Kenneth Kuttler

376 APPENDIX B. INTEGRATION ON ROUGH PATHS∗

≤ supP

(∑

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

)1/p

≤ supP

(∑i|Z (ti+1)−Z (ti)|p

)1/p

(∑

i|Y (ti+1)−Y (ti)|p

)1/p

This is by

≤ supP

(∑

i|Z (ti+1)−Z (ti)|p

)1/p

+ supP

(∑

i|Y (ti+1)−Y (ti)|p

)1/p

= ∥Z∥p,[0,T ]+∥Y∥p,[0,T ]

Thus ∥·∥V p clearly is a norm. What about those inclusions? Let F ∈ V p. Is it also in V q?Suppose F ∈V p and ∥F∥p,[0,T ] = 1. What about ∥F∥q,[0,T ]? For any P(

∑i|F (ti+1)−F (ti)|q

)1/p

≤

(∑

i|F (ti+1)−F (ti)|p

)1/p

≤ 1

and so ∥F∥q/pq,[0,T ] ≤ 1. Therefore, if ∥F∥p,[0,T ] < ∞,

∥∥∥ F∥F∥p,[0,T ]

∥∥∥q/p

q,[0,T ]≤ 1 and so

∥∥∥∥∥ F∥F∥p,[0,T ]

∥∥∥∥∥q,[0,T ]

≤ 1, ∥F∥q,[0,T ] ≤ ∥F∥p,[0,T ]

as claimed. Thus V p ⊆V q and the inclusion map is continuous. There is nothing to verifyfor the uniform norm part of ∥F∥V p . How about the embedding into C0?∥F∥C0 ≤ ∥F∥V pbydefinition.

In the above, there is nothing sacred about the interval [0,T ]. You could use any otherinterval. Then we write ∥F∥V p(I) , ∥F∥p,I to denote the above with respect to the interval I.

Lemma B.1.5 Suppose a = t0 < t1 < · · ·< tn = b. Then ∑n−1i=0 ∥F∥

pp,[ti,ti+1]

≤ ∥F∥pp,[a,b]

Proof: It is sufficient to verify this with two intervals. Say a < b < c. Then let P1be a dissection for [a,b] and P2 a dissection for [b,c]. Then P = P1 ∪P2 is clearly adissection for [a,c] . Then from definition,

∑P1

|F (ti+1)−F (ti)|p +∑P2

|F (ti+1)−F (ti)|p = ∑P

|F (ti+1)−F (ti)|p ≤ ∥F∥pp,[a,c]

Then taking the sup over all such P1 one gets

∥F∥pp,[a,b]+∑

|F (ti+1)−F (ti)|p ≤ ∥F∥pp,[a,c]

Now take sup over all such P2. Note that in finding ∥F∥pp,[a,c] there is no guarantee that b

will be in any of the dissections. That is, although P1∪P2 is a disection of [a,c] it mightnot be any of the dissections needed to obtain ∥F∥p

p,[a,c]. Thus we can’t expect to have thisinequality an equation.

376 APPENDIX B. INTEGRATION ON ROUGH PATHS*1/p< sup (E(2teon) 200+ Gon) —¥ 00")i1/p \/p< sup | |) 12 (t+1) — zt) + (Eire) eor)DP "lThis is byI/p 1/p< sup (E200) 210) + sup (Eira reat)= IZ|l, 0,7) +1¥ llpyo.7Thus ||-||) clearly is a norm. What about those inclusions? Let F € V”. Is it also in V7?Suppose F € V? and ||F'|,, (9.7) = 1. What about ||F'||q (9,7)? For any 71/p 1/p(Eira Foal) < (Eien Fol) <1. q/Pand so Thal ‘(o.r) < !- Therefore, if ||F', (0,7) < Tlion lq.o.r < 1 and soF| IFloon <1, llFllgjo7) S (0,7)POT) [0,7]as claimed. Thus V? C V4 and the inclusion map is continuous. There is nothing to verifyfor the uniform norm part of ||F'||». How about the embedding into C°? ||F||-o < ||F ||ypbydefinition. §fIn the above, there is nothing sacred about the interval [0,7]. You could use any otherinterval. Then we write ||F'||yo(y) , ||F'||,,, to denote the above with respect to the interval /.Lemma B.1.5 Suppose a = 19 < t) < +++ < tf =b. Then Y") ||F\I? |F ||?Paltetisa) Fl a,b)Proof: It is sufficient to verify this with two intervals. Say a<b<c. Then let Y;be a dissection for [a,b] and #2 a dissection for [b,c]. Then Y = Y; UP, is clearly adissection for [a,c]. Then from definition,LIF (+1) ~FN + LIF (ist) -F = LIF tit) —F (ti)? SFM) ae)PyThen taking the sup over all such Y; one getsFI ta +y IF (tis1) —F (ti) |? < FI tagPyNow take sup over all such #). Note that in finding ||F Ir. lac] there is no guarantee that bwill be in any of the dissections. That is, although vr U Py is a disection of [a,c] it mightnot be any of the dissections needed to obtain ||F IF lac . Thus we can’t expect to have thisinequality an equation. J