B.3. THE YOUNG INTEGRAL 379

Lemma B.2.4 Let F,Y ∈V p ([0,T ]) and let q > p≥ 1. Then

∥F−Y∥V q ≤ 2(q−p)/q ∥F−Y∥q−p

q∞ ∥F−Y∥p/q

p,[0,T ]+∥F−Y∥∞

Note that 2q−p

q ≤ 2.Proof: Let P be a dissection. Then

∑i|F (ti+1)−Y (ti+1)− (F (ti)−Y (ti))|q

≤ supi|F (ti+1)−Y (ti+1)− (F (ti)−Y (ti))|q−p

∑i|F (ti+1)−Y (ti+1)− (F (ti)−Y (ti))|p

≤ 2q−p ∥F−Y∥q−p∞ ∑

i|F (ti+1)−Y (ti+1)− (F (ti)−Y (ti))|p

Then taking the sup over all dissections,

∥F−Y∥qq,[0,T ] ≤ 2q−p ∥F−Y∥q−p ∥F−Y∥p

p,[0,T ]

∥F−Y∥q,[0,T ] ≤ 2(q−p)/q ∥F−Y∥(q−p)/q ∥F−Y∥p/qp,[0,T ]

Hence ∥F−Y∥V q ≤ 2(q−p)/q ∥F−Y∥(q−p)/q ∥F−Y∥p/qp,[0,T ]+∥F−Y∥

∞

In all the above, one can replace [0,T ] with [a,b] through simple modifications.

B.3 The Young IntegralThis dates from about 1936. Basically, you can do

∫ T0 Y dF if Y is continuous and F is of

bounded variation or the other way around. This is the old Stieltjes integral. However, thisintegral has to do with Y ∈V q and F ∈V p and of course, these functions are not necessarilyof bounded variation although they are continuous. First, here is a simple lemma which isused a little later.

Lemma B.3.1 Let F be piecewise linear with respect to a dissection P and let Y becontinuous. Then t→

∫ t0 Y dF is continuous.

Proof: Say P is given by {t0, t1, · · · , tn} where 0 = t0 < · · · < tn = T . Let G(t) ≡∫ t0 Y dF. Then on [0, t1] ,

G(t) =∫ t

0Y (s)

F (t1)−F (0)t1−0

ds =F (t1)−F (0)

t1−0

∫ t

0Y (s)ds

which is clearly continuous. Then on [t1, t2] you have

G(t) = G(t1)+∫ t

t1Y (s)

F (t2)−F (t1)t2− t1

ds

which is again continuous. Continuing this way shows the desired conclusion.