The Cauchy–Schwarz master class - An Introduction to the Art of Mathematical Inequalities - J. Michael Steele

∫ ∞
0
xtφ(x) dx
is called the tth moment of φ. Show that if t ∈ (t0, t1) then
µt ≤ µ1−αt0 µαt1 where t = (1− α)t0 + αt1 and 0 < α < 1.
In other words, the linearly interpolated moment is bounded by the
geometric interpolation of two extreme moments.
Exercise 9.5 (Complex Ho¨lder — and the Case of Equality)
Ho¨lder’s inequality for real numbers implies that for complex numbers
a1, a2, . . . , an and b1, b2, . . . , bn one has the bound∣∣∣∣ n∑
k=1
akbk
∣∣∣∣ ≤ ( n∑
k=1
|ak|p
)1/p( n∑
k=1
|bk|q
)1/q
(9.30)
when p > 1 and q > 1 satisfy 1/p + 1/q = 1. What conditions on
the complex numbers a1, a2, . . . , an, and b1, b2, . . . , bn are necessary and
sufficient equality to hold in the bound (9.30)? Although this exercise is
easy, it nevertheless offers one useful morsel of insight that should not
be missed.
150 Ho¨lder’s Inequality
Exercise 9.6 (Jensen Implies Minkowski)
By Jensen’s inequality, we know that for a convex φ and positive
weights w1, w2, . . . , wn one has
φ
(
w1x1 + w2x2 + · · ·+ wnxn
w1 + w2 + · · ·+ wn
)
≤ w1φ(x1) + w2φ(x2) + · · ·+ wnφ(xn)
w1 + w2 + · · ·+ wn . (9.31)
Consider the concave function φ(x) = (1 + x1/p)p on [0,∞], and show
that by making the right choice of the weights wk and the values xk in
Jensen’s inequality (9.31) one obtains Minkowski’s inequality.
Exercise 9.7 (Ho¨lder’s Inequality for Integrals)
Naturally there are integral versions of Ho¨lder’s inequality and, in
keeping with the more modern custom, there is no cause for a name
change when one switches from sums to integrals.
Let w : D → [0,∞) be given, and reinforce your mastery of Ho¨lder’s
inequality by checking that our earlier argument (page 137) also shows
that, for all suitably integrable functions f and g from D to R,∫
D
f(x)g(x)w(x) dx ≤
(∫
D
|f(x)|pw(x) dx
)1/p(∫
D
|g(x)|qw(x) dx
)1/q
where, as usual, 1 < p <∞ and p−1 + q−1 = 1.
Exercise 9.8 (Legendre Transforms and Young’s Inequality)
If f : (a, b)→ R, then the function g : R→ R defined by
g(y) = sup
x∈(a,b)
{xy − f(x)} (9.32)
is called the Legendre transform of f . It is used widely in the theory of
inequalities, and part of its charm is that it helps us relate products to
sums. For example, the definition (9.32) gives us the immediate bound
xy ≤ f(x) + g(y) for all (x, y) ∈ (a, b)× R. (9.33)
(a) Find the Legendre transform of f(x) = xp/p for p > 1 and compare
the general bound (9.33) to Young’s inequality (9.6).
(b) Find the Legendre transforms of f(x) = ex and φ(x) = x log x−x.
(c) Show that for any function f the Legendre transform g is convex.
Ho¨lder’s Inequality 151
Exercise 9.9 (Self-Generalizations of Ho¨lder’s Inequality)
Ho¨lder’s inequality is self-generalizing in the sense that it implies sev-
eral apparently more general inequalities. This exercise address two of
the most pleasing of these generalizations.
(a) Show that for positive p, q, bigger than r one has
1
p
+
1
q
=
1
r
⇒
{ n∑
j=1
(ajbj)r
}1/r
≤
{ n∑
j=1
apj
}1/p{ n∑
j=1
bqj
}1/q
.
(b) Given p, q, and r are bigger than 1, show that if
1
p
+
1
q
+
1
r
= 1,
then one has the triple produce inequality
n∑
j=1
ajbjcj ≤
{ n∑
j=1
apj
}1/p{ n∑
j=1
bqj
}1/q{ n∑
j=1
crj
}1/r
.
Exercise 9.10 (The Historical Ho¨lder Inequality)
The inequality which Ho¨lder actually proved in his 1889 article asserts
that for wk ≥ 0, yk ≥ 0, and p > 1 one has
n∑
k=1
wkyk ≤
{ n∑
k=1
wk
}(p−1)/p{ n∑
k=1
wky
p
k
}1/p
. (9.34)
Show, as Ho¨lder did, that this inequality follows from the weighted ver-
sion (9.31) of Jensen’s inequality. Finally close the loop by showing that
the historical version (9.34) of Ho¨lder’s inequality is equivalent to the
modern version that was introduced by F. Riesz. That is, check that
inequality (9.34) implies inequality (9.1), and vice versa.
Exercise 9.11 (Minkowski Implies Ho¨lder)
The triangle inequality implies Cauchy’s inequality, so it surely seems
reasonable to guess that Minkowski’s inequality might also imply Ho¨lder’s
inequality. The guess is true, but the confirmation is a bit subtle. As
a hint, consider what Minkowski’s inequality (9.12) for s says for the
vectors θ(ap/s1 , a
p/s
2 , . . . , a
p/s
n ) and (1 − θ)(bq/s1 , bq/s2 , . . . , bq/sn ) when s is
very large.
152 Ho¨lder’s Inequality
∑m
j=1
∏n
k=1 a
wk
jk ≤
∏n
k=1
(∑m
j=1 ajk
)wk
S
cols
G
rows A
GS
Fig. 9.2. Ho¨lder’s inequality for an array (9.35) is easier to keep in mind if one
visualizes its meaning. In fact, it asserts a natural commutativity relationship
between the summation operation S and the geometric mean operation G. As
the figure suggests, if we let G act on rows and let S act on columns, then the
inequality (9.35) tells us that by acting first with the geometric mean G we
get a smaller number than if we act first with S.
Exercise 9.12 (Ho¨lder’s Inequality for an Array)
Any formula that generalizes Ho¨lder’s inequality to an array is likely
to look complicated but, as Figure 9.2 suggests, it is still possible for
such a formula to be conceptually simple.
Show that for nonnegative real numbers ajk, 1 ≤ j ≤ m, 1 ≤ k ≤ n
and positive weights w1, . . . , wn that sum to 1, we have the bound
m∑
j=1
n∏
k=1
awkjk ≤
n∏
k=1
( m∑
j=1
ajk
)wk
. (9.35)
Prove this inequality, and use it to prove the mixed mean inequality
which asserts that for nonnegative x, y, z one has
x+ (xy)
1
2 + (xyz)
1
3
3
≤
(
x · x+ y
2
· x+ y + z
3
)1/3
. (9.36)
Exercise 9.13 (Rogers’s Inequality — the Proto-Ho¨lder)
The inequality that L.C. Rogers proved in this 1888 article asserts
that for 0 < r < s < t < ∞ and for nonnegative ak, bk, k = 1, 2, . . . , n,
one has the bound( n∑
k=1
akb
s
k
)t−r
≤
( n∑
k=1
akb
r
k
)t−s( n∑
k=1
akb
t
k
)s−r
,
Ho¨lder’s Inequality 153
which we may write more succinctly as
(Ss)t−r ≤ (Sr)t−s(St)s−r where Sp =
n∑
k=1
akb
p
k for p > 0. (9.37)
Rogers gave two proofs of his bound (9.37). In the first of these he
called on the Cauchy–Binet formula [see (3.7), page 49], and the second
he used the AM-GM inequality which he wrote in the form
xw11 x
w2
2 · · ·xwnn ≤
(
w1x1 + w2x2 + · · ·+ wnxn
w1 + w2 + · · ·+ wn
)w1+w2+···+wn
where the values w1, w2,. . . ,wn are assumed to be positive but which
are otherwise arbitrary.
Now, follow in Rogers’s footsteps and use the very clever substitutions
wk = akbsk and xk = b
t−s
k to deduce the bound(
b
a1b
s
1
1 b
a2b
s
2
2 · · · banb
s
n
n
)t−s
≤ (St/Ss)Ss , (9.38)
and use the substitutions wk = akbsk and xk = b
r−s
k to deduce the bound(
b
a1b
s
1
1 b
a2b
s
2
2 · · · banb
s
n
n
)r−s
≤ (Sr/Ss)Ss . (9.39)
Finally, show how these two relations imply Rogers’s inequality (9.37).
Exercise 9.14 (Interpolation for Positive Matrices)
Let 1 ≤ s0, t0, s1, t1 ≤ ∞ be given and consider an m × n matrix T
with nonnegative real entries cjk, 1 ≤ j ≤ m, 1 ≤ k ≤ n. Show that if
there exist constants M0 and M1 such that
‖Tx‖t0 ≤M0‖x‖s0 and ‖Tx‖t1 ≤M1‖x‖s1 (9.40)
for all x ∈ Rm, then for each 0 ≤ θ ≤ 1, one has the bound
‖Tx‖t ≤Mθ‖x‖s for all x ∈ Rm (9.41)
where Mθ is defined by Mθ = Mθ1M
1−θ
0 and where s and t are given by
1
s
=
θ
s1
+
1− θ
s0
,
1
t
=
θ
t1
+
1− θ
t0
. (9.42)
This problem takes some time to absorb, but the result is important,
and it pays generous interest on all invested effort. Figure 9.3 should help
one visualize the condition (9.42) and the constraints on the parameters
1 ≤ s0, t0, s1, t1 ≤ ∞. One might also note that the bound (9.41) would
follow trivially from the hypotheses (9.40) if θ = 0 or θ = 1. Moreover,
154 Ho¨lder’s Inequality
Fig. 9.3. The constraints 1 ≤ s0, t0, s1, t1 ≤ ∞ mean that the reciprocals are
contained in the unit square S = [0, 1] × [0, 1], and the exponent relation
(9.42) tells us that (1/s, 1/t) is on the line from (1/s1, 1/t1) to (1/s0, 1/t0).
The parameter θ is then determined by the explicit interpolation formula
(1/s, 1/t) = θ(1/s1, 1/t1) + (1− θ)(1/s0, 1/t0).
the bound (9.41) automatically recaptures the inequality (9.24) from
Challenge Problem 9.6; one only needs to set t1 = 1, s1 = 1, M1 = A,
t0 =∞, s0 =∞, M0 = B, and θ = 1/p.
Despite the apparent complexity of Exercise 9.14, one does not need
to look far to find a plan for proving the interpolation formula (9.41).
The strategy which worked for Problem 9.6 (page 146) seems likely to
work here, even though it may put one’s skill with the splitting trick to
the test.
Finally, for anyone who may still be hesitant to take up the challenge
of Exercise 9.14, there is one last appeal: first think about proving the
more concrete inequality (9.43) given below. This inequality is typical of
a large class of apparently tough problems which crumble quickly after
one calls on the interpolation formula (9.41).
Exercise 9.15 (An 2 Interpolation Bound)
Let cjk, 1 ≤ j ≤ m, 1 ≤ k ≤ n be an array of nonnegative real
numbers for which one has the implication
Xj =
n∑
k=1
cjkxk for all j = 1, 2, . . . ,m ⇒
n∑
j=1
|Xj |2 ≤
n∑
k=1
|xk|2.
Show that for all 1 ≤ p ≤ 2 one then has the bound( m∑
j=1
∣∣Xj∣∣q)1/q ≤M (2−p)/p( n∑
k=1
|xk|p
)1/p
(9.43)
where and q = p/(p− 1) and M = max |cjk|.
10
Hilbert’s Inequality
and Compensating Difficulties
Some of the most satisfying experiences in problem solving take place
when one starts out on a natural path and then bumps into an unex-
pected difficulty. On occasion this deeper view of the problem forces us
to look for an entirely new approach. Perhaps more often we only need
to find a way to press harder on an appropriate variation of the original
plan.
This chapter’s introductory problem provides an instructive case; here
we will discover two difficulties. Nevertheless, we manage to achieve our
goal by pitting one difficulty against the other.
Problem 10.1 (Hilbert’s Inequality)
Show that there is a constant C su