Let $Q \in \mathbb{C}[X]$ be a non-zero polynomial. We define the function: $$\begin{aligned} \varphi : [0,+\infty[ &\rightarrow \mathbb{R} \\ p &\mapsto \begin{cases} \ln\left(M_p(Q)\right) & \text{if } p > 0 \\ 0 & \text{if } p = 0 \end{cases} \end{aligned}$$ where $M_p(Q) = \frac{1}{2\pi}\int_0^{2\pi}\left|Q(e^{i\theta})\right|^p d\theta$. Show carefully that $\varphi$ is differentiable on $]0,+\infty[$ and compute its derivative on this interval.

Let $Q \in \mathbb{C}[X]$ be a non-zero polynomial. We define the function: $$\begin{aligned} \varphi : [0,+\infty[ &\rightarrow \mathbb{R} \\ p &\mapsto \begin{cases} \ln\left(M_p(Q)\right) & \text{if } p > 0 \\ 0 & \text{if } p = 0 \end{cases} \end{aligned}$$ where $M_p(Q) = \frac{1}{2\pi}\int_0^{2\pi}\left|Q(e^{i\theta})\right|^p d\theta$ and $M(Q) = \exp\left(\frac{1}{2\pi}\int_0^{2\pi}\ln\left|Q(e^{i\theta})\right|d\theta\right)$. Compute the limit of $\varphi'$ at $0^+$ then deduce that: $$M_p(Q)^{1/p} \underset{p \rightarrow 0^+}{\longrightarrow} M(Q).$$

For each complex number $w$, we denote by $\operatorname{Re}(w)$ the real part of $w$. Show that, for all $z \in \stackrel{\circ}{\mathbb{D}}$: $$\ln|1-z| = -\operatorname{Re}\left(\sum_{n=1}^{\infty} \frac{z^n}{n}\right)$$ To do this, one may write $z = re^{i\theta}$ with $0 \leq r < 1$ and $\theta \in \mathbb{R}$, then study the function: $$\begin{aligned} F : [0,1[ &\rightarrow \mathbb{R} \\ \rho &\mapsto \ln\left|1 - \rho e^{i\theta}\right| \end{aligned}$$

Let $z \in \stackrel{\circ}{\mathbb{D}}$. Using the result that for all $z \in \stackrel{\circ}{\mathbb{D}}$, $\ln|1-z| = -\operatorname{Re}\left(\sum_{n=1}^{\infty}\frac{z^n}{n}\right)$, deduce that the Mahler measure of the polynomial $X - z$ is 1 and, in the case where $z \neq 0$, that of the polynomial $X - z^{-1}$ is $|z|^{-1}$.

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 $\left|e^{i\theta} - re^{i\psi}\right| \geq |\sin(\theta - \psi)|$.

Let $\lambda$ be the leading coefficient of $Q$ and let $\alpha_1, \ldots, \alpha_n$ be the roots of $Q$ counted with multiplicity. Deduce from the previous questions that: $$M(Q) = |\lambda| \prod_{i=1}^n \max\left\{1, \left|\alpha_i\right|\right\}$$

Deduce that, for all $n \in \mathbb{N}$ and $P \in \mathbb{C}_n[X]$, the function from $\mathbb{R}$ to $\mathbb{C}$, $\theta \mapsto P(\cos\theta)$ is in $\mathcal{S}_n$.
Recall that $\mathcal{S}_n$ is the $\mathbb{C}$-vector space of functions $f : \mathbb{R} \rightarrow \mathbb{C}$ satisfying $$\exists (a_0, \ldots, a_n) \in \mathbb{C}^{n+1}, \quad \exists (b_1, \ldots, b_n) \in \mathbb{C}^n, \quad \forall t \in \mathbb{R}, \quad f(t) = a_0 + \sum_{k=1}^{n}\left(a_k \cos(kt) + b_k \sin(kt)\right)$$

Deduce that, for all $n \in \mathbb{N}$ and $P \in \mathbb{C}_n[X]$, the function from $\mathbb{R}$ to $\mathbb{C}$, $\theta \mapsto P(\cos\theta)$ is in $\mathcal{S}_n$.
Recall that $\mathcal{S}_n$ is the $\mathbb{C}$-vector space of functions $f : \mathbb{R} \rightarrow \mathbb{C}$ satisfying $$\exists (a_0, \ldots, a_n) \in \mathbb{C}^{n+1}, \quad \exists (b_1, \ldots, b_n) \in \mathbb{C}^n, \quad \forall t \in \mathbb{R}, \quad f(t) = a_0 + \sum_{k=1}^{n}\left(a_k \cos(kt) + b_k \sin(kt)\right)$$

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 $C_{n,m}^I$. Deduce from question 4.32 that there exists a polynomial $Q_3$ whose roots are all in $[-1, +\infty[$ and such that the pair $(Q_3, R_2)$ forms 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. Show that: $$Q = \prod_{k=m+1}^{n+m}\left(X - x_k\right) \quad \text{and} \quad R = \prod_{k=1}^{m}\left(X - x_k\right)$$

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, with $Q = \prod_{k=m+1}^{n+m}(X-x_k)$ and $R = \prod_{k=1}^{m}(X-x_k)$.
Verify that for all $x \in ]-\infty, -1[$, we have $|Q(x)| > |Q(-1)|$.

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$. By contradiction, show that $|P(-1)| = \|P\|_I$.
To do this, one may choose a real number $\epsilon > 0$, introduce the segment $I_\epsilon = [-1-\epsilon, 1]$ and bound the quantity: $$\frac{\|Q\|_{I_\epsilon} \|R\|_{I_\epsilon}}{\|P\|_{I_\epsilon}}$$ using question 4.37.

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 $|P(y)| = \|P\|_I$.

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$. Show that $P$ satisfies the differential equation: $$\|P\|_I^2 - P^2 = \frac{1}{(n+m)^2}\left(1 - X^2\right)P'^2$$

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$. Deduce from question 4.41 that: $$\left(1 - X^2\right)P'' - XP' + (n+m)^2 P = 0$$

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$. By considering the function: $$\begin{aligned} f : \mathbb{R} &\rightarrow \mathbb{R} \\ y &\mapsto P(\cos y), \end{aligned}$$ verify that for all $x \in [-1,1]$, $$P(x) = \|P\|_I \cos\left((n+m)\operatorname{Arccos} x\right).$$

We choose $I = [-1,1]$ and fix any very good extremal pair $(Q, R)$. We set $P = QR$. Deduce from question 4.43 that: $$C_{n,m} = 2^{n+m-1} \cdot \left[\prod_{k=1}^n \left(1 + \cos\left(\frac{2k-1}{2(n+m)}\pi\right)\right)\right] \cdot \left[\prod_{k=1}^m \left(1 + \cos\left(\frac{2k-1}{2(n+m)}\pi\right)\right)\right].$$

Show that, for all $n \in \mathbb{N}$, $\left\|T_n'\right\|_{L^\infty([-1,1])} = n^2$.
One may begin by establishing that, for all $n \in \mathbb{N}$ and $\theta \in \mathbb{R}$, $|\sin(n\theta)| \leqslant n|\sin\theta|$.
The sequence of polynomials $\left(T_n\right)_{n \in \mathbb{N}}$ is defined by $T_0 = 1, T_1 = X$ and $\forall n \in \mathbb{N}, T_{n+2} = 2X T_{n+1} - T_n$.

Show that, for all $n \in \mathbb{N}$, $\left\|T_n'\right\|_{L^\infty([-1,1])} = n^2$.
One may begin by establishing that, for all $n \in \mathbb{N}$ and $\theta \in \mathbb{R}$, $|\sin(n\theta)| \leqslant n|\sin\theta|$.
The sequence of polynomials $\left(T_n\right)_{n \in \mathbb{N}}$ is defined by $T_0 = 1, T_1 = X$ and $\forall n \in \mathbb{N}, T_{n+2} = 2X T_{n+1} - T_n$.

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 $$\forall j \in \llbracket 0, n \rrbracket, \quad L_i(x_j) = \begin{cases} 0 & \text{if } j \neq i, \\ 1 & \text{if } j = i. \end{cases}$$