Let $F$ be a non-empty closed set of $E$ and $u$ in $E$. Show that there exists $v$ in $F$ such that
$$\forall w \in F, \quad \|u - v\| \leqslant \|u - w\|$$

Let $F$ be a non-empty closed set of $E$ and $u$ in $E$. Show that there exists $v$ in $F$ such that
$$\forall w \in F, \quad \|u - v\| \leqslant \|u - w\|$$

Suppose that there exists $f : \mathbb{N} \rightarrow \mathcal{P}(\mathbb{N})$ bijective. By considering $A = \{x \in \mathbb{N} \mid x \notin f(x)\}$, establish a contradiction.

6. Show that for all $n \in \mathbb{N}$, for every function $f$ of $S_n$ satisfying $\lim_{x \rightarrow \pm\infty} |f(x)| = \pm\infty$, there exists an element $g \in \mathscr{P}_{n+1}$ such that $f \sim g$ (where $\sim$ is the relation defined in I.2).

We denote by $\widehat{S}$ the set of $f \in S_*$ satisfying $\lim_{x \rightarrow -\infty} f(x) = -\infty$ and $\lim_{x \rightarrow +\infty} f(x) = +\infty$. We denote by $\operatorname{Mi}(f)$ the set of minima of $f$ and by $\operatorname{Ma}(f)$ the set of maxima of $f$, so $E(f) = \operatorname{Mi}(f) \cup \operatorname{Ma}(f)$.
1. Let $f \in \widehat{S}$. a. Verify that $\operatorname{Card} \operatorname{Mi}(f) = \operatorname{Card} \operatorname{Ma}(f)$ and that for $y \in \mathbb{R}$, $f^{-1}([-\infty, y])$ is the union of non-empty open intervals that are pairwise disjoint. We denote by $\mathscr{I}(y)$ their set. b. Show that for every element $M$ of $\operatorname{Ma}(f)$, there exists a unique element $m$ of $\operatorname{Mi}(f)$ such that $f(m) < f(M)$ and $m > M$.

We define, for all $x \in \mathbb{R}^{+*}$, $$\Gamma(x) = \int_0^{+\infty} t^{x-1} \mathrm{e}^{-t} \, \mathrm{d}t$$ Justify that we thus define a function on $\mathbb{R}^{+*}$.

Let $n \in \mathbb{N}$. We consider $n+1$ distinct points in $I$, denoted $x_0 < x_1 < \cdots < x_n$.
Show that, for all $i \in \llbracket 0, n \rrbracket$, there exists a unique polynomial $L_i \in \mathbb{R}_n[X]$ such that $$\forall j \in \llbracket 0, n \rrbracket, \quad L_i(x_j) = \begin{cases} 0 & \text{if } j \neq i, \\ 1 & \text{if } j = i. \end{cases}$$

Let $K$ be a closed, bounded and infinite subset of $\mathbb{C}$. Show that, if the inequality $\|Q\|_K \|R\|_K \geq \|QR\|_K$ is an equality, then there exists $z_0 \in K$ such that: $$\left|Q\left(z_0\right)\right| = \|Q\|_K, \quad \left|R\left(z_0\right)\right| = \|R\|_K \quad \text{and} \quad \left|Q\left(z_0\right)R\left(z_0\right)\right| = \|QR\|_K.$$

Let $K$ be a closed, bounded and infinite subset of $\mathbb{C}$. We fix two non-zero natural integers $n$ and $m$. We introduce the $\mathbb{C}$-vector space $V = \mathbb{C}_n[X] \times \mathbb{C}_m[X]$ as well as the set: $$E = \left\{(Q,R) \in V \mid \|Q\|_K = \|R\|_K = 1\right\}.$$ Show that there exists a pair $\left(Q_0, R_0\right) \in E$ such that: $$\left\|Q_0 R_0\right\|_K = \inf\left\{\|QR\|_K \mid (Q,R) \in E\right\}.$$ To do this, one may equip $V$ with the norm defined by $$\|(Q,R)\| = \|Q\|_K + \|R\|_K$$ for $(Q,R) \in V$, then study the application $$\begin{aligned} f : E &\rightarrow \mathbb{R} \\ (Q,R) &\mapsto \|QR\|_K. \end{aligned}$$

Let $Q \in \mathbb{C}[X]$ be a non-zero polynomial. Verify that the integral: $$\int_0^{2\pi} \ln\left|Q\left(e^{i\theta}\right)\right| d\theta$$ converges absolutely in the sense of Definition 1.
One may use the d'Alembert-Gauss theorem.

Using the result of question 2.14, show that, for all $z \in \partial\mathbb{D}$, the Mahler measure of $X - z$ is 1.
To do this, one may be interested in the function: $$\begin{aligned} g : [0,1[ &\rightarrow \mathbb{R} \\ r &\mapsto M(X - rz) \end{aligned}$$ and note that, for all $r \in [0,1[$ and $\theta, \psi \in \mathbb{R}$, we have the inequality $\left|e^{i\theta} - re^{i\psi}\right| \geq |\sin(\theta - \psi)|$.

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. Let $(Q_2, R_2)$ be a good extremal pair. Let $w$ be a root of $Q_2$ and let $S \in \mathbb{C}[X]$ be such that: $$Q_2(X) = (X - w)S(X)$$ By setting: $$S_2(X) = (X + 1 - |w+1|)S(X)$$ show that $(S_2, R_2)$ is a good extremal pair.

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. Show that there exists a polynomial $Q_4$ whose roots are all in $I$ and such that the pair $(Q_4, R_2)$ forms a good extremal pair.
To do this, given a root $w$ of $Q_3$ that is not in $I$, one may introduce the polynomial: $$S_3(X) = \frac{X-1}{X-w} Q_3(X)$$ then one may follow the method used in the two previous questions.

We choose $I = [-1,1]$ and write $C_{n,m}$ instead of $C_{n,m}^I$. Briefly explain why there exists a polynomial $R_4$ such that the pair $(Q_4, R_4)$ forms a very good extremal pair, i.e., a good extremal pair in which all complex roots of $Q_4$ and $R_4$ are contained in $I$.

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$ and denote by $x_1 \leq \ldots \leq x_{n+m}$ the roots of $P$ counted with multiplicity. We are given an integer $k \in \{m+1, m+2, \ldots, m+n-1\}$ and we set: $$S(X) = \left(X - x_k\right)\left(X - x_{k+1}\right).$$ Show that for all $\epsilon > 0$, there exists a polynomial $T \in \mathbb{R}[X]$ such that $S - T$ has degree 1 and: $$\begin{gathered} \|S - T\|_I \leq \epsilon \\ |T(-1)| = |S(-1)| \\ \forall x \in ]-1,1] \backslash ]x_k - \epsilon, x_{k+1} + \epsilon[, |T(x)| < |S(x)|. \end{gathered}$$ Deduce that there exists $y \in ]x_k, x_{k+1}[$ such that $|P(y)| = \|P\|_I$.
To handle this last point, one may proceed by contradiction, write $Q$ in the form $SU$ for a certain polynomial $U$, then verify that if $\epsilon$ is chosen appropriately, the pair $(TU, R)$ forms a very good extremal pair.

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$ and denote by $x_1 \leq \ldots \leq x_{n+m}$ the roots of $P$ counted with multiplicity. Following the method used in question 4.39, show that there exists an element $y \in ]x_m, x_{m+1}[$ such that $|P(y)| = \|P\|_I$.

The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$.
Show that $\int _ { 1 } ^ { + \infty } \frac { q ( u ) } { e ^ { t u } - 1 } \mathrm {~d} u$ is well-defined for all real $t > 0$.

Show that $\int_{1}^{+\infty} \frac{q(u)}{e^{tu}-1} \mathrm{~d}u$ is well defined for all real $t > 0$.

Let $g : [ 0,1 ] \rightarrow [ 0,1 ]$ be an increasing function (but not necessarily continuous). Show that $g$ has at least one fixed point. Hint: one may consider the set $$E = \{ x \in [ 0,1 ] ; x \leqslant g ( x ) \} .$$

Let $[ a , b ]$ be a closed bounded interval of $\mathbb { R }$. If $\phi : [ a , b ] \rightarrow [ a , b ]$ is continuous, show that $\phi$ has at least one fixed point.

If $\phi : \mathbb { R } \rightarrow \mathbb { R }$ is of class $\mathcal { C } ^ { 1 }$ and satisfies $$\sup \left\{ \left| \phi ^ { \prime } ( x ) \right| ; x \in \mathbb { R } \right\} < 1$$ show that $\phi$ has at least one fixed point (one may study the sign of $x - \phi ( x )$ for $| x |$ sufficiently large). Show that this fixed point is unique.

Let $\ell$ be a strictly positive integer. Let $F$ be a closed subset of $\mathbb { R } ^ { \ell }$ and let $\phi : F \rightarrow F$ be a map. We assume that there exists $k \in [ 0,1 [$ such that $$\forall x \in F , \forall y \in F , \quad \| \phi ( y ) - \phi ( x ) \| \leqslant k \| y - x \| .$$
(a) We choose a point $x _ { 0 } \in F$. Show that the formula $x _ { n + 1 } = \phi \left( x _ { n } \right)$ defines a sequence $\left( x _ { n } \right) _ { n \geqslant 0 }$ of elements of $F$, and that this sequence is convergent in $F$.
(b) Deduce that $\phi$ has a unique fixed point in $F$.
(c) This fixed point being denoted $x ^ { * }$, bound $\left\| x _ { n } - x ^ { * } \right\|$ as a function of $\left\| x _ { 0 } - x ^ { * } \right\|$.
(d) In what precedes, we assume that $$\phi = \underbrace { \theta \circ \cdots \circ \theta } _ { m \text { times } } ,$$ where $\theta : F \rightarrow F$ is a map and $m \geqslant 2$ is an integer. Show that $\theta$ has a fixed point, and a unique one, in $F$.

Let $g : [ 0,1 ] \rightarrow [ 0,1 ]$ be an increasing function (but not necessarily continuous). Show that $g$ has at least one fixed point. Hint: one may consider the set $$E = \{ x \in [ 0,1 ] ; x \leqslant g ( x ) \} .$$

Let $C$ be a non-empty, convex and closed subset of $\mathbb{R}^d$ and $x \in \mathbb{R}^d$, consider: $$\inf_{y \in C} \|x - y\|^2. \tag{1}$$ Show that (1) has a unique solution (that is, there exists a unique $y \in C$ such that $\|x - y\|^2 \leqslant \|x - z\|^2$ for all $z \in C$) which we will call the projection of $x$ onto $C$ and denote $\operatorname{proj}_C(x)$. Show that $x = \operatorname{proj}_C(x)$ if and only if $x \in C$.

Let $C$ and $D$ be two non-empty convex subsets of $\mathbb{R}^d$ such that $C$ is closed and bounded, $D$ is closed, and $C \cap D = \emptyset$. Show that there exist $p \in \mathbb{R}^d$ and $\varepsilon > 0$ such that $$p \cdot x \leqslant p \cdot y - \varepsilon, \forall (x, y) \in C \times D$$ (we say that $C$ and $D$ can be strictly separated).