SECTION 3.4 Holder Inequality ̈ 157
true forn= 2. We may assume that 0≤tn<1; set
x 0 =
t 1 x 1 +···+tn− 1 xn− 1
1 −tn;
note thatx 0 ∈[a,b]. We have
f(t 1 x 1 +t 2 x 2 +···+tnxn) = f((1−tn)x 0 +tnxn)
≥ (1−tn)f(x 0 ) +tnf(xn) (by induction)
≥ (1−tn)(
t 1
1 −tn
·f(x 1 ) +···+tn− 1
1 −tn
·f(xn− 1 ))+ tnf(xn) (induction again)
= t 1 f(x 1 ) +t 2 f(x 2 ) +···+tnf(xn),and we’re finished.
3.4 The H ̈older Inequality
Extending the notion of quadratic mean, we can define, for any real
numberp≥1 the “p-mean” of positive real numbersx 1 ,...,xn:
pM(x 1 ,x 2 ,...,xn) = pÃ
xp 1 +xp 2 +···+xpn
n.
We shall show that if 1≤p≤qthat for positive real numbersx 1 ,...,xn
one has
pM(x 1 ,x 2 ,...,xn)≤qM(x 1 ,x 2 ,...,xn).The proof is not too difficult—a useful preparatory result isYoung’s
inequality, below.
Lemma 2. (Young’s Inequality) Given real numbers 0 ≤ a, b and
0 < p, qsuch that
1
p+
1
q= 1, one hasab≤ap
p+
bq
q,
with equality if and only ifap=bq.