170B Note
Sangchul Lee
January 27, 2016
1
Solution of HW #2
Exercise 6.7. Suppose f is a real function on (0, 1] and f ∈ R on [c, 1] for every c > 0. Define
1
Z
0+
f (x) dx = lim
c→0
1
Z
f (x) dx
c
if the limit exists (and is finite).
(a) If f ∈ R on [0, 1], show that this definition of the integral agrees with the old one.
(b) Construct a function f such that the above limit exists, although it fails to exist with | f | in place of f .
Solution. (a) Since f ∈ R on [0, 1], it is bounded and thus we can pick M > 0 such that | f (x)| ≤ M for all
x ∈ [0, 1]. Then
Z 1
Z 1
Z c
f (x) dx −
f (x) dx = f (x) dx ≤ Mc.
c
0
0
R1
R1
Taking c → 0 shows that 0+ f (x) dx exists and is equal to 0 f (x) dx as desired.
(b) Choose f (x) = sin(1/x) − (1/x) cos(1/x). Then f has an antiderivative x sin(1/x) on (0, 1] and hence
1
Z
0+
On the other hand, we claim that
f
g x=1
f (x) dx = lim x sin(1/x)
= sin 1.
x=c
c→0
1
c
| f (x)| dx diverges to +∞ as c → 0. To this end, it is sufficient to find
R1
a particular sequence (cn ) of positive reals such that cn → 0 and c | f (x) dx → +∞. Integrating both
n
sides of the inequality | f (x)| ≥ (1/x)| cos(1/x)| − 1 on [c, 1], we have
R
1
Z
1
Z
| f (x)| dx ≥
c
c
| cos(1/x)|
dx −
x
1/π
Z
≥
c
1
1
Z
dx
c
| cos(1/x)|
dx − 1.
x
Now we pick cn = 1/(nπ) and plug c = cn . With the substitution u = 1/x,
Z
1/π
1/nπ
| cos(1/x)|
dx =
x
=
nπ
Z
π
n−1 Z (k+1)π
X
| cos u|
du =
u
k=1
n−1 Z π
X
k=1
cos u
du ≥
u + kπ
0
kπ
n−1 Z π
X
k=1
0
| cos u|
du
u
n−1
cos u
2X 1
du =
.
(k + 1)π
π k=1 k + 1
Since this lower bound diverges to +∞ as n → ∞, the proof is done by the comparison argument.
Exercise 6.8. Suppose f ∈ R on [a, b] for every b > a where a is fixed. Define
∞
Z
f (x) dx = lim
b→∞
a
b
Z
f (x) dx
a
if this limit exists (and is finite). In that case, we say that the integral on the left converges. If it also converges
after f has been replaced by | f | , it is said to converge absolutely.
Assume that f (x) ≥ 0 and that f decreases monotonically on [1, ∞) . Prove that
∞
Z
f (x) dx
1
converges if and only if
∞
X
f (n)
n=1
converges. (This is so-called “integral test” for convergence of series.)
Solution. Since f is non-negative on [1, ∞), we have the following dichotomy:
R∞
Rb
P
PN
• 1 f (x) dx (resp., ∞
n=1 f (n)) converges if b 7→ 1 f (x) dx (resp., N 7→
n=1 f (n)) is bounded.
R∞
Rb
P∞
PN
• 1 f (x) dx (resp., n=1 f (n)) diverges if b 7→ 1 f (x) dx (resp., N 7→ n=1 f (n)) is unbounded.
Rb
PN
We show that 1 f (x) dx is bounded if and only if n=1
f (n) is bounded. This immediately follows from the
inequality
N
X
N
Z
f (n) ≤
n=2
1
f (x) dx ≤
N
−1
X
f (n).
n=1
Indeed, this is true since we have
Z
n
f (n) ≤
f (x) dx ≤ f (n − 1)
n−1
for all n ≥ 2, which follows since f decreases monotonically.
2
Exercise 6.10. Let p and q be positive real numbers such that
1 1
+ = 1.
p q
Prove the following statements:
(a) If u ≥ 0 and v ≥ 0, then
uv ≤
up vq
+
.
p
q
Equality holds if and only if u p = v q .
(b) If f ∈ R (α) , g ∈ R (α) , f ≥ 0, g ≥ 0 and
b
Z
f α dα = 1 =
b
Z
g q dα,
a
a
then
b
Z
f g dα ≤ 1.
a
(c) If f and g are complex functions in R (α) , then
) 1/q
) 1/p (Z b
(Z b
Z b
q
p
|g| dα
.
| f | dα
f g dα ≤
a
a
a
This is Hölder’s inequality. When p = q = 2 it is usually called the Schwarz inequality.
(d) Show that the Hölder’s inequality is also true for the “improper” integrals described in Exercise 7 and 8.
Solution. Refer to the appendix if you need some knowledge on complex-valued Riemann-Stieltjes integrals.
(a) Fix v ≥ 0 and consider the function f : [0, ∞) → R defined by
f (u) =
up vq
+
− uv.
p
q
Then we have f 0 (u) = u p−1 − v. From the identity (1/p) + (1/q) = 1, we know that p > 1, q > 1 and
q(p − 1) = p. Let u0 ≥ 0 be the unique solution of f 0 (u0 ) = 0. By analyzing the sign of f 0, we know that
u = u0 is the unique global minimum of f . Moreover,
p−1
p
• Taking power to the q, we know that u0 = v is equivalent to u0 = v q .
• f (u) ≥ f (u0 ) = 0 and the equality holds if and only if u = u0 , which holds if and only if u p = v q .
This proves the inequality.
(b) We know from Theorem 6.13 that f g ∈ R (α) as well. Then by the inequality (a) above,
!
Z b
Z b
f p gq
f g dα ≤
+
dα = 1.
p
q
a
a
3
(c) We know (from the appendix, for instance) that f g ∈ R (α), | f | p ∈ R (α) and |g| q ∈ R (α). Assume first
Rb
Rb
that a | f | p dα > 0 and a |g| q dα > 0. Let
) 1/p
b
(Z
k f k p :=
kgkq :=
|g| q dα
a
Then it follows that
R
b
(| f |/k
a
) 1/q
b
(Z
and
| f | p dα
.
a
f k p ) p dα = 1 and
R
b
(|g|/kgkq ) q
a
dα = 1. Consequently by (b),
Z b
Z b
Z b
f g dα ≤
| f g| dα = k f k p kgkq
(| f |/k f k p )(|g|/kgkq ) dα ≤ k f k p kgkq .
a
a
a
Motivated by this argument, we prove the statement in full generality as follows: let ε > 0 and write
Z b
Z b
Z b
|g|
|f|
·
dα.
f g dα ≤
| f g| dα = (k f k p + ε)(kgkq + ε)
k
f
k
+
ε
kgk
p
q +ε
a
a
a
Then by the inequality (a) we have
!p
!q)
(
Z b
k f kp
kgkq
1
1
+
f g dα ≤ (k f k p + ε)(kgkq + ε)
p k f kp + ε
q kgkq + ε
a
≤ (k f k p + ε)(kgkq + ε).
Take ε → 0 to conclude the desired inequality.
(d) We first consider the improper integrals as in Exercise 7. We formulate the improper version of the Hölder’s
inequality as follows:
Theorem. Assume that f ∈ R (α) and g ∈ R (α) on [c, 1] for any 0 < c < 1 and that both
Z
1
0+
Z
| f | p dα
and
1
0+
|g| q dα
converge. Then
1
Z
f g dα
0+
also converges and we have
) 1/p (Z 1
) 1/q
Z 1
(Z 1
f g dα ≤
| f | p dα
|g| q dα
.
0+
0+
0+
Indeed, letting c → 0 to the following Hölder’s inequality
1
Z
| f g| dα ≤
c
) 1/p (Z
1
(Z
1
p
| f | dα
c
|g| dα
c
4
) 1/q
q
and taking advantage of the fact that the integrand | f g| is non-negative, we find that
and satisfies
(Z 1
) 1/p (Z 1
) 1/q
Z 1
p
q
| f g| dα ≤
| f | dα
|g| dα
.
0+
0+
R
1
0+
| f g| dα converges
0+
R1
This is already close to the actual statement, but we need to relate this to the integral 0+ f g dα. To establish
the convergence of this improper integral, write f g = (h+ − h− ) + i(k+ − k− ), where
| Re( f g)| ± Re( f g)
= max{± Re( f g), 0},
2
| Im( f g)| ± Im( f g)
= max{± Im( f g), 0}.
k± =
2
Then h± and k ± are also in R (α) on [c, 1] for any 0 < c < 1, and we have 0 ≤ h± ≤ | f g| and 0 ≤ k ± ≤ | f g|.
R1
R1
Thus by comparison argument, we know that 0+ h± dα and 0+ k± dα converge as well. This is enough to
R1
conclude that 0+ f g dα converges. Finally, the inequality follows by taking limit as c → 0 to
Z 1
Z 1
f g dα ≤
| f g| dα.
c
c
This argument applies to the Hölder’s inequality for the improper integrals as in Exercise 8 in exactly the
same way.
h± =
Exercise 6.11. Let α be a fixed increasing function on [a, b]. For u ∈ R (α) , define
b
(Z
kuk2 =
) 1/2
2
.
|u| dα
a
Suppose f , g, h ∈ R (α) , and prove the triangle inequality
k f − hk2 ≤ k f − gk2 + kg − hk2
as a consequence of the Schwarz inequality, as in the proof of Theorem 1.37.
Solution. We have
k f − hk22 =
b
Z
a
| f − h| 2 dα ≤
b
Z
(| f − g| + |g − h|) 2 dα
a
= k f − gk22 + 2
From the Hölder’s inequality with p = q = 2, we have
back gives
R
b
a
b
Z
a
| f − g||g − h| dα + kg − hk22 .
| f − g||g − h| dα ≤ k f − gk2 kg − hk2 . Plugging this
k f − hk22 ≤ (k f − gk2 + kg − hk2 ) 2 .
Taking square root to both sides gives the desired inequality.
5
Exercise 6.12. With the notations of Exercise 11, suppose f ∈ R (α) and ε > 0. Prove that there exists a
continuous function g on [a, b] such that k f − gk2 < ε .
Hint. Let P = {x 0, · · · , x n } be a suitable partition of [a, b], define
g(t) =
xi − t
t − x i−1
f (x i−1 ) +
f (x i )
∆x i
∆x i
if x i−1 ≤ t ≤ x i .
Solution. Let P and g be as in the hint. Pick M > 0 such that | f (x)| ≤ M for all x ∈ [a, b]. We first prove the
following lemma:
Lemma. Let Mi = supt ∈[x i−1, x i ] f (t) and mi = inf t ∈[x i−1, x i ] f (t) as usual. Then
sup
t ∈[x i−1, x i ]
| f (t) − g(t)| 2 ≤ 2M (Mi − mi ).
Indeed, notice that we always have mi ≤ f (t) ≤ Mi and mi ≤ g(t) ≤ Mi for t ∈ [x i−1, x i ]. Combining these two
inequalities, we have
−(Mi − mi ) ≤ f (t) − g(t) ≤ Mi − mi ,
which is equivalent to | f (t) − g(t)| ≤ Mi − mi . Now using yet another easy inequality |g(t)| ≤ M, we get
| f (t) − g(t)| 2 ≤ (| f (t)| + |g(t)|)| f (t) − g(t)| ≤ 2M (Mi − mi )
and we are done.
Let us return to the original problem. Choose a partition P such that U (P, f , α) − L(P, f , α) < ε 2 /2M. Then
by the lemma above,
k f − gk22 ≤ U (P, | f − g| 2, α)
n
X
=
sup | f (t) − g(t)| 2 ∆α i
i=1 t ∈[x i−1, x i ]
≤
n
X
2M (Mi − mi )∆α i
i=1
= 2M (U (P, f , α) − L(P, f , α))
< ε2 .
Taking square root to both sides completes the proof.
6
2
Appendix: complex-valued integrals
If f is a complex-valued function on [a, b], then we say that f ∈ R (α) if both Re( f ) ∈ R (α) and Im( f ) ∈ R (α)
on [a, b], and in this case we define
Z b
Z b
Z b
f dα =
Re( f ) dα + i
Im( f ) dα.
a
a
a
It turns out that most of the properties of (real-valued) Riemann-Stientjes integrals carry out for these complexvalued integrals. To be precise, we find that Theorem 6.12 (except for the property (b)) and Theorem 6.13
continue to hold mutatis mutandis for this integral. Indeed,
• For Theorem 6.12.(a), only the second part is non-trivial. Let c = c1 + ic2 and f = f 1 + i f 2 , where
c1, c2 ∈ R and f 1, f 2 ∈ R (α) are real-valued. Then
Re(c f ) = c1 f 1 − c2 f 2 ∈ R (α)
and
and thus c f ∈ R (α) as well. Then
Z b
Z b
Z
c f dα =
(c1 f 1 − c2 f 2 ) dα + i
a
a
= c1
b
Z
f 1 dα − c2
a
b
Z
= (c1 + ic2 )
=c
b
Z
a
a
Im(c f ) = c1 f 2 + c2 f 1 ∈ R (α)
b
(c1 f 2 + c2 f 1 ) dα
!
Z b
Z
f 2 dα + i c1
f 2 dα + c2
a
a
f 1 dα + i
b
Z
b
a
!
f 1 dα
!
f 2 dα
a
b
Z
f dα
a
and hence the property (a) for complex-valued integral is still true.
• Theorem 6.12.(c) and (e) follow immediately by considering real part and imaginary part separately.
• Theorem 6.12.(d) and Theorem 6.13.(b) follow from Theorem 6.25 with k = 2.
• Theorem 6.13.(a) is true since both Re( f g) = Re( f ) Re(g) − Im( f ) Im(g) and Im( f g) = Re( f ) Im(g) +
Re(g) Im( f ) are in R (α) on [a, b] whenever f ∈ R (α) and g ∈ R (α) on [a, b].
We summarize these observations into the following theorems:
Theorem 6.12’ and 6.13’. Every integrand and scalar multiplication appearing in this theorem is complexvalued unless stated otherwise.
(a) If f 1 ∈ R (α) and f 2 ∈ R (α) on [a, b], then
f 1 + f 2 ∈ R (α),
c f ∈ R (α) for every constant c ∈ C, and
b
Z
a
( f 1 + f 2 ) dα =
b
Z
c f dα = c
a
b
Z
a
b
Z
f dα.
a
7
f 1 dα +
b
Z
a
f 2 dα
(b) If f ∈ R (α) on [a, b] and if a < c < b, then f ∈ R (α) on [a, c] and on [c, b], and
c
Z
f dα +
a
b
Z
f dα =
b
Z
c
f dα.
a
(c) If f ∈ R (α) on [a, b], then | f | ∈ R (α) and
Z b
Z b
f dα ≤
| f | dα.
a
a
In particular, if | f (x)| ≤ M on [a, b], then
Z b
f dα ≤ M[α(b) − α(a)].
a
(d) If f ∈ R (α1 ) and f ∈ R (α2 ), then f ∈ R (α1 + α2 ) and
b
Z
a
f d(α1 + α2 ) =
b
Z
a
f dα1 +
b
Z
a
if f ∈ R (α) and c is a positive constant, then f ∈ R (cα) and
b
Z
f d(cα) = c
b
Z
a
f dα.
a
(e) If f ∈ R (α) and g ∈ R (α) on [a, b], then f g ∈ R (α).
8
f dα2 ;