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I encountered this problem when trying a French agregation subject. I will only mention the relevant parts for my question. I'm not even sure how to tag it, since it contains real analysis, algebra, and it's used to prove a probability result. :)

Denote $L_{1\times 1}$ the space of positive linear forms on the set of measurable functions $h : [0,1]^2 \to \Bbb{R}$ which are bounded bounded (we also know that $\Pi(1)=1$). For each such linear form $\Pi$ we can consider the forms $\Pi_1,\Pi_2$ defined on the set of measurable functions $h:[0,1]\to \Bbb{R}$, which are bounded such that

$$ \Pi_1(f)=\Pi(h) \text{ if }h(x,y)=f(x) $$

and similar for $\Pi_2$. If there exist densities $\ell_1,\ell_2$ (measurable with integral one) such that $$ \Pi_i(f)=\int_0^1 \ell_i(x)f(x)dx$$ for every $f$ then we call $\ell_1,\ell_2$ the marginal densities of $\Pi$.

Consider now two densities $q,r$ and $L(q,r)$ the subspace of $L_{1\times 1}$ of forms $\Pi$ with marginal positive densities $q,r$ (recall that the densities are integrable on $[0,1]$ with integral one).

Consider $1_{x\neq y}$ the characteristic function of $\Bbb{R}^2\setminus \{(x,x) : x \in \Bbb{R}\}$.

It is asked to prove that $$ \frac{1}{2} \int_0^1 |q-r| \leq \Pi(1_{x\neq y})$$

I'm not sure how can I relate $\Pi(1_{x \neq y})$ to the fact that $\Pi$ has marginal densities $q,r$.

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Answers to agregation subjects are found in the RMS (Revue de Mathématiques Spéciales), and there are also other good "annales". Can you tell us what year it is ? – Ewan Delanoy Nov 26 '12 at 4:33
@EwanDelanoy: It is Analyse et Probabilites 2003. – Beni Bogosel Nov 26 '12 at 10:39
@EwanDelanoy: Thank you for the info. I think I might find the respective magazine at the library of the university – Beni Bogosel Nov 26 '12 at 10:45
Yes, if the year is so recent it shouldn't be too difficult to find it. – Ewan Delanoy Nov 27 '12 at 16:39
@EwanDelanoy: I searched it, but I found only the statement of the problem in RMS. Here is the link:… For the section Mathematiques generales I've found the solutions, but for Analyse et Probabilites It seems that there are no solutions presented. – Beni Bogosel Nov 27 '12 at 16:56
up vote 3 down vote accepted

Your question as stated is somewhat misleading, because in summing things up you forgot to mention an important related question elsewhere in the examination text. Also, your integral on $[0,1]$ should actually be on $[0,1]^2$.

The golden rule in such competitive exams : NEVER answer a single question before having read the WHOLE exam text.

In question I.3b) of the text, it is shown that

$$ \frac{1}{2}\int_{[0,1]^n} |q(x)-r(x)|dx= {\sup}_{0 \leq f \leq 1} \bigg| \int_{[0,1]^n} f(x)q(x)dx - \int_{[0,1]^n} f(y)r(y)dy \bigg| $$

where the sup is taken over borelian functions $f$. Once you have this identity, everything becomes simpler. It suffices to show that for any borelian $f$ with $0 \leq f \leq 1$, we have

$$ \bigg| \int_{[0,1]^n} f(x)q(x)dx - \int_{[0,1]^n} f(y)r(y)dy \bigg| \leq \Pi ({\bf 1}_{x \neq y}) $$

But this is easy : let $A(x,y)=f(x), B(x,y)=f(y)$. Then the right-hand side above is exactly $|\Pi (A-B)|$. But $A-B$ is zero when $x=y$, and we also have $$ |A(x,y)-B(x,y)| \leq {\sf max}(A(x,y),B(x,y)) = {\sf max}(f(x),f(y)) \leq 1 $$ So in any case, we have $|A-B| \leq {\bf 1}_{x \neq y}$ and the result follows by the positivity of $\Pi$.

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I did solve all the points up to this one, but I got stuck here... Thank you for the answer. – Beni Bogosel Nov 28 '12 at 18:05

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