Kenneth Kuttler

2328 CHAPTER 68. A DIFFERENT KIND OF STOCHASTIC INTEGRATION

Lemma 68.3.4 The following holds for f ∈ Em∥∥ f̃∥∥

L2(T m)≤ ∥ f∥L2(T m)

alsoIm ( f ) = Im

(f̃)

Proof: This follows because, thanks to the properties of Lebesgue measure,∫T m| f (t1, · · · , tm)|2 dt1 · · ·dtm =

∫T m| f (tσ1 , · · · , tσm)|

2 dt1 · · ·dtm

≡∫

T m| fσ |2 dt1 · · ·dtm

therefore, ∥∥ f̃∥∥

L2(T m)≤ 1

m! ∑σ∈Sm

∥ fσ∥L2(T m) =1

m! ∑σ∈Sm

∥ f∥L2(T m) = ∥ f∥L2(T m)

The next claim follows because on the right, the terms making up the sum just happen in adifferent order for each σ .

More generally, here is a lemma about off diagonal things. It uses sets Ai rather thanintervals [a,b).

Lemma 68.3.5 Let {A1, · · · ,Am} be pairwise disjoint sets in B (T ) each having finite mea-sure. Then the products Ai1 ×·· ·×Ain are pairwise disjoint. Also to say that the function

(t1, · · · , tn)→∑i

ciXAi1×···×Ain(t1, · · · , tn)

equals 0 whenever some t j = ti, i ̸= j is to say that ci = 0 whenever there is a repeated indexin i.

Proof: Suppose the condition that the Ak are pairwise disjoint holds and consider twoof these products, Ai1 × ·· · ×Ain and A j1 × ·· · ×A jn . If the two ordered lists (i1, · · · , in)and ( j1, · · · , jn) are different, then since the Ak are disjoint the two products have emptyintersection because they differ in some position.

Now suppose that ci = 0 whenever there is a repeated index. Then the sum is taken overall permutations of n things taken from {1, · · · ,m} and so if some tr = ts for r ̸= s, all termsof the sum equal zero because XAi1×···×Ain

̸= 0 only if t ∈ Ai1 ×·· ·×Ain and since tr = tsand the sets {Ak} are disjoint, there must be the same set in positions r and s so ci = 0.Hence the function equals 0.

Conversely, suppose the sum ∑i ciXAi1×···×Ainequals zero whenever some tr = ts for

s ̸= r. Does it follow that ci = 0 whenever some tr = ts? The value of this function att ∈ Ai1 ×·· ·×Ain is ci because for any other ordered list of indices, the resulting producthas empty intersecton with Ai1 ×·· ·×Ain . Thus, since tr = ts, it is given that this functionequals 0 which equals ci.

2328 CHAPTER 68. A DIFFERENT KIND OF STOCHASTIC INTEGRATIONLemma 68.3.4 The following holds for f € EnIF llecrm) < IF lln2crmalsoIn (f) =In (f)Proof: This follows because, thanks to the properties of Lebesgue measure,[a le tees stm) Pty an = Jif looys-*+ stom) P dt +++ dtmI. fol dty -+- dtmtherefore,1 1IF llczcrmy <7 , Il follz2¢rm) = mal Y WF llezcrm, = lf llezcrm* O€Sin * oESinThe next claim follows because on the right, the terms making up the sum just happen in adifferent order foreacho. JMore generally, here is a lemma about off diagonal things. It uses sets A; rather thanintervals |[a,b).Lemma 68.3.5 Let {A1,--- ,Am} be pairwise disjoint sets in B(T) each having finite mea-sure. Then the products Aj, X «++ X Aj, are pairwise disjoint. Also to say that the function(t1,-°° tn) => Ya%,, Xe XAjn (t1,-°° stn)iequals 0 whenever some t; =t;,i # j is to say that cj = 0 whenever there is a repeated indexini.Proof: Suppose the condition that the A; are pairwise disjoint holds and consider twoof these products, Aj, x --- x Aj, and Aj, x --- x Aj,. If the two ordered lists (i),--+ ,in)and (j1,--+,jn) are different, then since the A, are disjoint the two products have emptyintersection because they differ in some position.Now suppose that cj = 0 whenever there is a repeated index. Then the sum is taken overall permutations of n things taken from {1,--- ,m} and so if some t, = t; forr #5, all termsof the sum equal zero because XA, x XA;n # 0 only if t € Aj, x --- x Aj, and since t, = fsand the sets {A,} are disjoint, there must be the same set in positions r and s so cj = 0.Hence the function equals 0.Conversely, suppose the sum Yj cj Ey, Xo XAlp equals zero whenever some f, = ts fors#r. Does it follow that cj = 0 whenever some t, = f;? The value of this function att € Aj, x --- x Aj, is cj because for any other ordered list of indices, the resulting producthas empty intersecton with A;, x --- x A;,. Thus, since ¢, = f,, it is given that this functionequals 0 which equals cj.In