B.3. THE YOUNG INTEGRAL 381

Also let P be a dissection 0 = t0 < t1 < · · ·< tr = T .Claim: Then there exists ti such that ω (ti−1, ti+1)≤ 5

r ω (0,T ).Proof of claim: If r = 2 so there are only two intervals in the dissection,

∥F∥pp,[0,T ] ≤ ∥F∥

pp,[0,T ]+∥Z∥

pp,[0,T ] = ω (0,T )<

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ω (0,T ) .

So assume r > 2. Then how many disjoint intervals [ti−1, ti+1] are there for either i odd or ieven? Certainly fewer than r and at least as many as r

2 . Then we have

∑i even

ω (ti−1, ti+1)+ ∑i odd

ω (ti−1, ti+1)≤ 2ω (0,T )

Now if it is not true that there exists such an i, then for each i,ω (ti−1, ti+1)>5r ω (0,T ) .

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ω (0,T ) =5r

ω (0,T )r2≤ ∑

i even

5r

ω (0,T )< ∑i even

ω (ti−1, ti+1)≤ ω (0,T )

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ω (0,T ) =5r

ω (0,T )r2≤ ∑

i odd

5r

ω (0,T )< ∑i odd

ω (ti−1, ti+1)≤ ω (0,T )

Hence 5ω (0,T )< 2ω (0,T ) which is absurd. Therefore, there exists such a ti. This showsthe claim.

Next points of P are deleted beginning with ti where ω (ti−1, ti+1)≤ 5r ω (0,T ) .∫

PZdF−

∫P\{ti}

ZdF = Z (ti−1)(F (ti)−F (ti−1))

+Z (ti)(F (ti+1)−F (ti))−Z (ti−1)(F (ti+1)−F (ti−1))

This equals Z (ti−1)F (ti)+Z (ti)F (ti+1)−Z (ti)F (ti)−Z (ti−1)F (ti+1)

= Z (ti)(F (ti+1)−F (ti))+Z (ti−1)(F (ti)−F (ti+1))

= (Z (ti)−Z (ti−1))(F (ti+1)−F (ti))

It follows that∣∣∣∣∫P

ZdF−∫

P\{ti}ZdF

∣∣∣∣≤ |(Z (ti)−Z (ti−1))| |F (ti+1)−F (ti)|

≤ ω (ti−1, ti)1/q

ω (ti, ti+1)1/p ≤ ω (ti−1, ti+1)

1/p+1/q ≤(

5r

ω (0,T ))1/p+1/q

In particular, this yields∣∣∣∣∫P(Y −Y (0))dF−

∫P\{ti}

(Y −Y (0))dF∣∣∣∣≤ (5

rω (0,T )

)1/p+1/q

=

(5r

(∥F∥p

p,[0,T ]+∥Y∥qq,[0,T ]

))1/p+1/q