Show that $$\forall x , y \in \mathbf { R } _ { + } , \quad x y \leq \frac { x ^ { p } } { p } + \frac { y ^ { q } } { q }$$ where $p , q \in ] 1 , + \infty [$ such that $\frac { 1 } { p } + \frac { 1 } { q } = 1$.

What inequality do we recover when $p = q = 2$ ? Give a direct proof of it.

Prove that $\varphi$ admits an extension $\psi$ to $V$ compatible with $u$.

Prove that there exists an injective linear application $\varphi : W \rightarrow \mathcal{D}$ such that $\xi \circ \varphi = \varphi \circ u_W$. According to Part III, this linear application $\varphi$ admits an extension $\psi : V \rightarrow \mathcal{D}$ compatible with $u$.

Verify that the image of $\psi$ is contained in the kernel of $\xi^n$.

Prove that the kernel of $\psi$ is a complement of $W$ stable by $u$.

Let $n \in \mathbb { N } ^ { * }$. Verify that
$$\frac { 2 n \ln ( 2 ) } { \ln ( 2 n ) } - 1 \geqslant \frac { n \ln ( 2 ) } { \ln ( 2 n ) }$$
then deduce that
$$\pi ( 2 n ) \geqslant n \frac { \ln ( 2 ) } { \ln ( 2 n ) }$$

Let $n \geq 1$ be an integer. A non-empty compact subset $P \subset \mathbb{R}^n$ is a polytope if there exist a non-empty finite set $I$ and if for all $i \in I$ there exist a linear form $\ell_i : \mathbb{R}^n \rightarrow \mathbb{R}$ and a real number $a_i \in \mathbb{R}$ such that $P = \{x \in \mathbb{R}^n : \ell_i(x) \leq a_i\ \forall i \in I\}$. A face $F$ of $P$ is a non-empty subset such that there exists $J \subset I$ with $F = F_J = \{x \in P : \ell_j(x) = a_j\ \forall j \in J\}$.
Verify that every face $F$ of $P$ is a polytope and that $\operatorname{dim} F < \operatorname{dim} P$ if $F \neq P$.

Let $x \geqslant 3$. Show that
$$\pi ( x ) \geqslant \frac { \ln ( 2 ) } { 6 } \frac { x } { \ln ( x ) }$$
One may set $n = \lfloor x / 2 \rfloor$ and use Q12.

Let $c > 0$, and let $\left(a_n\right)_{n \in \mathbb{N}}$ be a sequence of positive real numbers such that $a_{n+1} \leq a_n - c(a_n)^2$ for all $n \in \mathbb{N}$. Show $a_n \leq a_0/(1 + nca_0)$ for all $n \in \mathbb{N}$. Hint: adapt the reasoning from question 10.c)

Let $V \subset \mathbb{R}^n$ be a non-empty finite set. We assume that $\operatorname{Conv}(V)$ is not contained in a hyperplane of $\mathbb{R}^n$ and contains 0 in its interior. Show that the set $Q$ defined by $$Q = \left\{\ell \in \mathbb{R}^n : \langle \ell, x \rangle \leq 1 \quad \forall x \in V\right\}$$ is a polytope of $\mathbb{R}^n$.

Let $V \subset \mathbb{R}^n$ be a non-empty finite set. We assume that $\operatorname{Conv}(V)$ is not contained in a hyperplane of $\mathbb{R}^n$ and contains 0 in its interior, and that the set $Q = \{\ell \in \mathbb{R}^n : \langle \ell, x \rangle \leq 1\ \forall x \in V\}$ is a polytope of $\mathbb{R}^n$. Deduce that $\operatorname{Conv}(V)$ is a polytope.

We will admit that for every non-empty closed convex set $C \subset \mathbb{R}^n$ and every $x \in \mathbb{R}^n \backslash C$, there exists a unique $y \in C$ such that $\langle x - y, z - y \rangle \leq 0$ for all $z \in C$.
Let $V \subset \mathbb{R}^n$ be a non-empty finite set. Prove that every vertex of $\operatorname{Conv}(V)$ belongs to $V$.

Let $\mathcal{F}_n$ be the $\mathbb{R}$-vector space of functions $f : \mathbb{R}^n \rightarrow \mathbb{R}$. For all $X \subset \mathbb{R}^n$, we denote $\mathbb{1}_X$ the indicator function of $X$. Let $\mathcal{U}_n$ be the vector subspace of $\mathcal{F}_n$ generated by the functions $\mathbb{1}_P$ where $P$ is a polytope of $\mathbb{R}^n$.
Let $f \in \mathcal{U}_1$. Prove that for all $x \in \mathbb{R}$ the limit of $f(y)$ as $y$ tends to $x$ while satisfying $y > x$, denoted $\lim_{y \rightarrow x^+} f(y)$, exists and that there exist finitely many reals $x \in \mathbb{R}$ such that $f(x) \neq \lim_{y \rightarrow x^+} f(y)$.

Let $\mathcal{F}_n$ be the $\mathbb{R}$-vector space of functions $f : \mathbb{R}^n \rightarrow \mathbb{R}$. For all $X \subset \mathbb{R}^n$, we denote $\mathbb{1}_X$ the indicator function of $X$. Let $\mathcal{U}_n$ be the vector subspace of $\mathcal{F}_n$ generated by the functions $\mathbb{1}_P$ where $P$ is a polytope of $\mathbb{R}^n$.
We assume that $n > 1$. Let $f \in \mathcal{U}_n$. For $z \in \mathbb{R}$, we define the function $f_z : \mathbb{R}^{n-1} \rightarrow \mathbb{R}$ by $f_z(x_1, \ldots, x_{n-1}) = f(x_1, \ldots, x_{n-1}, z)$ for all $(x_1, \ldots, x_{n-1}) \in \mathbb{R}^{n-1}$. Prove that $f_z \in \mathcal{U}_{n-1}$.

Let $\mathcal{F}_n$ be the $\mathbb{R}$-vector space of functions $f : \mathbb{R}^n \rightarrow \mathbb{R}$. For all $X \subset \mathbb{R}^n$, we denote $\mathbb{1}_X$ the indicator function of $X$. Let $\mathcal{U}_n$ be the vector subspace of $\mathcal{F}_n$ generated by the functions $\mathbb{1}_P$ where $P$ is a polytope of $\mathbb{R}^n$.
Prove that the following definition allows us to define a linear form $\chi_n : \mathcal{U}_n \rightarrow \mathbb{R}$. We define $\chi_1(f)$ by the sum $\chi_1(f) = \sum_{x \in \mathbb{R}} \left(f(x) - \lim_{y \rightarrow x^+} f(y)\right)$, then for $f \in \mathcal{U}_n$ with $n > 1$, we set $$\chi_n(f) = \chi_1(g) \text{ with } g \text{ defined by } g(z) = \chi_{n-1}(f_z) \text{ for } z \in \mathbb{R}.$$ We will show at the same time the formula $\chi_n(\mathbb{1}_P) = 1$ for every polytope $P$ of $\mathbb{R}^n$ and we will justify that $\chi_n$ is independent of the coordinate system, namely that for every invertible linear map $A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ we have $\chi_n(f \circ A) = \chi_n(f)$ for all $f \in \mathcal{U}_n$.

Let $\mathcal{U}_n$ be the vector subspace of functions $\mathbb{R}^n \rightarrow \mathbb{R}$ generated by indicator functions of polytopes of $\mathbb{R}^n$. For a polytope $P$ of $\mathbb{R}^n$, the relative interior $P^\circ$ is defined as $P^\circ = \{x \in P : \ell_i(x) = a_i \Leftrightarrow i \in S_F\}$ where $S_F = \{i \in I, \ell_i(x) = a_i\ \forall x \in F\}$.
Show that for every polytope $P$ of $\mathbb{R}^n$ and for all $x \in P \backslash P^\circ$, there exists a face $F \subset P$ such that $F \neq P$ and $x \in F$.

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 $\chi_n(\mathbb{1}_{P^\circ}) = (-1)^{\operatorname{dim} P}$.

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, satisfying $\chi_n(\mathbb{1}_{P^\circ}) = (-1)^{\operatorname{dim} P}$ for every polytope $P$.
Deduce Euler's formula $\sum_F (-1)^{\operatorname{dim} F} = 1$ where $F$ ranges over the faces of $P$.

A complex is a non-empty finite set $\mathcal{C}$ of polytopes of $\mathbb{R}^n$ such that for all $P, Q \in \mathcal{C}$, $P \cap Q$ is either empty or simultaneously a face of both $P$ and $Q$. A face of $\mathcal{C}$ is a subset $F \subset |\mathcal{C}|$ that is a face of one of the $P \in \mathcal{C}$.
Show that if $P$ is a polytope of $\mathbb{R}^n$ of dimension $k > 0$, the set of its faces of dimension $k-1$ forms a complex.

A complex is a non-empty finite set $\mathcal{C}$ of polytopes of $\mathbb{R}^n$ such that for all $P, Q \in \mathcal{C}$, $P \cap Q$ is either empty or simultaneously a face of both $P$ and $Q$.
Let $P$ be a polytope of $\mathbb{R}^n$ of dimension $k > 0$ and $x \in P^\circ$. For each face $F$ of dimension $k-1$ of $P$ we denote $F_x = \operatorname{Conv}(F \cup \{x\})$. Show that the family of $F_x$ forms a complex whose realization equals $P$.

A triangulation of a polytope $P$ is a complex formed of simplices whose realization equals $P$. Show that every polytope admits a triangulation.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x, y \in \mathbb{R}$, $\tilde{x} := p_f(x)$, $\tilde{y} := p_f(y)$. We choose $v := \tilde{y} - \tilde{x}$ in inequality $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$ Show that the left-hand side is positive for all $t \in [0,1]$. Deduce that $$|\tilde{x} - \tilde{y}|^2 \leq (x-y)(\tilde{x} - \tilde{y}).$$

A complex $\mathcal{C}$ is a non-empty finite set of polytopes of $\mathbb{R}^n$ such that for all $P, Q \in \mathcal{C}$, $P \cap Q$ is either empty or simultaneously a face of both $P$ and $Q$. We denote $\chi(\mathcal{C}) = \sum_F (-1)^{\operatorname{dim} F}$ where $F$ ranges over the faces of $\mathcal{C}$.
Show that every complex $\mathcal{C}$ whose realization is convex satisfies $\chi(\mathcal{C}) = 1$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. Show that $\left|p_f(x) - p_f(y)\right| \leq |x-y|$ for all $x, y \in \mathbb{R}$. Deduce that the sequence $\left(\left|x_n - x_*\right|\right)_{n \in \mathbb{N}}$ is decreasing.