HW #1 SOLUTIONS
1. Let f : [a, b] → R and let v : I → R, where I is an interval containing f ([a, b]). Suppose that f is continuous at x ∈ [a, b]. Suppose that lims→f (x) v(s) = v(f (x)). Using the − δ definition of limit, prove that
limt→x v(f (t)) = v(f (x)).
Solution.
Take an > 0. Since lims→f (x) v(s) = v(f (x)), there exists 1 > 0 such that if s ∈ I and |s − f (x)| < 1
then |v(s) − v(f (x))| < . Now, f is continuous at x, so there exists δ > 0 such that if t ∈ [a, b] is such
that |t − x| < δ then |f (t) − f (x)| < 1 . Since I contains f ([a, b]), we automatically have f (t) ∈ I, and also
|f (t) − f (x)| < 1 from above, so |v(f (t)) − v(f (x))| < as desired. This shows that limt→x v(f (t)) = v(f (x)),
i.e. that the composition v ◦ f is continuous at x.
2. Let g : I → R be a function, where I is an interval containing 0 in its interior. Suppose limx→0
Define
g(x) sin x1 if x 6= 0,
f (x) =
0
if x = 0
g(x)
x
= 0.
Prove that f 0 (0) = 0.
Solution.
By definition, f 0 (0) = limx→0
f (x)−f (0)
x−0
= limx→0
g(x) sin
x
1
x
g(x) g(x) sin x1 ≤
. We have the inequalities 0 ≤ x
x g(x)
1
for x 6= 0, since | sin
x→0 x1 = 0, we
y| ≤ 1 for all y ∈ R, in particular for y = x . Since we are given that lim
g(x) sin x also have limx→0 g(x)
= 0, or
x = 0. Of course, limx→0 0 = 0, so by the Squeeze Theorem, limx→0 x
limx→0
g(x) sin
x
1
x
= 0.
Remark. We’re using the easy-to-check fact that for a function h : (a, b) → R and t ∈ (a, b), limx→t h(x) = 0 if and
only if limx→t |h(x)| = 0. To see this, we need only unravel the definition of limit in this context: limx→t h(x) = 0
says that for all > 0, there exists δ > 0 such that if |x − t| < δ then |h(x) − 0| < , whereas limx→t |h(x)| = 0
says that for all > 0, there exists δ > 0 such that if |x − t| < δ then ||h(x)| − 0| < . Note however that
||h(x)| − 0| = |h(x)| = |h(x) − 0| so the two statements are saying the same thing.
3. Let f, g : R → R be differentiable functions satisfying f 0 (x) > g 0 (x) for all x ∈ R and f (0) = g(0). Prove that
f (x) > g(x) for x > 0 and f (x) < g(x) for x < 0.
Solution.
Let h : R → R be defined by h(x) = f (x) − g(x) for x ∈ R. Then h(0) = f (0) − g(0) = 0 and h0 (x) =
1
2
HW #1 SOLUTIONS
f 0 (x) − g 0 (x) > 0 if x ∈ R . If x > 0, then by the Mean Value Theorem, h(x) = h(x) − h(0) = h0 (c)(x − 0) for
some c between 0 and x. But then h0 (c) = f 0 (c) − g 0 (c) > 0, so h(x) = h0 (c)x > 0.
If x < 0, then the same computation h(x) = h0 (c)x for some c between 0 and x shows that h(x) < 0 because h0 (c) > 0 and x < 0.
4. Rudin Ch 5. Exercise #7. Suppose f 0 (x), g 0 (x) exist, g 0 (x) 6= 0, and f (x) = g(x) = 0. Prove that
f (t)
f 0 (x)
= 0
.
t→x g(t)
g (x)
lim
This holds also for complex functions.
Solution.
f (t)
= lim
lim
t→x
t→x g(t)
f (t)−f (x)
t−x
g(t)−g(x)
t−x
, and we are given that limt→x
both exist, hence the ratio limt→x
f (t)
g(t)
=
f 0 (x)
g 0 (x)
f (t)−f (x)
t−x
= f 0 (x) and limt→x
g(t)−g(x)
t−x
= g 0 (x) 6= 0
(note that we need g 0 (x) = 0).
5. Rudin Ch 5. Exercise #13. Suppose a and c are real numbers, c > 0, and f is defined on [−1, 1] by
|x|a sin(|x|−c ) (if x 6= 0),
f (x) =
0
(if x = 0).
Prove the following statements:
(a) f is continuous if and only if a > 0.
(b) f 0 (0) exists if and only if a > 1.
(c) f 0 is bounded if and only if a ≥ 1 + c.
(d) f 0 is continuous if and only if a > 1 + c.
(e) f 00 (0) exists if and only if a > 2 + c.
(f) f 00 is bounded if and only if a ≥ 2 + 2c.
(g) f 00 is continuous if and only if a > 2 + 2c.
Solution.
1/c
1
(a) Suppose a ≤ 0. Then for each positive integer k, consider a real number xk = π +2πk
∈ (0, 1] ⊂ [−1, 1].
2
a/c
1
Then, f (xk ) = |xk |a sin(|xk |−c ) = π +2πk
sin( π2 + 2πk) = ( π2 + 2πk)−a/c . If a = 0, then limk→∞ f (xk ) = 1,
2
whereas if a < 0 then limk→∞ f (xk ) = ∞. In either case, we have limk→∞ f (xk ) 6= f (0), whereas limk→∞ xk = 0
(c > 0). Therefore f is not continuous at 0 (recall that a function f defined on a neighborhood of 0 is continuous
at 0 if and only if limk→∞ f (xk ) = f (0) for all sequences {xk }k such that limk→∞ xk = 0). In fact, we have shown
that f has neither the left nor the right limit at 0.
Suppose now that a > 0. Then if x 6= 0, we have |f (x)| = |x|a | sin(|x|−c ) ≤ |x|a , and for x = 0 of course
we have the same inequality (since a > 0, |0|a = 0 is defined also of course). Since limx→0 |x|a = 0, we have
limx→0 |f (x)| = 0 by the Squeeze Theorem.
HW #1 SOLUTIONS
(b) f 0 (0) = limx→0
f (x)−f (0)
x−0
=
|x|a sin(|x|−c )
x
3
if the limit exists.
1/c
1
If a ≤ 1, then limx→0+ xa−1 sin(|x|−c ) does not exist, as in part (a) by considering the sequence xk = π +2πk
.
2
a
−c )
Therefore we need a > 1. In this case, |x| sin(|x|
= |x|a−1 sin(|x|−c ) ≤ |x|a−1 and the latter tends to 0
x
as x → 0, so f 0 (0) exists by the Squeeze Theorem and f 0 (0) = 0 in fact in this case.
(c) Note first that f (x) = f (−x) if x ∈ [−1, 1], i.e. f is an even function. For x 6= 0, f is certainly differentiable at x since f is given by the expression xa sin(x−c ) if x > 0 on a neighborhood of x not containing 0 and
by (−x)a sin((−x)c ) on a neighborhood of x not containing 0 and both expressions are products of compositions
of differentiable functions, hence differentiable. We assume that a > 1 from part (b) to discuss the existence of
f 0 on [−1, 1] in the first place.
For x > 0, f 0 (x) = axa−1 sin(x−c ) + xa cos(x−c )(−cx−c−1 ) = axa−1 sin(x−c ) − cxa−c−1 cos(x−c ) (recall the
remark above that f is given by x 7→ xa sin(x−c ) in a neighborhood of x in [−1, 1] not containing 0). Also for
a > 1, f 0 (0) exists and f 0 (0) = 0.
For x < 0, f is again differentiable at x because f (x) = f (−x) expresses f as a composition of the differentiable function f |(0,1] (this is the restriction of f on (0, 1]) together with the differentiable function x 7→ −x on a
neighborhood of x not containing 0. Using the chain rule now, we have f 0 (x) = −f 0 (−x), so that

if x > 0
 axa−1 sin(x−c ) − cxa−c−1 cos(x−c )
0
0
if x = 0
f (x) =

c|x|a−c−1 cos(|x|−c ) − a|x|a−1 sin(|x|−c ) if x < 0
At any rate, we then need only analyze the boundedness of the expression axa−1 sin(x−c ) − cxa−c−1 cos(x−c )
on (0, 1] (for 0 < x ≤ 1). Since a > 1, the first term has absolute value |axa−1 sin(x−c )| ≤ a|x|a−1 , which tends to
0 as x → 0+ because a > 1. It is thus bounded. For the latter term, |cxa−c−1 cos(x−c )| ≤ cxa−c−1 , and this tends
1 1/c
∈ (0, 1],
to 0 as x → 0+ if a ≥ 1 + c. In the case a < 1 + c, we take the sequence {xk }k≥1 defined by xk = 2πk
1 (a−c−1)/c
which tends to 0 as k → ∞, and yet f (xk ) = −c 2πk
→ −∞ as k → ∞, showing that f is not bounded
on (0, 1] if a < 1 + c.
(d) We assume f 0 exists throughout [−1, 1], i.e. a > 1.
Assume first that a > 1 + c, then from f 0 (x) = −f 0 (−x) again, we need only show that limx→0+ f 0 (x) = f 0 (0) = 0
to check contiuity of f 0 (it is continuous on (0, 1]) because it is expressed on a neighborhood of a point in
(0, 1] by a combination of products, addition, and composition of differentiable functions). limx→0+ f 0 (x) =
limx→0+ (axa−1 sin(x−c ) − cxa−c−1 cos(x−c )), and the first term has limit limx→0+ axa−1 sin(x−c ) = 0 since a > 1
(Squeeze Theorem yet again here). For the latter term, note that limx→0+ −cxa−c−1 cos(x−c ) = 0 again by the
Squeeze Theorem because a > 1 + c.
Assume now that f 0 is continuous on [−1, 1]. Then f 0 ([−1, 1]) is compact hence bounded, so part (c) implies that
a ≥ 1 + c. We need only rule out the case a = 1 + c then to show that a > 1 + c. So suppose a = 1 + c. Then
for x > 0, f 0 (x) = axc sin(x−c ) − c cos(x−c ). Now take the sequence xk = (2πk)−1/c , which has the properly that
xk → 0+ as k → ∞, and f 0 (xk ) = −c for all k. That is, limk→∞ f 0 (xk ) = −c 6= 0 = f 0 (0), so f 0 is not continuous
at 0.
(e) We assume here that f 0 is continuous throughout [−1, 1], i.e. a > 1 + c by part (d).
f 00 (0) = limx→0
f 0 (x)−f 0 (0)
.
x−0
For x < 0,
f 0 (x)−f 0 (0)
x−0
=
−f 0 (−x)−f 0 (0)
x−0
=
f 0 (−x)
−x ,
and for x > 0,
f 0 (x)−f 0 (0)
x−0
=
4
HW #1 SOLUTIONS
f 0 (x)
x . Therefore
a−c−2
−c
0
f 00 (0) exists if and only if the right hand limit limx→0+ f x(x) = limx→0+ (axa−2 sin(x−c ) −
cx
cos(x )) exists and if this right hand limit exists then its limit is f 00 (0). Write g(x) = axa−2 sin(x−c ) −
a−c−2
cx
cos(x−c ) then.
Suppose a < 2 + c. Then take the sequence xk = (2πk)−1/c . Then xk → 0 as k → ∞ whereas g(xk ) =
−c(2πk)(c+2−a)/c , so that limk→∞ g(xk ) = −∞, so f 00 (0) does not exist.
Suppose a = 2 + c. Then take xk = (πk)−1/c , so g(xk ) = −c(−1)k , and limk→∞ g(xk ) does not exist.
Suppose a > 2 + c. Then the first term |axa−2 sin(x−c )| ≤ axa−2 and this tends to 0 as x → 0. For the
second term, |cxa−c−2 cos(x−c )| ≤ cxa−c−2 and again this tends to 0 as x → 0. So by the Squeeze Theorem,
f 00 (0) exists.
(f) Assume now that a > 2 + c. We know from the above computation in part (e) that f 00 (0) = 0. f 0 is expressed
by differentiable function on [−1, 0) ∪ (0, 1], so f 0 is differentiable on it. Therefore with the assumption thta
a > 2 + c, f 0 is differentiable on [−1, 1]. We have f 0 (x) = −f 0 (−x), so by the Chain Rule, f 00 (x) = f 00 (−x), i.e. f 00
d
(axa−1 sin(x−c ) − cxa−c−1 cos(x−c )) =
is an even function and it suffices to compute f 0 (x) for x > 0, i.e. dx
a−2
2 a−2c−2
−c
a−c−2
−c
(a(a − 1)x
−c x
) sin(x ) − c(−2a + c + 1)x
cos(x ). Note that both a(a − 1) > 0 and
(−2a + c + 1) > 0 from a > 2 + c and c > 0. All the terms appearing here except for xa−2c−2 sin(x−c )
tends to 0 as x → 0, so f 00 is bounded on [−1, 1] if and only if xa−2c−2 sin(x−c ) is bounded on (0, 1]. By the same
argument as in the previous parts, this happens exactly when a − 2c − 2 ≥ 0, or when a ≥ 2 + 2c.
(g) Suppose f 00 is continuous on [−1, 1]. Then f 00 ([−1, 1]) is compact since [−1, 1] is compact (closed and bounded,
Heine-Borel). So by part (f), f 00 is bounded on [−1, 1] and thus a ≥ 2 + 2c. We rule out the case a = 2 + 2c. If
a = 2 + 2c, we have for x > 0, f 00 (x) = (a(a − 1)xa−2 − c2 xa−2c−2 ) sin(x−c ) − c(−2a + c + 1)xa−c−2 cos(x−c ). Set
−1/c
xk = π2 + 2πk
as usual, then f 00 (xk ) = a(a − 1)( π2 + 2πk)−(a−2)/c − c2 , which does not tend to 0 as k → ∞.
00
Since f (0) = 0 and limk→∞ xk = 0, f 00 is not continuous at 0.
Suppose a > 2 + 2c. Then all the terms in the expression above for f 00 (x) for x > 0 tends to 0 as x → 0
by the Squeeze Theorem, so f 00 is continuous at 0. It’s continuous everywhere else so f 00 is continuous on [−1, 1].
6. Rudin Ch 5. Exercise #14. Let f be a differentiable real function defined in (a, b). Prove that f is convex if and only if f 0 is monotonically increasing. Assume next that f 00 (x) exists for every x ∈ (a, b), and prove
that f is convex if and only if f 00 (x) ≥ 0 for all x ∈ (a, b).
Solution.
Recall that a real function f defined on (a, b) is convex if and only if for all x < y < z in (a, b),
We will prove this equivalence at the end of this solution.
f (y)−f (x)
y−x
≤
f (z)−f (y)
.
z−y
Suppose that f is convex. Take x < y in (a, b). Take sequences {xn }n and {yn }n such that x < xn < yn < y for all
n, limn→∞ xn = x and limn→∞ yn = y. Since f is assumed to be differentiable, we then have f 0 (x) = limn→∞ an
(x)
(y)
(x)
(xn )
and f 0 (y) = limn→∞ bn if an = f (xxnn)−f
and bn = f (yynn)−f
. By the convexity, an = f (xxnn)−f
≤ f (yynn)−f
≤
−x
−y
−x
−xn
f (y)−f (yn )
y−yn
= bn for all n ≥ 1. Therefore f 0 (x) = limn→∞ an ≤ limn→∞ bn = f 0 (y). That is, f 0 is monotonically
increasing.
Suppose now that f 0 is monotonically increasing. Take x < y < z in (a, b). By the Mean Value Theorem,
HW #1 SOLUTIONS
5
f (y)−f (x)
y−x
(y)
= f 0 (c) for some c ∈ (x, y) and f (z)−f
= f 0 (d) for some d ∈ (y, z). Then x < c < y < d < z, and in
z−y
0
0
particular c < d. Therefore f (c) ≤ f (d), and so f is convex.
Assuming that f 00 (x) exists for all x ∈ (a, b), note only that f 0 is monotonically increasing if and only if f 00 (x) ≥ 0
for all x ∈ (a, b) by the Mean Value Theorem. One direction follows from Theorem 5.11. Suppose that f 0 is
monotonically increasing. Then fix y ∈ (a, b) and let h : [y, b) → R be defined by h(x) = f 0 (x) − f 0 (y). Then
h(y) = 0, h is differentiable since f 0 is differentiable, and h(x) = h(x) − 0 = h(x) − h(y) = h0 (c)(x − y) for some
c ∈ (y, x). But h0 (c) = f 00 (c) ≥ 0 (we’re fixing y), so that h(x) ≥ 0. That is, f 0 (x) − f 0 (y) ≥ 0, or f 0 (x) ≥ f 0 (y).
So f 0 is monotonically increasing.
Proof of equivalent definitions of convexity. We compare the two conditions on a function f : (a, b) → R: i)
For all t ∈ (0, 1) and x, y ∈ (a, b), f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y) and ii) For all x, y, z ∈ (a, b) such that
(y)
(x)
≤ f (z)−f
.
x < y < z, f (y)−f
y−x
z−y
i) implies ii). Take x < y < z in (a, b). We take here t =
f (y) ≤ tf (x)+(1−t)f (z) =
y−x
z−y
z−x f (x)+ z−x f (z).
z−y
z−x ,
so that t ∈ (0, 1). Then y = tx + (1 − t)z, and so
Rearranging this inequality gives precisely
f (y)−f (x)
y−x
≤
f (z)−f (y)
.
z−y
ii) implies i). If t ∈ (0, 1) and x < y are in (a, b), then x < tx + (1 − t)y < y. So using the condition ii),
(x)
(tx+(1−t)y)
we have f (tx+(1−t)y)−f
≤ f (y)−f
tx+(1−t)y−x
y−(tx+(1−t)y) . Rearranging this inequality gives precisely that f (tx + (1 − t)y) ≤
tf (x) + (1 − t)f (y).
7. Rudin Ch 5, Exercise #15. Suppose a ∈ R1 , f is twice-differentiable real function on (a, ∞), and M0 , M1 , M2
are the least upper bounds of |f (x)|, |f 0 (x)|, |f 00 (x)|, respectively, on (a, ∞). Prove that M12 ≤ 4M0 M2 . Does this
hold for vector valued functions too?
Solution.
If x ∈ (a, ∞) and h > 0, then by the theorem of Taylor, there is ζ ∈ (x, x + h) such that f (x + h) =
00
00
f (x) + f 0 (x)h + f 2(ζ) h2 , or f 0 (x)h = f (x + h) − f (x) − f 2(ζ) h2 . Therefore |f 0 (x)|h ≤ M0 + M0 + M22 h2 .
M2
2M0
M2
0
0
Or |f 0 (x)| ≤ 2M
h + 2 h. Fixing h > 0, we see here then that the set {|f (x)| | x ∈ (a, ∞)} has h + 2 h as an
2M0
M2
2M0 M2 2
upper bound, or M1 ≤ h + 2 h, or M1 h ≤ + 2 h . This inequality certainly holds for h ≤ 0 as well since
the RHS is nonnegative for h ∈ R.
We have so far that M22 h2 − M1 h + 2M0 ≥ 0 for all h ∈ R. Considering the real vlaued function g(h) =
M2 2
2 h − M1 h + 2M0 defined on R, a degree 2 polynomial, the fact that it is always nonnegative implies that it
has at most 1 root, i.e. that its discriminant M12 − 4( M22 )(2M0 ) ≤ 0, or M12 ≤ 4M0 M2 .
Let’s consider now a vector valued, twice differentiable function f : (a, ∞) → Rn . We reduce the problem to the
n
n
one-variable case by considering various projections of the curve
Pn f 2in R onto the 1-dimensional subspaces of R .
n−1
To that end, take a direction u ∈ S
= {(a1 , a2 , . . . , an ) | i=1 ai = 1}. We then consider the projection of the
curve f onto the line spanned by u, i.e. consider the function g : (a, ∞) → R by g(t) = u · f (t) = |u||f (t)| cos θ
where θ is the “angle” between u and f (t).
Using the one-variable case, we have (supt∈(a,∞) |g 0 (t)|)2 ≤ 4 supt∈(a,∞) |g(t)| supt∈(a,∞) |g 00 (t)|. Note that g 0 (t) =
u · f 0 (t) and g 00 (t) = u · f 00 (t), so by the Cauchy-Schwartz inequality we have |g 0 (t)| ≤ |u||f 0 (t)| = |f 0 (t)| and
likewise |g 00 (t)| ≤ |f 00 (t)|.
6
HW #1 SOLUTIONS
We have thus shown that for all u ∈ Sn−1 , (supt∈(a,∞) |u · f 0 (t)|)2 ≤ 4M0 M2 . Note that (supt∈(a,∞) |f 0 (t)|)2 =
supt∈(a,∞) |f 0 (t)|2 . Now take any > 0. Then there exists ζ ∈ (a, ∞) such that supt∈(a,∞) |f 0 (t)| − < |f 0 (ζ)|.
We now then choose u ∈ Sn−1 to maximize the magnitude of the projection f 0 (ζ) onto u, i.e. take u =
the direction parallel to f 0 (ζ). Then we have the inequalities
f 0 (ζ)
|f 0 (ζ)|
in
( sup |f 0 (t)| − )2 ≤ |f 0 (ζ)|2 = |u · f 0 (ζ)|2 ≤ ( sup |u · f 0 (t)|)2 ≤ 4M0 M2
t∈(a,∞)
t∈(a,∞)
.
Since > 0 was arbitrary, we have M12 ≤ 4M0 M2 .
Now take
f (x) =
2x2 − 1
x2 −1
x2 +1
if − 1 < x < 0
if x ≥ 0
We show that M0 = 1, M1 = 4, and M2 = 4 for the above f defined on (−1, ∞). It’s easy to verify the
following:
4x
if − 1 < x < 0
0
f (x) =
4x
if x ≥ 0
2
2
(x +1)
(
4
if − 1 < x < 0
f 00 (x) =
−12x2 +4
if x ≥ 0
2
3
(x +1)
The only care one needs to take in differentiating the above f is at 0, but this is handled in a straightfor2
2
ward manner by comparing the right-hand and the left-hand limits at 0. Writing xx2 −1
+1 = 1 − x2 +1 , it’s clear that
M0 = 1. Likewise for M1 = 4 and M2 = 4.
8. Rudin Ch 5, Exercise #18. Suppose f is a real function on [a, b], n is a positive integer, and f (n−1) exists for every t ∈ [a, b]. Let α, β, P be as in Taylor’s theorem (5.15). Define
Q(t) =
f (t) − f (β)
t−β
for t ∈ [a, b], t 6= β, differentiate
f (t) − f (β) = (t − β)Q(t)
n-1 times at t = α, and derive the following version of Taylor’s theorem:
f (β) = P (β) +
Q(n−1) (α)
(β − α)n .
(n − 1)!
Solution.
We’ll verify the conclusion by induction on n. Note that P (β) depends on n, as P (β) = f (α) + f 0 (α)(β −
α) +
f 00 (α)
2! (β
− α)2 + · · · +
f (n−1) (α)
(n−1)! (β
− α)n−1 .
If n = 1, then the conclusion reads f (β) = P (β) + Q(α)(β − α) = f (α) + Q(α)(β − α), which just comes
from the definition of Q(t) (and evaluating this at t = α).
HW #1 SOLUTIONS
7
Before we proceed to the inductive step, we develop a formula for derivatives of f (t)−f (β) = (t−β)Q(t). Differentiating this once, we have f 0 (t) = Q(t)+(t−β)Q0 (t), differentiating yet again, we have f 00 (t) = 2Q0 (t)+(t−β)Q00 (t).
It’s easy to see by (yet another) induction that f (n) (t) = nQ(n−1) (t) + (t − β)Q(n) (t) for n ≥ 1. Plugging in t = α,
we have f (n) (α) = nQ(n−1) (α) + (α − β)Q(n) (α).
(n−1)
Now we resume the induction. Assuming that the conclusion holds for n, we have f (β) = P (β)+ Q(n−1)!(α) (β −α)n .
From here, one simply substitutes Q(n−1) (α) =
paragraph.
f (n) (α)+(β−α)Q(n) (α)
n
from the equation we derived in the previous
9. Rudin Ch 5, Exercise #26. Suppose f is differentiable on [a, b], f (a) = 0, and there is a real number A
such that |f 0 (x)| ≤ A|f (x)| on [a, b]. Prove that f (x) = 0 for all x ∈ [a, b].
Solution.
Note that we may as well assume that A > 0, for if A ≤ 0 then 0 ≥ |f 0 (x)| ≤ 0, or f 0 (x) = 0 for all x ∈ [a, b], in
which case we have that for t ∈ [a, b], f (t) = f (t) − 0 = f (t) − f (a) = f 0 (c)(t − a) for some c ∈ [a, t], and since
f 0 (c) = 0 we have f (t) = 0, i.e. f (x) = 0 for all x ∈ [a, b].
1
+
is chosen such that b−a
Assuming now that A > 0, set ∆ = b−a
n , where n ∈ Z
n < A (note that we can arrange for this by the Archimedean property). For i ∈ {0, 1, 2, . . . , n}, set ti = a + i∆, so that [a, b] = ∪ni=1 [ti−1 , ti ].
We show by induction on i that f is 0 on [ti−1 , ti ].
Base case is when i = 1, or showing that f is 0 on [a, a + ∆]. Set M be the sup of the set {|f (x)| | x ∈ [a, a + ∆]}.
Taking x ∈ [a, a+∆], we have by the Mean Value Theorem there exists c ∈ [a, x] such that |f (x)| = |f (x)−f (a)| =
|f 0 (c)||x − a| ≤ A|f (c)||x − a| ≤ A · ∆ · M . This being true for all x ∈ [a, a + ∆], we have M ≤ A · ∆ · M . If M > 0,
then dividing the inequality by M , we have 1 ≤ A · ∆, contradicting ∆ < A1 . Therefore M = 0 and f (x) = 0 for
all x ∈ [a, a + ∆].
Assume now that f is 0 on [t0 , t1 ] ∪ [t1 , t2 ] ∪ · · · ∪ [ti−1 , ti ]. We now show that f is 0 on [ti , ti+1 ]. We know
that f (ti ) = 0. The same argument applies: Let M be the sup of the set {|f (x)| | x ∈ [ti , ti+1 ]}. For any
x ∈ [ti , ti+1 ], we have |f (x)| = |f (x) − f (ti )| = |f 0 (c)||x − ti | ≤ A|f (c)| · ∆ ≤ A · ∆ · M , for some c ∈ [ti , x]. This
being true for all x ∈ [ti , ti+1 ], we have M ≤ A · ∆ · M . Same argument as before, we must then have M = 0.
Therefore f = 0 on [ti , ti+1 ].
10. Rudin Ch 5, Exercise #27. Let φ be a real function defined on a rectangle R in the plane, given by
a ≤ x ≤ b, α ≤ y ≤ β. A solution of the initial-value problem
y 0 = φ(x, y),
y(a) = c (α ≤ c ≤ β)
is, by definition, a differentiable function f on [a, b] such that f (a) = c, α ≤ f (x) ≤ β, and
f 0 (x) = φ(x, f (x))
(a ≤ x ≤ b).
8
HW #1 SOLUTIONS
Prove that such a problem has at most one solution if there is a constant A such that
φ(x, y2 ) − φ(x, y1 )| ≤ A|y2 − y1 |
whenever (x, y1 ) ∈ R and (x, y2 ) ∈ R.
Solution.
Suppose g : [a, b] → R is yet another solution of the above initial-value problem. Consider a function h : [a, b] → R
defined by h(x) = f (x)−g(x) for x ∈ [a, b]. Then for x ∈ [a, b], |h0 (x)| = |f 0 (x)−g 0 (x)| = |φ(x, f (x))−φ(x, g(x))| ≤
A|f (x) − g(x)| = A|h(x)|. Also, h(a) = f (a) − g(a) = c − c = 0. By problem #26 then we have h = 0 on [a, b].
We now classify all solutions to y 0 = y 1/2 , y(0) = 0. Certainly zero function is a solution to this differential
equation. Suppose now that f is a solution
to the differential equation on an interval I containing 0. Certainly
p
f (x) ≥ 0 for all x ∈ I, so that f 0 (x) = f (x) implies that f is nondecreasing on I. For x ∈ I and x ≤ 0, we must
thus have 0 ≤ f (x) ≤ f (0) = 0, or f (x) = 0. We can then just (uniquely by monotonicity of f ) extend f to all of
(−∞, 0] ∪ I by defining it to be 0 on (−∞, 0], therefore we assume that our function is defined f : (−∞, a] → R,
where a ≥ 0. We can assume that a > 0 (if a = 0 the monotonicity argument above already implies that zero
function is the only possible solution on (−∞, 0]). Zero function defined on all of (−∞, ∞) is of course a solution
to the differential equation. Since f is nondecreasing, if f is not the zero function then f (a) = b > 0.
Now let c = sup{x | f (x) = 0} = S. Note that c exists because S has a as an upper bound and is nonempty
because it contains 0. Then there is a sequence {xn }n≥1 in S such that limn→∞ xn = c, and by continuity of f
we must have f (c) = 0. We have c < a. Take a point l such that c < l < a, and note that f (l) = s > 0. Now let
a−l
Kn = [c + l−c
n , l + n ] for n ≥ 2. Then Kn is compact, f is defined on Kn and f is never zero on Kn . Therefore
f (Kn ) is compact and does not contain 0. Considering the function φ(y) = y 1/2 defined on Kn , we see then
that φ0 (y) = 21 y −1/2 is defined and continuous on f (Kn ), so φ0 (f (Kn )) is compact, i.e. there is a real number
A (may depend on n) such that for all x ∈ f (Kn ), |φ0 (x)| ≤ A. Therefore if q, w ∈ f (Kn ) then by the Mean
Value Theorem, |φ(q) − φ(w)| = |φ0 (c)||q − w| ≤ A|q − w|, where c is between q and w. Also f (Kn ) is compact
and connected in R1 , so it is a closed interval (Theorem 2.47, say). If we use the rectangle R as in problem #27
as Kn × f (Kn ), then f 0 (x) = φ(f (x)) and |φ(y1 ) − φ(y2 )| ≤ A|y1 − y2 | for all y1 , y2 ∈ f (Kn ), so the conditions
of problem #27 are satisfied (note that φ(y) is independent of x here, an “autonomous” case). Therefore the
solution to this differential
equation with the initial condition f (l) = s is unique, and it’s readily checked
that
√
√
(x+2 s−l)2
(x+2 s−l)2
the mapping x 7→
is a solution to this initial differential equation, so we must have f (x) =
4
4
on Kn .
Now we take xn = c +
l−c
n
for n ≥ 2, so that xn ∈ Kn and f (xn ) =
so by continuity of f we must then have 0 = f (c) =
f (l) =
(l−c)2
4 .
√
(c+2 s−l)2
.
4
Note that limn→∞ xn = c, and
p
s = f (l), so we must have c + 2 f (l) − l = 0, or
Therefore
(
f (x) =
0
(l−c)2
4
√
(xn +2 s−l)2
.
4
if x ≤ c
if x ≥ c