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 $q,r$.
 A: 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 $\Pi$.
