The Mathematics of Arbitrage

13.4 Some Results on the Topology ofG 269

Theorem 13.2.5, the random variableu(x)isinG. For arbitraryxwe split
into the positive and the negative part. This defines a linear mapping from
∞intoG.ForeachQ∈Mewe have thatu(x)=

∑∞

k=1xkfk, where the sum
actually converges inL^1 (Q). Let us now calculate the norm ofu(x). For an
arbitrary measureQ∈Mewe find

‖u(x)‖L (^1) (Q)≤

∫ ∑∞

k=1

|xk|‖fk‖L (^1) (Q)
and hence we have
‖u(x)‖≤sup
k
|xk|.
Take now forε>0 given, an indexksuch that supk|xk|>‖x‖∞−ε.Take
the measureQk,εas above.
We find that
‖u(x)‖L (^1) (Qk,ε)≥

∫

Ak

|u(x)|φ(k, ε)dP

≥α(k, ε)|xk| 2 P[Ak].

Since clearlyα(k, ε)P[Ak]≥ 1 −εwe find that

‖u(x)‖L^1 (Qk,ε)≥(‖x‖∞−ε)2(1−ε).

Becauseε>0 was arbitrary we find that

‖u(x)‖=‖x‖∞.

The linear mapping is therefore an isometry. Furthermore it is easily seen that
for eachx∈∞we haveu(x)∈G∞.

Theorem 13.4.7.In the setting of the above example, there is a contingent
claimfinGsuch that for eachQ∈Mewe haveEQ[f]=0, but such thatf
is not in the closure ofG∞.

Proof.We will make use of the notation and proof of the preceding theorem.
So we take the same sequence (An)n≥ 1 as above. This time we introduce
stopping times
Tn=inf{t|Mt≥n+1}

and functions

fn=(MTn−1) (^1) An.
Exactly as in the previous proof one shows that the contingent claim∑ f=
∞
n=1fnis inGand has norm 1. Suppose now thathis a bounded variable in
G. We will show that‖f−h‖≥1. For eachnwe take an elementQn∈Me
such thatQ[An]≥ 1 −^1 n; such an element surely exists. BecauseQ[fn=n|
Fn]=^1 n (^1) Anwe find forn>‖h‖∞,that