh→0‖B(h)−Df(x)‖ = 0.

50. Suppose f : R2 → R2 is continuously di�erentiable with f(x, y) = (ψ(x, y), φ(x, y)).De�ne g : R2 → R2 by g(x, y) = 1

2[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 → Rand Df(x0, y0) is invertible, then f(x0, y0) = 0.

Solution. Let h(x, y) = 12[x2 + y2]. Then g(x, y) = (h ◦ f)(x, y). By the chain rule,

Dg(x, y) = Dh(f(x, y))Df(x, y) =(ψ(x, y) φ(x, y)

)(D1ψ(x, y) D2ψ(x, y)D1φ(x, y) D2φ(x, y)

)= f(x, y)Df(x, y).

The result follows when x = x0, y = y0 by multiplying both sides above on the right byDf(x0, y0)

−1 and using the fact that Dg(x0, y0) = 0 since (x0, y0) is a minimizer of g.

51. Suppose ψ : R2 → R is continuously di�erentiable and de�ne g : R2 → R by

g(s, t) = ψ(s2t, s).

Find ∂g/∂s(s, t) and ∂g/∂t(s, t).

Solution. By the chain rule,

∂g

∂s=∂g

∂x

∂x

∂s+∂g

∂y

∂y

∂s,

∂g

∂t=∂g

∂x

∂x

∂t+∂g

∂y

∂y

∂t.

That is,

∂g

∂s(s, t) = D1ψ(s

2t, s)2st+D2ψ(s2t, s),

∂g

∂t(s, t) = D1ψ(s

2t, s)s2.

52. Suppose g, h : R2 have continuous second derivatives and de�ne

u(s, t) = g(s− t) + h(s+ t).

18

Prove that∂2u

∂t2(s, t)− ∂2u

∂s2(s, t) = 0

for all (s, t) ∈ R2.

Solution. By the chain rule,

∂u(s, t)

∂s= g′(s− t) + h′(s+ t),

∂u(s, t)

∂t= −g′(s− t) + h′(s+ t)

and

∂2u

∂s(s, t) = g′′(s− t) + h′′(s+ t),

∂2u

∂s(s, t) = −(−g′′(s− t)) + h′′(s+ t).

The result follows.

53. Let U, V ⊂ R be open, suppose f : U → V is di�erentiable and bijective, and leta ∈ U be such that f ′(a) = 0. Prove that f−1 : V → U is not di�erentiable at b = f(a).

Solution. Suppose g := f−1 is di�erentiable at b. Then by the chain rule,

(g ◦ f)′(a) = g′(b)f ′(a) = 0.

But (g ◦ f)(x) = x for all x ∈ U , so that (g ◦ f)′(a) = 1, contradiction.

54. For the following mappings f, g : R2 → R2, apply the inverse function theorem atthe point (0, 0) and calculate the partial derivatives of the components of the inversemapping at the points f(0, 0) and g(0, 0).

a) f(x, y) = (x+ x2 + ex2y2 ,−x+ y + sin(xy)), b) g(x, y) = (ex+y, ex−y).

Solution. Since

Df(0, 0) =

(1 0−1 1

), Dg(0, 0) =

(1 11 −1

)and f(0, 0) = (1, 0) and g(0, 0) = (1, 1), by the inverse function theorem

Df−1(1, 0) =

(1 0−1 1

)−1=

(1 01 1

), Dg−1(1, 1) =

(1 11 −1

)−1=

(12

12

12−1

2

).

The �rst (resp. second) row of Df−1(1, 0) contains the partial derivatives of the �rst(resp. second) component of f−1 at the point (1, 0), and similarly for g−1.

19

55. De�ne f : R2 → R2 by f(x, y) = (ex cos y, ex sin y). Show that the inverse functiontheorem applies at every point (x, y) ∈ R2, but that f is not one-to-one. Does the latterstatement contradict the former?

Solution. Observe that

Df(x, y) =

(ex cos y −ex sin yex sin y ex cos y

)and so detDf(x, y) = ex 6= 0 for all (x, y) ∈ R2. Thus, the inverse function applies atall (x, y) ∈ R2. However, f is not one-to-one since f(x, y) = f(x, y + 2π). There is nocontradiction here: a function can be locally invertible at every point without having aglobal inverse.

56. For a pair of real numbers a, b, consider the system of nonlinear equations

x+ x2 cos y + xyex3y2 = a,

y + x5 + y3 − x2 cos(xy) = b.(8)

Use the inverse function theorem to show that there is some positive number r such thatif a2 + b2 < r2, then this system of equations has at least one solution.

Solution. Let f(x, y) = (x+ x2 cos y + xyex3y2 , y + x5 + y3 − x2 cos(xy)). Then

Df(x, y) =

(1 + 2x cos y + yex

3y2 + 3x3yex3y2 −x2 sin y + xex

3y2 + 2xy2ex3y2

5x4 − 2x cos(xy) + x2y sin(xy) 1 + 3y2 + x3 sin(xy)

)and so

Df(x, y) =

(1 00 1

)which is nonsingular. So by the inverse function theorem, there are neighborhoods U of(0, 0) and V of f(0, 0) = (0, 0) such that 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.

57. Suppose ψ : R2 → R is continuously di�erentiable and de�ne f : R2 → R2 by

f(x, y) = (ψ(x, y),−ψ(x, y)).

Explain both geometrically and analytically why the hypotheses of the inverse functiontheorem must fail for f at every point (x, y) ∈ R2.

Solution. The analytic explanation is that

detDf(x, y) =

∣∣∣∣ D1ψ(x, y) D2ψ(x, y)−D1ψ(x, y) −D2ψ(x, y)

∣∣∣∣ = 0

20

for every (x, y). The geometric explanation is that Im f is the line y = x, which isone-dimensional, and an injective continuous function de�ned on the plane cannot havea 1-dimensional image.

58. De�ne f(x) = xn for x ∈ R. Prove that f is bijective if n ≥ 1 is odd. Then showthat f is not stable if n > 1.

Solution. Let n ≥ 1 be odd. To see that f is onto, let y ∈ R. By the intermediate valuetheorem 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 valuetheorem there is c between a and b such that f ′(c) = ncn−1 = 0, a contradiction sincec > min{a, b} > 0. Similarly, f(a) = f(b) < 0 only if a = b. The other possibility is thatf(a) = f(b) = 0, but f(x) = xn = 0 only when x = 0. To see that f is not stable whenn > 1, we use the di�erence of powers formula

f(x)− f(y) = xn − yn = (x− y)n−1∑i=0

xn−1−iyi.

Suppose f were stable with constant c. If y = 2x where 0 < x < n−1√c/(2n), then∣∣∣∣f(x)− f(y)x− y

∣∣∣∣ =∣∣∣∣∣n−1∑i=0

xn−1−iyi

∣∣∣∣∣ = 2nxn−1 < c,

contradiction.

59. Let f : (a, b) → R be di�erentiable. Prove that f is stable with stability constant cif and only if |f ′(x)| ≥ c for all x ∈ (a, b).

Solution. Suppose that |f ′(x)| ≥ c for all x ∈ (a, b). If there were x, y ∈ (a, b) 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 ′(r)| = |f(x) − f(y)|/|x − y| < c, contradiction. Thus, f must be stable. Conversely,suppose f is stable with constant c. Then |f(x) − f(y)| ≥ c|x − y| for all x, y ∈ (a, b)and so

|f ′(y)| =∣∣∣∣limx→y f(x)− f(y)x− y

∣∣∣∣ ≥ c.

60. De�ne g(x, y) = (x2, y) for (x, y) ∈ R2. Show there is no neighborhood of (0, 0) suchthat g : R2 → R2 is stable.

21

Suppose g were stable with stability constant c. Then for x ∈ R such that 0 < |x| < c,

‖g(x, 0)− g(0, 0)‖‖(x, 0)− (0, 0)‖

=x2

|x|= |x| < c,

contradiction.

61. Is the sum of stable mappings also a stable mapping?

Solution. No. For example, f(x) = x and g(x) = −x, de�ned for x ∈ R, are both stable,but f + g ≡ 0 is not.

62. Let U ⊂ Rn be open and suppose f : U → Rn is continuously di�erentiable andstable. Prove for each x ∈ U , Df(x) is nonsingular.

Solution. Let x ∈ U . By the �rst order approximation theorem,

f(x+ h)− f(x) = Df(x)h+ o(h).

Suppose f has stability constant c. Then

‖Df(x)h‖ ≥ ‖Df(x)h+ o(h)‖ − ‖o(h)‖= ‖f(x+ h)− f(x)‖ − ‖o(h)‖ ≥ c‖h‖ − ‖o(h)‖.

Thus, ∥∥∥∥Df(x) h

‖h‖

∥∥∥∥ ≥ c− ‖o(h)‖‖h‖

→ c as h→ 0.

This shows that Df(x)u 6= 0 for all unit vectors u. It follows that Df(x) is nonsingular.

63. For a point (ρ, θ, φ) ∈ R3, de�ne

f(ρ, θ, φ) = (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ).

At what points does the inverse function apply to f?

Solution. Since

detDf(ρ, θ, φ) =

∣∣∣∣∣∣sinφ cos θ −ρ sinφ sin θ ρ cosφ cos θsinφ sin θ ρ sinφ cos θ ρ cosφ sin θcosφ 0 −ρ sinφ

∣∣∣∣∣∣= cosφ

∣∣∣∣−ρ sinφ sin θ ρ cosφ cos θρ sinφ cos θ ρ cosφ sin θ

∣∣∣∣− ρ sinφ ∣∣∣∣sinφ cos θ −ρ sinφ sin θsinφ sin θ ρ sinφ cos θ

∣∣∣∣= −ρ2 sinφ cos2 φ− ρ2 sin3 φ = −ρ2 sinφ,

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the inverse function theorem applies when ρ 6= 0 and φ is not an integer multiple of π.

64. De�ne f(x) = x2 for x ∈ R.a) Show that if U ⊂ R is open and 0 /∈ U , then f(U) is open.b) Show that if U ⊂ R is open and 0 ∈ U , then f(U) is not open.

Solution. a) Any open set U ⊂ R is a union of open intervals. Thus, we only have to showthat 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)).

b) Let U ⊂ R with 0 ∈ U . Then 0 ∈ f(U) but for every ε > 0, (−ε, ε) 6⊂ f(U) since theimage of f contains no negative numbers. Thus, f(U) is not open.

65. Give an example of a continuously di�erentiable mapping f : Rn → Rn with theproperty that there is no open subset U ⊂ Rn such that f(U) is open.

Solution. Let f(x) ≡ 0. f is continuously di�erentiable (since it is linear), but for everyopen U ⊂ Rn, f(U) = {0} is not open.

66. Consider the equation x2 + y2 = 0 for (x, y) ∈ R2. Show that the assumptions ofDini's Theorem do not hold at the point (0, 0). Explain geometrically why the conclusionof Dini's theorem does not hold at (0, 0).

Solution. Let f(x, y) = x2 + y2. Since D1f(0, 0) = D2f(0, 0) = 0, the assumptions ofDini's theorem do not hold at (0, 0). Note that the solution set for x2 + y2 = 0 consistsonly of the point (0, 0). Thus, y cannot be written as a function of x (or vice-versa) onan open set containing 0.

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 (0, 0) implicitlyde�ne x as a function of y or vice-versa? If so, compute the derivative of this function(s)at the point 0.

Solution. Let 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).

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Thus, D1f(0, 0) = 2 6= 0 and D2f(0, 0) = −3 6= 0. So the set of solutions de�nes x asa function of y, and vice-versa, in a neighborhood of (0, 0). Writing ψ and φ for thesefunctions, respectively, we have

D1f(ψ(x), x)ψ′(x) +D2f(ψ(x), x) = 0,

D1f(x, φ(x)) +D2f(x, φ(x))φ′(x) = 0,

for x in a neighborhood of 0. Thus,

ψ′(0) = −D2f(0, 0)

D1f(0, 0)=

3

2, φ′(0) = −D1f(0, 0)

D2f(0, 0)=

2

3.

68. Let O ⊂ R2 be open and f : O → R continuously di�erentiable. Suppose thatf(a, b) = 0 and ∇f(a, b) 6= (0, 0). Show that ∇f(a, b) is orthogonal to the tangent lineat (a, b) of the implicitly de�ned function.

Solution. Since∇f(a, b) 6= (0, 0), we can assume without loss of generality thatD2f(a, b) 6=0. So by Dini's theorem, in a neighborhood of (a, b) the equation f(x, y) = 0 implicitlyde�nes y = φ(x) and moreover

D1f(x, φ(x)) +D2f(x, φ(x))φ′(x) = 0.

Setting x = a, this rewrites as

〈∇f(a, b), (1, φ′(a))〉 = 0.

Observe that (1, φ′(a)) is the tangent line to the implicitly de�ned function at x = a,since in a neighborhood of (a, b) the graph of that function is the curve x 7→ (x, φ(x)).

69. Suppose the function φ : R → R is continuously di�erentiable and that φ′(a) 6= 0.Set b = φ(a) and de�ne f : R2 → R by f(x, y) = y − φ(x) for (x, y) ∈ R2. Apply Dini'stheorem to the function f : R2 → R at the point (a, b) and compare the result with theconclusion of the Inverse Function Theorem applied to φ at a.

Solution. Since D1f(a, b) = −φ′(a) 6= 0, by Dini's theorem, there is a continuouslydi�erentiable function ψ de�ned on a neighborhood of b such that x = ψ(y) uniquelysolves the equation f(x, y) = y − φ(x) = 0 in a neighborhood of (a, b). Thus, for y in aneighborhood of b, ψ(y) = φ−1(y) and

0 = D1f(ψ(y), y)ψ′(y) +D2f(ψ(y), y) = −φ′(φ−1(y))(φ−1)′(y) + 1. (9)

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Since φ′(a) 6= 0, the Inverse Function Theorem shows that in a neighborhood of b, φ hasa continuously di�erentiable inverse and for y in this neighborhood,

(φ−1)′(y) = φ′(φ−1(y))−1. (10)

Observe that equations (9) and (10) are simply rearrangements of one another.

70. Suppose f : R2 → R is continuously di�erentiable and c > 0 is such that

D2f(x, y) ≥ c for all (x, y) ∈ R2. (11)

Prove there is a continuously di�erentiable function g : R→ R such that f(x, g(x)) = 0for all x ∈ R and if f(x, y) = 0, then y = g(x).

Solution. Fix x and de�ne φ : R → R by φ(y) = f(x, y). The assumption (11) impliesφ′(y) ≥ c > 0. Thus, φ has a unique root (why?), call it g(x). That is,

φ(g(x)) = f(x, g(x)) = 0.

Uniqueness of the root (for arbitrary x) shows that if f(x, y) = 0 then y = g(x). To seethat g is continuously di�erentiable at an arbitrary point a ∈ R, note that by (11), Dini'stheorem holds at the point (a, g(a)). Since g is unique, it must agree with the implicitfunction furnished by Dini's theorem in a neighborhood of a. Since the latter function iscontinuously di�erentiable at a, so is g.

71. Consider the linear system of equations

a11x+ a12y + a13z = 0,

a21x+ a22y + a23z = 0, (x, y, z) ∈ R3.(12)

De�ne ν = (a11, a12, a13) and β = (a21, a22, a23). Show that if ν × β 6= (0, 0, 0), thenthe equations (12) de�nes two of the variables as a function of the remaining variable.Interpret this geometrically.

Solution. De�ne

f(x, y, z) = (a11x+ a12y + a13z, a21x+ a22y + a23z)

and observe that

ν × β =

∣∣∣∣a12 a13a22 a23

∣∣∣∣ (1, 0, 0)− ∣∣∣∣a11 a13a12 a23

∣∣∣∣ (0, 1, 0) + ∣∣∣∣a12 a12a21 a22

∣∣∣∣ (0, 0, 1)= |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).

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Since ν×β 6= (0, 0, 0), one of the determinants above must be nonzero. Thus, the ImplicitFunction Theorem shows that (12) implicitly de�nes two of the variables as a functionof the third. Geometrically, each equation of (12) represents a plane in R3, and theassumption ν × β 6= (0, 0, 0) guarantees that these two planes are not parallel. Thus, theplanes intersect in a line, which is why we can solve for two of the variables as a functionof the third.

72. Consider the solutions of the equation y3 − x2 = 0 for (x, y) ∈ R2. Does theImplicit Function Theorem apply at the point (0, 0)? Does this equation de�ne one ofthe components of a solution (x, y) as a function of the other component?

Solution. The hypotheses of the Implicit Function Theorem do not hold at (0, 0): lettingf(x, y) = y3− x2, we have D1f(0, 0) = D2f(0, 0) = 0. However, the equation f(x, y) = 0does de�ne y as a function of x, namely y = x2/3. (Note, however, that y is not acontinuously di�erentiable function of x.)

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