# Linear forms, “projections”, an inequality - French agregation

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: rms-math.com/… 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

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