MATH 16200-33: HONORS CALCULUS-2
HOMEWORK 4: SOLUTIONS
Problem 13-39. Suppose that f and g are integrable on [a, b]. The CauchySchwarz inequality states that
Z b ! Z b !
Z b !2
g2 .
f2
fg
≤
a
a
a
(a) Show that the Schwarz inequality is a special case of the Cauchy-Schwarz
inequality.
Proof. Recall from Problem 2-21 that the Schwarz inequality states
n
X
xi yi ≤
i=1
n
X
x2i
n
1/2 X
i=1
yi2
1/2
.
i=1
Let functions f : [0, n] → R and g : [0, n] → R be defined by
f (x) = xi
if i − 1 ≤ x < i;
g(x) = yi
if i − 1 ≤ x < i
for 1 ≤ i ≤ n and define f (n) = xn and g(n) = yn . Direct computation yields
n
X
i=1
xi yi
2
=
(f g)(x) = xi yi if i − 1 ≤ x < i
Z b !2
Z b ! Z b !
n
n
X
CS
X
≤
fg
f2
x2i
yi2 .
g2 =
a
a
a
i=1
i=1
Taking square roots gives the desired result.
(b) Give a proof of the Cauchy-Schwarz inequality by imitating the proof of the
Schwarz inequality in Problem 2-21.
Proof. If g = 0, then all integrals are 0, so equality holds. If g 6= 0, then
Z b
Z b
Z b
Z b
Z b
0≤
(f − λg)2 =
(f 2 − 2λf g + λ2 g) =
f 2 − 2λ
f g + λ2
g.
a
a
a
a
a
The right-hand side is a non-negative quadratic in λ, so the discriminant is less
than or equal to 0. That is,
Z b !2
Z b Z b
4
fg − 4
f2
g 2 ≤ 0,
a
a
Z
!2
b
fg
Z
≤
a
a
1
a
b
f2
Z
a
b
g2 .
(c) If equality holds, is it necessarily true that f = λg for some λ? What if f
and g are continuous?
Solution. It is not necessarily true that f = λg for some λ if equality holds. For
a counterexample, let f : [0, 1] → R and g : [0, 1] → R be defined by f (x) = 1
for all x and g(x) = 1 for x < 1 and g(1) = 0. Then
2
1
Z
1
Z
=1=1·1=
fg
2
f
g
2
,
0
0
0
1
Z
but there is no λ such that 1 = f (1) = λg(1) = 0. If f and g are continuous,
then equality in the Cauchy-Schwarz inequality implies f = λg for some λ. To
see this, note that equality implies that the quadratic in part (b) has a zero at
Rb
some λ, and at this λ we have 0 = a (f −λg)2 . Since (f −λg)2 is a non-negative
continuous function, we have f = λg.
(d) Prove that
by a and b?
R
1
0
f
2
≤
R
1
0
f 2 . Is this result true if 0 and 1 are replaced
Proof. Let g : [0, 1] → R be defined by g(x) = 1 for all x. By the CauchySchwarz inequality,
2
1
Z
≤
f
f
0
Conversely, suppose
R
1
Z
2
Z
f
2
≤
Z
1
0
b
a
1
=
f
0
R
b
a
f2
b
!2
1
2
.
0
was true for all f . Then in particular
it holds for f (x) = 1, so
2
Z
(b − a) =
Z
≤
1
a
!
b
= b − a.
1
a
Thus b − a ≤ 1 is a necessary condition.
Problem 14-21. Suppose that f 0 is integrable on [0, 1] and f (0) = 0. Prove
that for all x ∈ [0, 1] we have
s
Z 1
|f (x)| ≤
|f 0 |2 .
0
Show also that the hypothesis f (0) = 0 is needed.
Proof. Applying the Cauchy-Schwarz inequality,
sZ
sZ
Z 1
Z 1
1
0
0
0
2
|f | =
|f | · 1 ≤
|f |
0
0
0
0
2
1
sZ
12
=
0
1
|f 0 |2 ,
and by the Fundamental Theorem of Calculus,
Z
|f (x)| = |f (x) − f (0)| = x
0
x
Z
0
f ≤
|f 0 | ≤
s
Z
1
Z
|f 0 | ≤
0
0
Remark. If f (x) = 1 for all x ∈ [0, 1], then |f (x)| = 1, but
1
|f 0 |2 .
0
R1
0
|f 0 |2 = 0.
Problem 14-22. Suppose that f is a differentiable function with f (0) = 0 and
0 < f 0 ≤ 1. Prove that for all x ≥ 0 we have
Z x 2
Z x
f .
f3 ≤
0
0
Proof. Equality is clear when x = 0, so assume x > 0. Since 0 < f 0 (x) and
f (x) = 0, we have f (x) > 0, and since 0 < f 0 (x) ≤ 1, we have
2f (x)f 0 (x) ≤ 2f (x).
Integrating both sides yields
f (x)2 ≤ 2
Z
x
f.
0
Note that f (x) > 0 since f 0 > 0 and f (0) = 0, so
Z x
f (x)3 ≤ 2f (x)
f
0
Integrating both sides again,
Z
x
f3 ≤
Z
f
0
1
Note that
N r+1
r+1
.
0
Problem 14-25. The limit limN →∞
called an “improper integral.”
R∞
(a) Determine 1 xr dx, if r < −1.
Solution. For any N > 1,
Z N
xr dx =
2
x
RN
a
f , if it exists, is denoted by
N
1
N r+1
1
xr+1 1 =
−
.
r+1
r+1
r+1
→ 0 since r + 1 < 0, so
Z
∞
Z
r
x dx = lim
1
N →∞
1
3
N
xr dx = −
1
.
r+1
R∞
a
f and
R∞
(c) Suppose that f (x) ≥ 0 for x ≥ 0 and that 0 f exists. Prove that if
≤ g(x) ≤ f (x) for all x ≥ 0, and g is integrable on each interval [0, N ], then
R0 ∞
g also exists.
0
RN
Proof. It is enough to show that H(N ) = 0 g is non-decreasing and bounded
R∞
above, for then the least upper bound axiom would imply 0 g := lim H(N )
N →∞
RN
exists. But this is immediate: since 0 ≤ g(x), the integral 0 g grows monotonically with N , and since g(x) ≤ f (x),
Z N
Z N
Z ∞
g≤
f≤
f < ∞.
0
0
0
R∞
1/(1 + x2 ) dx exists.
R1
R∞
Solution. Note that 0 1/(1 + x2 ) dx exists, so 0 1/(1 + x2 ) dx exists if and
R∞
only if 1 1/(1 + x2 ) dx exists. From 0 ≤ 1/(1 + x2 ) ≤ 1/x2 and
N
Z N
Z ∞
1
−1
−1
1
= lim
dx = lim
= lim
+1 =1
N →∞ 1
N →∞
N →∞
x2
x2
x 1
N
1
R∞
part (c) implies 1 1/(1 + x2 ) dx exists.
(d) Explain why
0
Ra
Problem 14-27. The improper integral −∞ f is defined in the obvious way,
Ra
R∞
as limN →−∞ N f . But another kind of improper integral −∞ f is defined in
R∞
R0
a nonobvious way: it is 0 f + −∞ f , provided these improper integrals both
exist.
R∞
(a) Explain why −∞ 1/(1 + x2 ) dx exists.
R∞
Solution. As shown in Problem 14-25(d), 0 1/(1 + x2 ) dx exists, and since
R0
f (x) = 1/(1 + x2 ) is an even function, −∞ 1/(1 + x2 ) dx exists as well.
(b) Explain why
R∞
−∞
x dx does not exist.
Solution. For any N > 0,
N
N
N2
,
2
0
0
R∞
so the integral is unbounded as N → ∞. Then 0 x dx does not exist and
R∞
hence −∞ x dx does not exist.
Z
x dx =
x2
2
=
Problem 14-27. There is another kind of “improper integral” in which the
interval is bounded, but the function is unbounded:
4
Ra √
Ra √
(a) If a > 0, find limε→0+ ε 1/ x√dx. This limit is denoted by 0 1/ x dx,
even though the function f (x) = 1/ x is not bounded on [0, a], no matter how
we define f (0).
Solution. For any ε > 0,
Z a
√ a
√
√
x−1/2 dx = 2 x ε = 2( a − e).
ε
Letting ε → 0+ , we see that
(b) Find
Ra
0
Ra
0
√
√
1/ x dx = 2 a.
xr dx if −1 < r < 0.
Solution. Let ε > 0 be given
Z a
xr dx =
ε
Since 0 < r + 1,
Z
ε
ar+1
εr+1
1 r+1 a
x
=
−
.
ε
1+r
1+r 1+r
a
xr dx →
ar+1
1+r
5
as ε → 0+ .