382 APPENDIX B. INTEGRATION ON ROUGH PATHS∗

It follows that removing the special point from the claim out of those ti which remain,denoted as t1, t2, t3, · · · ,∣∣∣∣∣

∫P(Y −Y (0))dF−

∫P\{t1}

(Y −Y (0))dF

∣∣∣∣∣+

∣∣∣∣∣∫

P\{t1}(Y −Y (0))dF−

∫P\{t1,t2}

(Y −Y (0))dF

∣∣∣∣∣++ · · ·+

∣∣∣∣∣∫

P\{t1,t2,···tr−2}(Y −Y (0))dF−

∫P\{t1,t2,···tr−1}

(Y −Y (0))dF

∣∣∣∣∣≤ 51/p+1/q (ω (0,T ))1/p+1/q

((1r

)1/p+1/q

+

(1

r−1

)1/p+1/q

+ · · ·+1

)

≤ 51/p+1/q (ω (0,T ))1/p+1/q∞

∑r=1

(1r

)1/p+1/q

The triangle inequality applied to that sum of terms inside || implies∣∣∣∣∣∫

P(Y −Y (0))dF−

∫P\{t1,t2,···tr−1}

(Y −Y (0))dF

∣∣∣∣∣=

∣∣∣∣∫P(Y −Y (0))dF

∣∣∣∣≤ 51/p+1/q (ω (0,T ))1/p+1/q∞

∑r=1

(1r

)1/p+1/q

since∫P\{t1,t2,···tr−1} (Y −Y (0))dF = 0. This is true because by definition it equals

(Y (0)−Y (0))(Y (T )−Y (0))

Of course all of this works for F replaced with F̃t ≡ Ft∥F∥p,[0,T ]

, and Y replaced with Ỹ where

Ỹt ≡ Yt∥Y∥q,[0,T ]

. Then in this case it is very convenient because ω (0,T )≤ 2. Then the abovereduces to ∣∣∣∣∫

P

(Ỹ − Ỹ (0)

)dF̃∣∣∣∣≤ 51/p+1/q (2)1/p+1/q

∞

∑r=1

(1r

)1/p+1/q

≡ Ĉpq

Then also, ∣∣∣∣∫P(Y −Y (0))dF

∣∣∣∣≤ Ĉpq ∥Y∥q,[0,T ] ∥F∥p,[0,T ]

Then this implies∣∣∣∣∫P

Y dF∣∣∣∣ ≤ Ĉpq ∥Y∥q,[0,T ] ∥F∥p,[0,T ]+∥Y (0)∥

∞∥FT −F0∥

= Ĉpq ∥Y∥q,[0,T ] ∥F∥p,[0,T ]+∥Y (0)∥∞(∥FT −F0∥p)1/p

≤ Ĉpq ∥Y∥V q([0,T ]) ∥F∥p,[0,T ]+∥Y∥V q([0,T ]) ∥F∥p,[0,T ]

=(Ĉpq +1

)∥Y∥V q([0,T ]) ∥F∥p,[0,T ] ≡Cpq ∥Y∥V q([0,T ]) ∥F∥p,[0,T ]