The Fundamental Theorem of Asset Pricing in L^{p}
In the stochastic discount factor notebook we proved the equivalence PNA \Leftrightarrow \exists\, m > 0 in a finite state space using a direct separation argument in \mathbb{R}^{S}. That proof exploits the fact that the positive orthant \mathbb{R}^{S}_{++} is open, which allows a global application of the Separating Hyperplane Theorem.
In an infinite-dimensional L^{p} space this shortcut fails: the cone of non-negative random variables has empty interior, so global separation is unavailable. This notebook presents the correct proof for the L^{p} setting, following the argument of Kabanov and Stricker (2001) and Clark (1993), both of which build on the classical results of Kreps (1981) and Yan (1980). The companion notebook FTAP with d Risky Assets presents the simpler case where the strategy space is finite-dimensional (\theta \in \mathbb{R}^{d}), in which simple no-arbitrage suffices.
The L^{p} Payoff Space
Let p \in [1, \infty) and let q denote the conjugate exponent defined by 1/p + 1/q = 1 (with q = \infty when p = 1). Let (\Omega, \mathcal{F}, P) be a probability space and let L^{p} = L^{p}(\Omega, \mathcal{F}, P) denote the space of p-integrable random variables, equipped with the norm \lVert{x}\rVert_{p} = \operatorname{E}(|x|^{p})^{1/p}. By the Riesz representation theorem, the dual space is (L^{p})^{*} = L^{q}. The positive cone is L^{p}_{+} = \{x \in L^{p} : x \geq 0 \text{ a.s.}\}, and we write x > 0 to mean x \geq 0 a.s. and P(x > 0) = 1, i.e., x is strictly positive almost surely. We also write L^{p}_{++} = \{x \in L^{p} : x > 0 \text{ a.s.}\}.
Unlike in \mathbb{R}^{S}, the positive cone L^{p}_{+} has empty interior: every open ball around a non-negative function contains functions that dip below zero on sets of positive measure. For example, if x \in L^{p}_{+} and A has positive probability, then x - \varepsilon \mathbf{1}_{A} is negative on A for large enough \varepsilon > 0. This single fact is the source of all the additional difficulty in the infinite-dimensional proof.
When p = 2, L^{2} is a Hilbert space under the inner product \langle{x, y}\rangle = \operatorname{E}(xy), with (L^{2})^{*} = L^{2} (self-duality). For p \neq 2, L^{p} is a Banach space with (L^{p})^{*} = L^{q}, q \neq p. The proof below applies uniformly to all p \in [1, \infty), though certain steps simplify for p = 2.
The Trading Model
Let X \subseteq L^{p} be a closed subspace of traded payoffs, and let \pi : X \rightarrow \mathbb{R} be a continuous linear pricing functional (the law of one price). Define the cone of claims attainable at non-positive cost by K = \{y \in L^{p} : \exists\, x \in X \text{ with } \pi(x) \leq 0 \text{ and } y \leq x \text{ a.s.}\}. Economically, K is the set of payoffs that can be super-replicated with initial cost at most zero. Any payoff that is dominated by a zero-cost traded payoff belongs to K. In particular, -L^{p}_{+} \subseteq K: every non-positive payoff is in K (take x = 0). A free lunch is a sequence (y_{n}) \subseteq K that converges in L^{p} to some y \in L^{p}_{+} with y \neq 0: the payoffs y_{n} approximate an arbitrage in the L^{p} sense, even if no exact arbitrage exists. Denote by \bar{K} the closure of K in L^{p}; then \bar{K} is a closed convex cone satisfying -L^{p}_{+} \subseteq \bar{K}.
Assumption 1 (No Free Lunch (L^{p})) \bar{K} \cap L^{p}_{+} = \{0\}.
This says no sequence of dominated zero-cost payoffs can converge (in L^{p}) to a non-negative, non-zero payoff. The weaker condition K \cap L^{p}_{+} = \{0\} (no exact free lunch in K) is not sufficient in infinite dimensions: one can construct examples of markets where no exact arbitrage in traded assets exists yet there are free lunches — sequences of strategies whose downside vanishes while their upside remains bounded. The No Free Lunch condition rules out both.
Clark (1993) formulates a slightly weaker no-free-lunch axiom (his A.5): only strictly positive payoffs (x > 0 a.s.) are required to lie outside \bar{K}, i.e., \bar{K} \cap L^{p}_{++} = \emptyset. The condition above is stronger: it rules out all non-zero non-negative payoffs. The stronger form simplifies Step 2 of the proof, which takes x = \mathbf{1}_{A} for sets A of positive probability; since \mathbf{1}_{A} is not strictly positive when P(A^{c}) > 0, Clark’s A.5 alone does not guarantee \mathbf{1}_{A} \notin \bar{K}. Clark resolves this via a supporting set argument for each x \in L^{p}_{++}; the condition above instead allows Step 2 to proceed directly.
The Fundamental Theorem
Property 1 (Fundamental Theorem of Asset Pricing) Assume:
- X \subseteq L^{p} is a closed subspace of traded payoffs,
- \pi : X \to \mathbb{R} is a continuous linear pricing functional (law of one price),
- no free lunch holds: \bar{K} \cap L^{p}_{+} = \{0\},
- there exists a traded numeraire x_{0} \in X with x_{0} > 0 a.s. and \pi(x_{0}) > 0.
Then there exists m \in L^{q} with m > 0 a.s. such that \pi(x) = \operatorname{E}(mx) \quad \text{for all } x \in X.
The random variable m is a strictly positive stochastic discount factor. The measure \tilde{P} defined by d\tilde{P}/dP = m/\operatorname{E}(m) satisfies \tilde{P} \sim P (the two measures share the same null sets). In this static setting, \tilde{P} is an equivalent pricing measure; in dynamic models it plays the role of an equivalent martingale measure. The SDF lies in the dual space L^{q}: when p = 1, m \in L^{\infty} is bounded; when p = 2, m \in L^{2} is square-integrable.
Proof Tools
The proof uses two results. The first holds in any L^{p} space; the second is a measure-theoretic fact about families of measures.
Roadmap: Hahn-Banach gives point-wise separators in L^{q} = (L^{p})^{*}; Halmos-Savage reduces the resulting uncountable family to a countable subfamily that preserves null sets.
Lemma 1 (Hahn-Banach Separation in L^{p}) Let C \subseteq L^{p} be a non-empty closed convex set, and let x \notin C. Then there exists z \in L^{q} such that \operatorname{E}(z\, y) \leq 0 < \operatorname{E}(z\, x) \quad \text{for all } y \in C, provided C is a cone containing 0.
This follows from the Hahn-Banach separation theorem applied in the locally convex space L^{p}, together with the duality (L^{p})^{*} = L^{q}. The separator is a bounded linear functional on L^{p}; by the Riesz representation theorem it is represented as f(\cdot) = \operatorname{E}(z\,\cdot) for a unique z \in L^{q}. When p = 2, the self-duality (L^{2})^{*} = L^{2} means the separator is already an element z \in L^{2}: the Hilbert-space projection theorem and Hahn-Banach collapse into one step, and no separate identification via Riesz is needed. This is the key advantage of the L^{2} setting: the dual is the same space, whereas for p \neq 2 the separator lives in a different space L^{q} \neq L^{p}.
Lemma 2 (Halmos-Savage Theorem) Let \{\mu_{\alpha} : \alpha \in I\} be a family of measures on (\Omega, \mathcal{F}), each absolutely continuous with respect to P. If for every measurable A with P(A) > 0 there exists some \alpha with \mu_{\alpha}(A) > 0, then there exists a countable subfamily \{\mu_{\alpha_{n}}\}_{n \geq 1} with the same null sets as the full family.
This theorem reduces an uncountable family of measures to a countable one without losing information about null sets. It is the step that has no finite-dimensional analogue: in \mathbb{R}^{S} a single global separator suffices, but in L^{p} we can only separate point-by-point and must then stitch the resulting separators together.
Proof of Property 1
The proof has four steps. Step 1 applies Hahn-Banach to each indicator function \mathbf{1}_{A}, building a family of non-negative separators indexed by positive-measure sets that collectively charges every non-null event. Step 2 uses Halmos-Savage to extract a countable equivalent subfamily. Step 3 averages that subfamily with geometrically-decaying weights to form a strictly positive density \rho \in L^{q}, normalized to m. Step 4 verifies that m prices correctly.
Step 1: Build a covering family of separators. For each measurable A with P(A) > 0, apply Lemma 1 to \mathbf{1}_{A} \in L^{p}_{+}: since \mathbf{1}_{A} \notin \bar{K} by no free lunch, there exists z_{A} \in L^{q} with \operatorname{E}(z_{A}\, y) \leq 0 \quad \text{for all } y \in \bar{K}, \qquad \operatorname{E}(z_{A}\, \mathbf{1}_{A}) > 0. Since -L^{p}_{+} \subseteq \bar{K}, for any w \geq 0 we have \operatorname{E}(z_{A}\, w) \geq 0. Taking w = \mathbf{1}_{\{z_{A} < 0\}} gives \operatorname{E}(z_{A}\, \mathbf{1}_{\{z_{A} < 0\}}) \geq 0, but z_{A} < 0 on \{z_{A} < 0\}, so P(z_{A} < 0) = 0, i.e., z_{A} \geq 0 a.s. Normalize to \lVert{z_{A}}\rVert_{q} = 1 (when q < \infty) or \lVert{z_{A}}\rVert_{\infty} = 1 (when q = \infty). Define the measure \mu_{A}(B) := \operatorname{E}(z_{A}\,\mathbf{1}_{B}). Then \mu_{A}(A) = \operatorname{E}(z_{A}\,\mathbf{1}_{A}) > 0, so the family \{\mu_{A} : P(A) > 0\} charges every positive-P-measure set by construction.
Step 2: Extract a countable equivalent subfamily. Each individual separator z_{A} may vanish on most of \Omega; what we need is a single density that is positive everywhere. By Lemma 2, there exists a countable sequence of sets \{A_{n}\}_{n \geq 1} (with P(A_n) > 0 for each n) such that the measures \{\mu_{A_{n}}\} have the same null sets as the full family \{\mu_{A}\}. In particular, if P(B) > 0 then \operatorname{E}(z_{A_{n}}\, \mathbf{1}_{B}) > 0 for some n.
Step 3: Construct the SDF. We average the countable family with geometrically-decaying weights to produce a single density. Define \rho = \sum_{n=1}^{\infty} 2^{-n} z_{A_{n}}. When q < \infty: the triangle inequality gives \lVert{\rho}\rVert_{q} \leq \sum_{n=1}^{\infty} 2^{-n}\lVert{z_{A_{n}}}\rVert_{q} = 1, so \rho \in L^{q}. When q = \infty (i.e., p = 1): since 0 \leq z_{A_{n}} \leq 1 a.s., we have 0 \leq \rho \leq 1 a.s. By Step 2, \rho > 0 a.s.: if P(\rho = 0) > 0 then there is a set B with P(B) > 0 where every summand 2^{-n} z_{A_{n}} vanishes, so \operatorname{E}(z_{A_{n}}\,\mathbf{1}_{B}) = 0 for all n, contradicting the null-set equivalence. Since \rho \in L^{q} and P is a probability measure, 0 < \operatorname{E}(\rho) \leq \lVert{\rho}\rVert_{q} \leq 1. Set m = \rho / \operatorname{E}(\rho) \in L^{q}; then m > 0 a.s. and \operatorname{E}(m) = 1.
Step 4: m prices all assets correctly. Define \ell : X \to \mathbb{R} by \ell(x) := \operatorname{E}(mx), well-defined by Hölder’s inequality.
Claim 1: \ell(y) \leq 0 for all y \in \bar{K}. Each z_{A_{n}} satisfies \operatorname{E}(z_{A_{n}}\, y) \leq 0 for all y \in \bar{K}. By Hölder, |\operatorname{E}(z_{A_{n}}\, y)| \leq \lVert{z_{A_{n}}}\rVert_{q}\lVert{y}\rVert_{p} = \lVert{y}\rVert_{p}, so the series \sum_{n=1}^{\infty} 2^{-n}\operatorname{E}(z_{A_{n}}\, y) is dominated by \lVert{y}\rVert_{p} and converges. Swapping sum and expectation by dominated convergence: \operatorname{E}(\rho\, y) = \sum_{n=1}^{\infty} 2^{-n} \operatorname{E}(z_{A_{n}}\, y) \leq 0. Dividing by \operatorname{E}(\rho) > 0 gives \ell(y) = \operatorname{E}(my) \leq 0 for all y \in \bar{K}.
Proportionality. For any x \in \ker \pi, both x and -x lie in K \subseteq \bar{K}, so Claim 1 gives \ell(x) \leq 0 and \ell(-x) \leq 0, hence \ell(x) = 0. Thus \ker \pi \subseteq \ker \ell. Since \pi \not\equiv 0 (assumption 4 gives \pi(x_{0}) > 0), the quotient X / \ker \pi \cong \mathbb{R} is one-dimensional, and any functional vanishing on \ker \pi is a scalar multiple of \pi. Hence \ell = c\pi for some c \in \mathbb{R}.
c > 0. The numeraire x_{0} satisfies x_{0} > 0 a.s. and m > 0 a.s., so \ell(x_{0}) = \operatorname{E}(mx_{0}) > 0. Therefore c = \ell(x_{0})/\pi(x_{0}) > 0.
Rescale. Set \hat{m} = m/c. Then \hat{m} \in L^{q}, \hat{m} > 0 a.s. and \operatorname{E}(\hat{m}\,x) = \frac{\ell(x)}{c} = \frac{c\,\pi(x)}{c} = \pi(x) \qquad \text{for all } x \in X.
Why the Finite-Dimensional Proof Is Simpler
The proof above has three steps that vanish entirely in \mathbb{R}^{S}:
No need for point-wise separation. In \mathbb{R}^{S}, the positive orthant \mathbb{R}^{S}_{++} is open, so the relevant pricing set and \mathbb{R}^{S}_{++} are disjoint convex sets with one of them open. The Separating Hyperplane Theorem applies globally and produces a single separator \phi in one shot. In L^{p}, the positive cone has empty interior, so global separation is not available and one must separate each point x \in L^{p}_{+} individually.
No need for Halmos-Savage. The global separator in \mathbb{R}^{S} is already a single vector \phi \in \mathbb{R}^{S}. In L^{p}, the point-wise separators \{z_{x}\} form an uncountable family, and Halmos-Savage is the mechanism that stitches them into a single density \rho.
Simple no-arbitrage is not enough. In \mathbb{R}^{S}, every linear subspace is automatically closed, so K is automatically closed and no-arbitrage (K \cap \mathbb{R}^{S}_{+} = \{0\}) suffices. In L^{p}, the cone K need not be closed: one can construct markets with no exact arbitrage yet with approximate arbitrages — sequences of dominated zero-cost payoffs converging to a non-negative, non-zero payoff (a free lunch). The correct infinite-dimensional hypothesis is therefore the stronger no free lunch condition \bar{K} \cap L^{p}_{+} = \{0\}.
In this L^{p} model, the no free lunch (NFL) condition is genuinely stronger than simple no arbitrage (NA). The precise difference stems from the possible non-closedness of K. Simple no-arbitrage requires only K \cap L^{p}_{+} = \{0\}, i.e., no exact non-negative non-zero payoff is attainable at non-positive cost. No free lunch imposes the strictly stronger requirement \bar{K} \cap L^{p}_{+} = \{0\}, where \bar{K} denotes the closure of K in the L^{p} topology. When K is not closed, it is possible to satisfy NA while violating NFL: there exist sequences of strategies in K that converge in L^{p} to a strictly positive payoff, even though no single strategy in K delivers a non-negative non-zero payoff exactly.
The companion notebook FTAP with d Risky Assets presents the simpler proof for finite-dimensional strategy spaces. That notebook also explains why the two settings differ: the finite dimensionality of \mathbb{R}^{d} supplies compactness (via Bolzano-Weierstrass) that forces NA to imply closedness, and the L^{1}/L^{\infty} duality (when p = 1) yields a bounded density d\tilde{P}/dP \in L^{\infty}, whereas the present L^{p} proof gives only m \in L^{q}.