Let $n \in \mathbb{N}^*$, $W$ be a monic polynomial of degree $n$, $Q = \frac { 1 } { 2 ^ { n - 1 } } T _ { n } - W$, and for all $k \in \llbracket 0 , n \rrbracket$, $z _ { k } = \cos \left( \frac { k \pi } { n } \right)$. In this question, we prove $$\sup _ { x \in [ - 1,1 ] } | W ( x ) | \geqslant \frac { 1 } { 2 ^ { n - 1 } }$$ by contradiction.

• If we assume that $\sup _ { x \in [ - 1,1 ] } | W ( x ) | < \frac { 1 } { 2 ^ { n - 1 } }$, show that, for all $k \in \llbracket 0 , n - 1 \rrbracket , Q \left( z _ { k } \right) Q \left( z _ { k + 1 } \right) < 0$.
• Deduce a contradiction and conclude.

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p)$, with $q$ a natural integer such that $q > p$ and $\left| a _ { n + 1 } \right| \geqslant \frac { \left| a _ { n } \right| } { 2 ( n + 1 ) }$ for all $n \geqslant q$. Deduce that, for all integer $n \geqslant q , \left| a _ { n } \right| \geqslant \frac { q ! \left| a _ { q } \right| } { 2 ^ { n - q } n ! }$.

Let $\mathcal{C}^{1}$ be the space of functions of class $C^{1}$ from $[-\pi, \pi]$ to $\mathbf{C}$. For $f \in \mathcal{C}^{1}$, we set $$\|f\|_{\infty} = \max\{|f(t)|; t \in [-\pi, \pi]\} \quad \text{and} \quad V(f) = \int_{-\pi}^{\pi} |f^{\prime}|.$$
Let $f \in \mathcal{C}^{1}$ with real values. We assume that the set $C(f)$ of points in $]-\pi, \pi[$ where the function $f^{\prime}$ vanishes is finite. We denote by $\ell$ the cardinality of $C(f)$ and, if $\ell \geq 1$, we denote by $t_{1} < \cdots < t_{\ell}$ the elements of $C(f)$. We set $t_{0} = -\pi$ and $t_{\ell+1} = \pi$. For $0 \leq j \leq \ell$, let $\psi_{j}$ be the function from $\mathbf{R}$ to $\{0,1\}$ equal to 1 on $\left[f\left(t_{j}\right), f\left(t_{j+1}\right)\right[$ and to 0 on $\mathbf{R} \backslash \left[f\left(t_{j}\right), f\left(t_{j+1}\right)\right[$.
If $y \in \mathbf{R}$, show that the set $f^{-1}(\{y\}) \cap [-\pi, \pi[$ is finite with cardinality bounded by $\ell+1$; we denote this cardinality by $N(y)$. If $y \in \mathbf{R}$, express $N(y)$ in terms of $\psi_{0}(y), \ldots, \psi_{\ell}(y)$. Deduce the inequality $$V(f) \leq 2\max\{N(y); y \in \mathbf{R}\}\|f\|_{\infty}$$

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. We call $\tilde{f}$ the application from $\mathbf{R}_+$ to $\mathbf{R}$, defined by: $$\forall x \in \mathbf{R}^+, \tilde{f}(x) = \ln(f(2x))$$
Suppose here that $x \in \mathbf{R}_+^*$, $(n,p) \in (\mathbf{N}^*)^2$ and $x \leq p$. Verify that $$\tilde{f}(n) - \tilde{f}(n-1) \leq \frac{\tilde{f}(n+x) - \tilde{f}(n)}{x} \leq \frac{\tilde{f}(n+p) - \tilde{f}(n)}{p}$$ and that $(\tilde{f}(n+x) - \tilde{f}(n))$ has a limit as $n$ tends to $+\infty$.

We define a sequence of polynomials $\left( F _ { i } \right) _ { i \geqslant 0 }$ by the following recurrence formula: $$\begin{aligned} F _ { 0 } & = f _ { 0 } + f _ { 1 } X + \cdots + f _ { d } X ^ { d } \\ \text { for } i \geqslant 0 , \quad F _ { i + 1 } & = F _ { i } + R _ { i } \end{aligned}$$ where $R _ { i }$ denotes the remainder of the Euclidean division of $P$ by $F _ { i }$. We denote by $Q _ { i }$ the quotient of the Euclidean division of $P$ by $F _ { i }$. We are furthermore given $r , s \in \mathbb { R } _ { + } ^ { * }$ with $r < \rho$ and we set, for $i \in \mathbb { N }$: $$\alpha _ { i } = s ^ { - d } \cdot \left\| F _ { i } - X ^ { d } \right\| _ { r , s } ; \quad \beta _ { i } = \left\| 1 - Q _ { i } \right\| _ { r , s } ; \quad \varepsilon _ { i } = s ^ { - d } \cdot \left\| R _ { i } \right\| _ { r , s } .$$
Show that we can choose $r$ and $s$ such that $\alpha _ { 0 } + 2 \varepsilon _ { 0 } \leqslant \frac { 1 } { 3 }$ and $\beta _ { 0 } + \varepsilon _ { 0 } \leqslant \frac { 1 } { 3 }$.

We define a sequence of polynomials $\left( F _ { i } \right) _ { i \geqslant 0 }$ by the following recurrence formula: $$\begin{aligned} F _ { 0 } & = f _ { 0 } + f _ { 1 } X + \cdots + f _ { d } X ^ { d } \\ \text { for } i \geqslant 0 , \quad F _ { i + 1 } & = F _ { i } + R _ { i } \end{aligned}$$ where $R _ { i }$ denotes the remainder of the Euclidean division of $P$ by $F _ { i }$. We denote by $Q _ { i }$ the quotient of the Euclidean division of $P$ by $F _ { i }$. We set, for $i \in \mathbb { N }$: $$\alpha _ { i } = s ^ { - d } \cdot \left\| F _ { i } - X ^ { d } \right\| _ { r , s } ; \quad \beta _ { i } = \left\| 1 - Q _ { i } \right\| _ { r , s } ; \quad \varepsilon _ { i } = s ^ { - d } \cdot \left\| R _ { i } \right\| _ { r , s } .$$
Show that, for all $i \in \mathbb { N }$, we have $\alpha _ { i + 1 } \leqslant \alpha _ { i } + \varepsilon _ { i }$ and if $\alpha _ { i + 1 } < 1$ then: $$\beta _ { i + 1 } \leqslant \beta _ { i } + \frac { \beta _ { i } \varepsilon _ { i } } { 1 - \alpha _ { i + 1 } } \quad \text { and } \quad \varepsilon _ { i + 1 } \leqslant \frac { \beta _ { i } \varepsilon _ { i } } { 1 - \alpha _ { i + 1 } } .$$

We define a sequence of polynomials $\left( F _ { i } \right) _ { i \geqslant 0 }$ by the following recurrence formula: $$\begin{aligned} F _ { 0 } & = f _ { 0 } + f _ { 1 } X + \cdots + f _ { d } X ^ { d } \\ \text { for } i \geqslant 0 , \quad F _ { i + 1 } & = F _ { i } + R _ { i } \end{aligned}$$ where $R _ { i }$ denotes the remainder of the Euclidean division of $P$ by $F _ { i }$. We denote by $Q _ { i }$ the quotient of the Euclidean division of $P$ by $F _ { i }$. We set, for $i \in \mathbb { N }$: $$\alpha _ { i } = s ^ { - d } \cdot \left\| F _ { i } - X ^ { d } \right\| _ { r , s } ; \quad \beta _ { i } = \left\| 1 - Q _ { i } \right\| _ { r , s } ; \quad \varepsilon _ { i } = s ^ { - d } \cdot \left\| R _ { i } \right\| _ { r , s } .$$
Deduce that, for all $i \in \mathbb { N }$, we have:

• $\alpha _ { i } \leqslant \alpha _ { 0 } + 2 \cdot \left( 1 - 2 ^ { - i } \right) \cdot \varepsilon _ { 0 }$,
• $\beta _ { i } \leqslant \beta _ { 0 } + \left( 1 - 2 ^ { - i } \right) \cdot \varepsilon _ { 0 }$,
• $\varepsilon _ { i } \leqslant 2 ^ { - i } \cdot \varepsilon _ { 0 }$.

Show that every function bounded in absolute value by a polynomial function in $|x|$ has slow growth.

Show that any function bounded in absolute value by a polynomial function in $|x|$ has slow growth.

We denote by $\mathbf{e}$ the matrix of $\mathcal{M}_{n,1}(\mathbb{R})$ whose coefficients are all equal to 1, $P = I_n - \frac{1}{n}\mathbf{e}\cdot\mathbf{e}^T$, and $\Delta_n$ the set of EDM of order $n$.
Show that a symmetric matrix $D$ of order $n$ with non-negative coefficients and zero diagonal is EDM if and only if $-\frac{1}{2}PDP$ is positive.

We denote by $\mathbf{e}$ the matrix of $\mathcal{M}_{n,1}(\mathbb{R})$ whose coefficients are all equal to 1, $P = I_n - \frac{1}{n}\mathbf{e}\cdot\mathbf{e}^T$, and $\Delta_n$ the set of EDM of order $n$.
Show that every non-zero symmetric matrix with non-negative coefficients and zero diagonal, having a unique strictly positive eigenvalue with eigenspace of dimension 1 and eigenvector $\mathbf{e}$, is EDM.

Show that if $v_n(k)$ is non-zero and $n \geq kr$, then $$k! \leq |D^n v_n(k)| \leq c_2 (AD)^n C^k.$$

We assume that $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence of strictly positive real numbers. We denote by $f$ the step function which, for all $k \in \mathbb { N } ^ { * }$, equals $a _ { k }$ on the interval $[ k - 1 , k [$.
Deduce Carleman's inequality in the case where $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence.

Explain how one can remove the decreasing hypothesis in Carleman's inequality.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
Prove that, for all $\left( x _ { 1 } , \ldots , x _ { n } \right) \in U _ { n } \cap X _ { s } , \left( \prod _ { i = 1 } ^ { n } x _ { i } \right) ^ { 1 / n } \leqslant \frac { 1 } { n } \sum _ { i = 1 } ^ { n } x _ { i }$ and deduce the arithmetic-geometric inequality
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \left( \mathbb { R } _ { + } \right) ^ { n } , \quad \left( \prod _ { i = 1 } ^ { n } x _ { i } \right) ^ { 1 / n } \leqslant \frac { 1 } { n } \sum _ { i = 1 } ^ { n } x _ { i }$$

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$. We assume by contradiction that $\lambda > \mathrm { e }$, where $\lambda$ is the real number from Q17, and $\omega_k$ is as defined in Q18b.
Prove that $\omega _ { 1 } \leqslant \frac { 1 } { \mathrm { e } }$ and that, for all $k$ in $\llbracket 1 , n \rrbracket , \omega _ { k } \leqslant \frac { k } { k + 1 }$.
You may prove, for $k \in \llbracket 1 , n - 1 \rrbracket$, that $\omega _ { k + 1 } ^ { k + 1 } = \frac { 1 } { \lambda } \omega _ { k } ^ { k } \left( 1 - \frac { \omega _ { k } } { k } \right) ^ { - k }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$. We assume by contradiction that $\lambda > \mathrm { e }$, where $\lambda$ is the real number from Q17, and $\omega_k$ is as defined in Q18b.
Reach a contradiction on $\omega _ { n }$. Deduce that, for all $n$ in $\mathbb { N } ^ { * }$, for all $\left( x _ { 1 } , \ldots , x _ { n } \right) \in \left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$ such that $x _ { 1 } + \cdots + x _ { n } = 1$,
$$\sum _ { k = 1 } ^ { n } \left( x _ { 1 } x _ { 2 } \cdots x _ { k } \right) ^ { 1 / k } \leqslant \mathrm { e }$$

Deduce Carleman's inequality:
$$\sum _ { n = 1 } ^ { + \infty } \left( \prod _ { k = 1 } ^ { n } a _ { k } \right) ^ { 1 / n } \leqslant \mathrm { e } \sum _ { n = 1 } ^ { + \infty } a _ { n }$$
for any sequence $\left( a _ { k } \right) _ { k \in \mathbb { N } ^ { * } }$ of strictly positive real numbers such that $\sum a _ { n }$ converges.

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ and $\left( c _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ be two sequences of strictly positive real numbers. Prove that
$$\sum _ { n = 1 } ^ { + \infty } \left( \prod _ { k = 1 } ^ { n } a _ { k } \right) ^ { 1 / n } \leqslant \sum _ { k = 1 } ^ { + \infty } c _ { k } a _ { k } \sum _ { n = k } ^ { + \infty } \frac { 1 } { n } \left( \prod _ { i = 1 } ^ { n } c _ { i } \right) ^ { - 1 / n }$$

We define the sequence $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ by
$$\left\{ \begin{array} { l } b _ { 0 } = - 1 \\ \forall n \in \mathbb { N } ^ { * } , \quad b _ { n } = - \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \frac { 1 } { k + 1 } b _ { n - k } \end{array} \right.$$
The Carleman-Yang inequality states:
$$\sum _ { n = 1 } ^ { + \infty } \left( \prod _ { k = 1 } ^ { n } a _ { k } \right) ^ { 1 / n } \leqslant \mathrm { e } \sum _ { n = 1 } ^ { + \infty } \left( 1 - \sum _ { k = 1 } ^ { + \infty } \frac { b _ { k } } { ( n + 1 ) ^ { k } } \right) a _ { n }$$
Prove that, for all $n$ in $\mathbb { N } ^ { * } , b _ { n } \geqslant 0$. In what way is the previous inequality a refinement of Carleman's inequality?

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(A) $a _ { n } < b _ { n }$ for all $n > 1$
(B) $a _ { n } > b _ { n }$ for all $n > 1$
(C) $a _ { n } = b _ { n }$ for infinitely many $n$
(D) None of the above

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(a) $a _ { n } < b _ { n }$ for all $n > 1$
(b) $a _ { n } > b _ { n }$ for all $n > 1$
(c) $a _ { n } = b _ { n }$ for infinitely many $n$
(d) none of the above.

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(a) $a _ { n } < b _ { n }$ for all $n > 1$
(b) $a _ { n } > b _ { n }$ for all $n > 1$
(c) $a _ { n } = b _ { n }$ for infinitely many $n$
(d) none of the above.

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(A) $a _ { n } < b _ { n }$ for all $n > 1$
(B) $a _ { n } > b _ { n }$ for all $n > 1$
(C) $a _ { n } = b _ { n }$ for infinitely many $n$
(D) None of the above

Define $a _ { n } = \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } \right) ^ { n }$ and $b _ { n } = n ^ { n } ( n ! ) ^ { 2 }$. Recall $n !$ is the product of the first $n$ natural numbers. Then,
(A) $a _ { n } < b _ { n }$ for all $n > 1$
(B) $a _ { n } > b _ { n }$ for all $n > 1$
(C) $a _ { n } = b _ { n }$ for infinitely many $n$
(D) None of the above