374 APPENDIX B. INTEGRATION ON ROUGH PATHS∗
Theorem B.0.3 Let ai,bi ∈ R. Then for p≥ 1,(n
∑i=1|ai +bi|p
)1/p
≤
(n
∑i=1|ai|p
)1/p
+
(n
∑i=1|bi|p
)1/p
Proof: First note that from the definition, p−1 = p/p′.
n
∑i=1|ai +bi|p ≤
n
∑i=1|ai +bi|p−1 (|ai|+ |bi|)
≤n
∑i=1|ai +bi|p/p′ |ai|+
n
∑i=1|ai +bi|p/p′ |bi|
Now from Lemma B.0.2,
≤
(n
∑i=1
(|ai +bi|p/p′
)p′)1/p′( n
∑i=1|ai|p
)1/p
+
(n
∑i=1
(|ai +bi|p/p′
)p′)1/p′( n
∑i=1|bi|p
)1/p
=
(n
∑i=1|ai +bi|p
)1/p′( n
∑i=1|ai|p
)1/p
+
(n
∑i=1|bi|p
)1/p
In case ∑ni=1 |ai +bi|p = 0 there is nothing to show in the inequality. It is obviously true. If
this is nonzero, then divide both sides of the above inequality by (∑ni=1 |ai +bi|p)1/p′ to get(
n
∑i=1|ai +bi|p
)1− 1p′
=
(n
∑i=1|ai +bi|p
)1/p
≤
(n
∑i=1|ai|p
)1/p
+
(n
∑i=1|bi|p
)1/p
B.1 Finite p VariationInstead of integrating with respect to a finite variation integrator function F , the functionwill be of finite p variation. This is more general than finite variation. Here it is assumedp > 0 rather than p > 1.
Definition B.1.1 Define for a function F : [0,T ]→ R
1. α Holder continuous if sup0≤s<t≤T|F(t)−F(s)||t−s|α <C < ∞. Then from this inequality, it
follows that |F (t)−F (s)| ≤C |t− s|α .
2. The function F has finite p variation if for some p > 0,
∥F∥p,[0,T ] ≡ supP
(m
∑i=1|F (ti+1)−F (ti)|p
)1/p
< ∞
where P denotes a partition of [0,T ] ,P = {t0, t1, · · · , tn} for
0 = t0 < t1 < · · ·< tn = T
also called a dissection in this subject. It was also called a division when discussingthe generalized Riemann integral. |P| denotes the largest length in any of the subintervals. It will be always assumed that actually p≥ 1.