Existence of functionals on $L^0$

Question

Studying a paper about risk measures by F. Delbaen, I bumped into this statement:

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space: if $\mathbb{P}$ is atomless, then there exists no functional $\rho:L^0\to\mathbb{R}$ such that:

• $\rho(X+a)=\rho(X)-a \quad \forall a \in \mathbb{R},$
• $\rho(X+Y)\le \rho(X)+\rho(Y),$
• $\rho(\lambda X)=\lambda\rho(X), \quad \forall \lambda>0,$
• $X\ge 0 \implies \rho(X)\le 0,$

for every $X\in L^0.$

Here we denote by $L^0$ the linear space of all random variables on $\Omega$ with the metric of the convergence in probability.

Then the author assesses that this is a consequence of the analytic Hahn-Banach theorem and of the fact that a continuous functional on $L^0$ must be necessarily null if $\mathbb{P}$ is atomless.

Now, I'm full of doubts: first of all I didn't know the statement about the linear functionals on the $L^0$ space: could you give me some reference where to read about it? I tried to google something but didn't find anything.

Secondly I didn't really undestand how to use in a clever way the Hahn-Banach theorem: this risk functionals were previously introduced on the space $L^\infty,$ where it is easy to check that they are continuous (wrt to $\|\cdot\|_\infty$), so I thought it was natural to use $L^\infty$ as subspace where to use Hahn-Banach, but I don't know which linear functional on $L^\infty$ I shoul use to be sure that it will be continuous on $L^0$ when extended.

Any help would be a lot appreciated. Thanks to everybody.

Answer 1

You can use either the geometric or the analytic Hahn-Banach theorem to prove the following proposition:

Proposition: Let $X$ be a topological vector space (real or complex). There exists nonzero continuous linear functional on $X$ if and only if there is a convex neighborhood of zero that is not the whole $X$.

We shall show that a convex neighborhood of zero in $L^0$ must be $L^0$ itself. Then, in light of the proposition, the only continuous linear functional on $L^0$ is the zero function.

As you mentioned, the space $L^0$ is endowed with the quasinorm $||X|| := \mathbb{E}[1 \wedge |X|]$. Let $V$ be a convex neighborhood of zero in $L^0$. $V$ must contain the centered ball $B_{\delta} := \{X: ||X|| < \delta\}$ for some $\delta > 0$. Now consider arbitrary random variable $X$. Since the probability measure is atomless, there exists a finite partition $A_1, A_2, \dots, A_n$ of $\Omega$, each set having probability measure less than $\delta$ (this is a basic result about finite atomless measure). We note that $X$ admits an expression as convex combination \begin{equation*} X = \frac{1}{n} \sum_{i=1}^n n X \textbf{1}_{A_i}, \end{equation*} where $\textbf{1}_{A_i}$ is the indicator function of $A_i$. The summands satisfy \begin{equation*} ||n X \textbf{1}_{A_i}|| \leq \mathbb{P}(A_i) < \delta, \end{equation*} whence $n X \textbf{1}_{A_i} \in B_\delta \subset V$. Now that $V$ is convex, $X$ must belong to $V$. This proves that $V = L^0$.

Finally, you can verify that the assumptions imposed on $\rho$ suggest that $\rho$ is a nonzero linear functional. Specifically, the second and the third assumptions imply linearity, while the first assumption implies that $\rho$ is nonzero. The above discussion suggests that $\rho$ can never be continuous.