Show, if $f$ and $g \in O_1$ have non-negative real coefficients and if $r \in [0, \infty]$, that $f \circ g(r) = f(g(r))$.

We consider a power series $f = \lambda z + F, F \in O_2, \lambda = (f)_1 \neq 0$, with $h$ the unique series in $O_1$ such that $h \circ f = I$ and $g$ the unique series in $O_1$ such that $f \circ g = I$. Show that $g = h$.

We fix two orthonormal families $u = (u_1, \ldots, u_p)$ of vectors of $V$ and $u^{\prime} = (u_1^{\prime}, \ldots, u_p^{\prime})$ of vectors of $V^{\prime}$ satisfying the conditions of question (1).
(a) Show that there exist $0 \leqslant \theta_1 \leqslant \cdots \leqslant \theta_p \leqslant \pi/2$ such that $\cos(\theta_k) = \langle u_k, u_k^{\prime}\rangle$ for all $k \in \llbracket 1, p \rrbracket$.
(b) Calculate the value of $\det(\operatorname{Gram}(u, u^{\prime}))$ as a function of the $\cos(\theta_k)$.
(c) Deduce that $\det(\operatorname{Gram}(u, u^{\prime})) \leqslant 1$. What can be said about $V$ and $V^{\prime}$ in the case of equality?

Let $C$ be a non-empty closed convex subset of $\mathbb{R}^d$ and let $\sigma_C : \mathbb{R}^d \rightarrow \mathbb{R} \cup \{+\infty\}$ be defined by: $$\sigma_C(p) := \sup\{p \cdot x, x \in C\}$$ show that $$C = \left\{x \in \mathbb{R}^d : p \cdot x \leqslant \sigma_C(p), \forall p \in \mathbb{R}^d\right\}$$ (so that $C$ is an intersection of closed half-spaces).

Let $K$ be a non-empty, convex, closed and bounded subset of $\mathbb{R}^d$. Show that $K = \operatorname{co}(\operatorname{Ext}(K))$.

For all $x = (x_1, \ldots, x_d) \in \mathbb{R}^d$, we set $$\|x\|_1 := \sum_{i=1}^d |x_i|, \quad \|x\|_\infty := \max\{|x_i|, i = 1, \ldots, d\}.$$ Show that for all $x \in \mathbb{R}^d$, we have $$\|x\|_1 = \max\left\{x \cdot y, y \in \mathbb{R}^d, \|y\|_\infty \leqslant 1\right\}$$ and $$\|x\|_\infty = \max\left\{x \cdot y, y \in \mathbb{R}^d, \|y\|_1 \leqslant 1\right\}.$$

Let $C$ be a non-empty closed convex subset of $\mathbb{R}^d$ and let $\sigma_C : \mathbb{R}^d \rightarrow \mathbb{R} \cup \{+\infty\}$ be defined by: $$\sigma_C(p) := \sup\{p \cdot x, x \in C\}$$ show that $$C = \left\{x \in \mathbb{R}^d : p \cdot x \leqslant \sigma_C(p), \forall p \in \mathbb{R}^d\right\}$$ (so that $C$ is an intersection of closed half-spaces).

Let $K$ be a non-empty, convex, closed and bounded subset of $\mathbb{R}^d$. Show that $K = \operatorname{co}(\operatorname{Ext}(K))$.

For all $x = (x_1, \ldots, x_d) \in \mathbb{R}^d$, we set $$\|x\|_1 := \sum_{i=1}^d |x_i|, \quad \|x\|_\infty := \max\{|x_i|, i = 1, \ldots, d\}.$$ Show that for all $x \in \mathbb{R}^d$, we have $$\|x\|_1 = \max\left\{x \cdot y, y \in \mathbb{R}^d, \|y\|_\infty \leqslant 1\right\},$$ and $$\|x\|_\infty = \max\left\{x \cdot y, y \in \mathbb{R}^d, \|y\|_1 \leqslant 1\right\}.$$

Let $u \in \operatorname { Symp } _ { \omega } ( E )$ be a symplectic endomorphism of $E$. Let $\lambda , \mu$ be real eigenvalues of $u$, and let $E _ { \lambda } ( u ) , E _ { \mu } ( u )$ be the associated eigenspaces. Show that, if $\lambda \mu \neq 1$, then the subspaces $E _ { \lambda } ( u )$ and $E _ { \mu } ( u )$ are $\omega$-orthogonal, that is:
$$\forall x \in E _ { \lambda } ( u ) , \quad \forall y \in E _ { \mu } ( u ) , \quad \omega ( x , y ) = 0$$

Show that, if $\omega$ is a symplectic form on $E$, then for every vector $x$ in $E$, $\omega ( x , x ) = 0$.

Let $u \in \operatorname { Symp } _ { \omega } ( E )$ be a symplectic endomorphism of $E$. Let $\lambda , \mu$ be real eigenvalues of $u$, and let $E _ { \lambda } ( u ) , E _ { \mu } ( u )$ be the associated eigenspaces. Show that, if $\lambda \mu \neq 1$, then the subspaces $E _ { \lambda } ( u )$ and $E _ { \mu } ( u )$ are $\omega$-orthogonal, that is:
$$\forall x \in E _ { \lambda } ( u ) , \quad \forall y \in E _ { \mu } ( u ) , \quad \omega ( x , y ) = 0$$

Show that, for all $i$ and $k$ in $\llbracket 1 , n \rrbracket$, $$L _ { i } \left( a _ { k } \right) = \begin{cases} 1 & \text { if } k = i \\ 0 & \text { otherwise } \end{cases}$$ where $L_i(X) = \prod _ { \substack { j = 1 \\ j \neq i } } ^ { n } \frac { X - a _ { j } } { a _ { i } - a _ { j } }$.

Show that, for all $i \in \llbracket 1 , n \rrbracket$ and all $P \in \mathbb { R } _ { n - 1 } [ X ]$, $$\left\langle L _ { i } , P \right\rangle = P \left( a _ { i } \right).$$

Show that, for any polynomial $P$ of degree at most $n - 2$, $$\sum _ { i = 1 } ^ { n } \frac { P \left( a _ { i } \right) } { \prod _ { \substack { j = 1 \\ j \neq i } } ^ { n } \left( a _ { i } - a _ { j } \right) } = 0 .$$

Let $\mu_1$ and $\mu_2$ be two probabilities on $\mathbb{N}^*$. We assume that $\forall r \in \mathbb{N}^*, \mu_1(\mathbb{N}^* r) = \mu_2(\mathbb{N}^* r)$. We want to show that $\mu_1 = \mu_2$.
We recall that we denote by $(p_i)_{i \in \mathbb{N}^*}$ the sequence of prime numbers, ordered in increasing order. Show that for all $r \in \mathbb{N}^*$ and all integer $n \geqslant 1$: $$\bigcup_{i=1}^{n+1} \mathbb{N}^* r p_i = \left(\bigcup_{i=1}^{n} \mathbb{N}^* r p_i\right) \cup \left(\mathbb{N}^* r p_{n+1} \backslash \bigcup_{i=1}^{n} \mathbb{N}^* r p_{n+1} p_i\right).$$

Let $\mu_1$ and $\mu_2$ be two probabilities on $\mathbb{N}^*$. We suppose that $\forall r \in \mathbb{N}^*, \mu_1(\mathbb{N}^* r) = \mu_2(\mathbb{N}^* r)$, where $\mathbb{N}^* r$ denotes the set of strictly positive multiples of $r$. We recall that $(p_i)_{i \in \mathbb{N}^*}$ denotes the sequence of prime numbers, ordered in increasing order. Show that for all $r \in \mathbb{N}^*$ and all integer $n \geqslant 1$: $$\bigcup_{i=1}^{n+1} \mathbb{N}^* r p_i = \left(\bigcup_{i=1}^{n} \mathbb{N}^* r p_i\right) \cup \left(\mathbb{N}^* r p_{n+1} \backslash \bigcup_{i=1}^{n} \mathbb{N}^* r p_{n+1} p_i\right)$$

In this part we consider a map $\varphi$ from $\mathbf { R } ^ { n }$ to $\mathbf { R } ^ { n }$ of class $\mathcal { C } ^ { 1 }$ such that $\varphi ( 0 ) = 0$, and denoting $a = d \varphi ( 0 )$, such that all eigenvalues of $a$ have strictly negative real part. Let $b(x,y) = \int_0^{+\infty} \langle e^{ta}(x) \mid e^{ta}(y) \rangle\, dt$ be the inner product on $\mathbf{R}^n$, and $q$ the associated quadratic form, i.e., $q(x) = b(x,x)$ for all $x \in \mathbf{R}^n$.
Prove then that: $$\forall x \in \mathbf { R } ^ { n } , \quad d q ( x ) ( a ( x ) ) = 2 b ( x , a ( x ) ) = - \| x \| ^ { 2 }$$

In this part we consider a map $\varphi$ from $\mathbf { R } ^ { n }$ to $\mathbf { R } ^ { n }$ of class $\mathcal { C } ^ { 1 }$ such that $\varphi ( 0 ) = 0$, and denoting $a = d \varphi ( 0 )$, such that all eigenvalues of $a$ have strictly negative real part. Let $b(x,y) = \int_0^{+\infty} \langle e^{ta}(x) \mid e^{ta}(y) \rangle\, dt$ and $q(x) = b(x,x)$. Let $x_0 \in \mathbf{R}^n$ and $f_{x_0}$ the solution of $y' = \varphi(y),\ y(0) = x_0$. For any function $y$ defined on $\mathbf{R}_+$, define: $$\varepsilon ( y ) : \mathbf { R } _ { + } \rightarrow \mathbf { R } ^ { n }, \quad t \mapsto \varphi ( y ( t ) ) - a ( y ( t ) )$$
Verify the equality $$\forall t \in \mathbf { R } _ { + } , \quad q \left( f _ { x _ { 0 } } \right) ^ { \prime } ( t ) = - \left\| f _ { x _ { 0 } } ( t ) \right\| ^ { 2 } + 2 b \left( f _ { x _ { 0 } } ( t ) , \varepsilon \left( f _ { x _ { 0 } } ( t ) \right) \right)$$

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function of class $C^{\infty}$ with compact support. For all $n \in \mathbb{N}$, we set $$M_{n} = \sup_{x \in \mathbb{R}} \left|f^{(n)}(x)\right| = \left\|f^{(n)}\right\|_{\infty}$$ In this part we assume that $f$ is not identically zero, and $x_{0} \in \operatorname{Supp}(f)$ is such that for all integers $n \geqslant 0$, $f^{(n)}(x_{0}) = 0$.
Show that for all $x \in \mathbb{R}$ and all $n \in \mathbb{N}$, we have
$$f(x) = \int_{x_{0}}^{x} \frac{(x-t)^{n}}{n!} f^{(n+1)}(t)\, dt$$

Let $T$ be a closed subset of $\mathcal{C}$, such that there exist $x, y \in T$ with $y \notin \{-x, x\}$. We assume that for all $a, b \in T$ with $b \notin \{-a, a\}$, we have that $\dfrac{b-a}{\|b-a\|_2}$ and $\dfrac{b+a}{\|b+a\|_2}$ belong to $T$. Show that $T = \mathcal{C}$.

Let $(f^0, g^0) \in \mathbb{R}^{I \times J}$. For all $k \geq 0$, we consider $$g^{k+1} = g_*(f^k) \text{ and } f^{k+1} = f_*(g^{k+1})$$ Assume that there exist $f^\infty = (f_i^\infty)_{i \in I}$ and $g^\infty = (g_j^\infty)_{j \in J}$ such that $|f_i^k - f_i^\infty| \rightarrow 0$ and $|g_j^k - g_j^\infty| \rightarrow 0$ for all $i \in I$ and $j \in J$. We denote $G_* = \sup\{G(f,g) \mid (f,g) \in \mathbb{R}^I \times \mathbb{R}^J\}$.
(a) Show that $G(f^\infty, g^\infty) = G_*$.
(b) Show that $G(f(\epsilon), g(\epsilon)) = G_*$.
(c) Show that there exists a constant $a \in \mathbb{R}$ such that $f(\epsilon)_i = f_i^\infty + a$ and $g(\epsilon)_j = g_j^\infty - a$ for all $(i,j) \in I \times J$.
(d) Deduce that $q(f^k, g^k) \rightarrow q(\epsilon)$.

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$.
Compare the real numbers $- x _ { n , k }$ and $x _ { n , n - k }$.

Let $G$ be a graph and $G ^ { \prime }$ a copy of $G$. Justify that $\chi _ { G } = \chi _ { G ^ { \prime } }$.

Let $\mathcal{U}_n$ be the vector subspace of functions $\mathbb{R}^n \rightarrow \mathbb{R}$ generated by indicator functions of polytopes of $\mathbb{R}^n$, and $\chi_n : \mathcal{U}_n \rightarrow \mathbb{R}$ the linear form defined recursively. For a polytope $P$ of $\mathbb{R}^n$, let $P^\circ$ denote its relative interior.
Show that for every polytope $P$ of $\mathbb{R}^n$, $\mathbb{1}_{P^\circ} \in \mathcal{U}_n$ and $\chi_n(\mathbb{1}_{P^\circ}) = (-1)^{\operatorname{dim} P}$.