9.3 No Free Lunch with Vanishing Risk 163Corollary 9.3.4.If the semi-martingaleSsatisfies (NFLVR) then the set
{(H·S)∞|His 1 -admissible}is bounded inL^0.
Proof.This follows immediately from the existence of the limit (H·S)∞and
from Proposition 9.3.1.
Remark 9.3.5.The convergence theorem shows in particular that in the defi-
nition ofK 0 the requirement that the limit exists is superfluous. We also want
to point out that to derive the above results 9.3.1 to 9.3.4, we only used the
condition(NFLVR)for integrands with bounded support, i.e. for integrands
that are zero outside a stochastic interval [[0,k]] f o r s o m e r e a l n u m b e rk.
The next result only uses the (very weak) assumption of no-arbitrage.
We emphasize that the property(NA), as we defined it, refers to general
integrands.
Proposition 9.3.6. (compare [S 94, Proposition 4.2])If the semi-martin-
galeSsatisfies (NA) then for every admissible integrandH, such that(H·
S)∞= limt→∞(H·S)texists, we have for eacht∈R+:
‖(H·S)−t‖∞≤‖(H·S)−∞‖∞.Proof.If‖(H·S)−t‖∞>‖(H·S)−∞‖∞then we define the setA∈Ftas
A={(H·S)t<−‖(H·S)−∞‖∞}.The integrandK = (^1) A (^1) ]]t,∞[[ is admissible, the random variable (K·S)∞
exists, is non-negative andP[(K·S)∞>0]>0. This violates(NA).
The next result may be seen as a sharpening of [S 94, Proposition 1.5]. It
combines the property(NA)with the conclusion of Proposition 9.3.1.
Proposition 9.3.7.If the semi-martingale S fails the property (NFLVR)
then eitherSfails (NA) or there existsf 0 :Ω→[0,∞]not identically 0 ,ase-
quence of variables(fn)n≥ 1 =((Hn·S)∞)n≥ 1 inK 0 withHna^1 n-admissible
integrand and such thatlimn→∞fn=f 0 in probability.
Proof.It is clear that the existence of such sequences violates(NFLVR). Indeed
the set{n(Hn·S)∞;n≥ 1 }is unbounded inL^0 , whereas the integrands
(nHn)n≥ 1 are 1-admissible. This contradicts Proposition 9.3.1.
The converse is less obvious. Suppose thatSsatisfies(NA)and suppose
that (gn)n≥ 1 is a sequence inCsuch thatg 0 = limn→∞gninL∞,g 0 ≥0,
P[g 0 >α]>α>0. From the hypothesis on the sequence (gn)n≥ 1 we deduce
that‖g−n‖∞tends to 0. By passing to a subsequence, if necessary, we may
suppose that‖g−n‖∞≤n^1 .Foreachnwe take a functionhninK 0 such that