Let $p$ and $q$ be two strictly positive reals such that $\frac{1}{p} + \frac{1}{q} = 1$. Show that, for all non-negative reals $a$ and $b$,
$$ab \leqslant \frac{a^{p}}{p} + \frac{b^{q}}{q}$$
You may use the concavity of the logarithm.

Justify that $\forall k \in \llbracket 1 , n \rrbracket , 0 \leqslant X ^ { k } \leqslant 1 + X ^ { n }$.

Let $X : \Omega \rightarrow \mathbb { R }$ be a real-valued random variable. We assume that $X ( \Omega )$ is countable and we use the notation from question 2: $X ( \Omega ) = \left\{ x _ { n } , n \in \mathbb { N } \right\}$ with $a _ { n } = \mathbb { P } \left( X = x _ { n } \right)$. We also assume that, for all integer $n \in \mathbb { N } , X$ admits a moment of order $n$ and that there exists a real $R > 0$ such that $$\mathbb { E } \left( | X | ^ { n } \right) = O \left( \frac { n ^ { n } } { R ^ { n } } \right) \quad \text { when } n \rightarrow + \infty$$ Show that for all $n \in \mathbb { N }$ and all $y \in \mathbb { R } , \left| \mathrm { e } ^ { \mathrm { i } y } - \sum _ { k = 0 } ^ { n } \frac { ( \mathrm { i } y ) ^ { k } } { k ! } \right| \leqslant \frac { | y | ^ { n + 1 } } { ( n + 1 ) ! }$.

Consider a quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$ where $n \in \mathbb{N}$, $\lambda_0, \ldots, \lambda_n \in \mathbb{R}$ and $x_0 < x_1 < \cdots < x_n$ are $n+1$ distinct points in $I$. We assume that the coefficients $(\lambda_j)_{0 \leqslant j \leqslant n}$ are chosen as $$\forall j \in \llbracket 0, n \rrbracket, \quad \lambda_j = \int_I L_j(x) w(x)\,\mathrm{d}x,$$ where $(L_0, \ldots, L_n)$ is the Lagrange basis associated with the points $(x_0, \ldots, x_n)$. Thus, the formula $I_n(f)$ is of order $m \geqslant n$.
By reasoning with the polynomial $\prod_{i=0}^n (X - x_i)$, show that $m \leqslant 2n+1$.

Let $K$ be a closed, bounded and infinite subset of $\mathbb{C}$. Let $Q$ and $R$ be two non-zero polynomials in $\mathbb{C}[X]$. Show that: $$\|Q\|_K \|R\|_K \geq \|QR\|_K.$$

Let $K$ be a closed, bounded and infinite subset of $\mathbb{C}$. We fix two non-zero natural integers $n$ and $m$, and we set: $$C_{n,m}^K = \sup\left\{\left.\frac{\|Q\|_K \|R\|_K}{\|QR\|_K}\right\rvert\, Q \in \mathbb{C}_n[X]\backslash\{0\}, R \in \mathbb{C}_m[X]\backslash\{0\}\right\} \in \mathbb{R} \cup \{+\infty\}$$ Show that $C_{n,m}^K > 1$.
To do this, one may choose two distinct elements $a$ and $b$ in $K$ and verify that, for $\rho \in \mathbb{R}$ sufficiently large, we have $\left\|Q_\rho R_\rho\right\|_K < \left\|Q_\rho\right\|_K \left\|R_\rho\right\|_K$ with $Q_\rho(X) = X - (\rho(b-a)+a)$ and $R_\rho(X) = X - (\rho(a-b)+b)$.

Let $Q \in \mathbb{C}[X]$ be a non-zero polynomial. We set for $p > 0$: $$M_p(Q) = \frac{1}{2\pi} \int_0^{2\pi} \left|Q\left(e^{i\theta}\right)\right|^p d\theta$$ Explain why $M_p(Q)$ is strictly positive for all $p > 0$.

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 $|Q(x)| > |Q(-1)|$.

If $f, g$ have non-negative real coefficients, $h, g \in O_1$, show that $h \prec g \Rightarrow f \circ h \prec f \circ g$.

Let $C$ be a non-empty, convex and closed subset of $\mathbb{R}^d$. Show that for all $(x_1, x_2) \in \mathbb{R}^d \times \mathbb{R}^d$, we have $$\left(\operatorname{proj}_C(x_1) - \operatorname{proj}_C(x_2)\right) \cdot \left(x_1 - x_2\right) \geqslant \left\|\operatorname{proj}_C(x_1) - \operatorname{proj}_C(x_2)\right\|^2$$ and deduce that $\operatorname{proj}_C$ is continuous.

Let $M \in \mathcal{M}_{k \times d}(\mathbb{R}), b = (b_1, \ldots, b_k) \in \mathbb{R}^k$ and $p \in \mathbb{R}^d$. Set $$\alpha := \inf\left\{p \cdot x : x \in \mathbb{R}^d, x \geqslant 0, Mx \leqslant b\right\}$$ and $$\beta := \sup\left\{b \cdot q : q \in \mathbb{R}^k, q \leqslant 0, M^T q \leqslant p\right\}$$ (adopting the convention: $\inf \emptyset = +\infty$ and $\sup \emptyset = -\infty$). Show that $\alpha \geqslant \beta$.

Let $C$ be a non-empty, convex and closed subset of $\mathbb{R}^d$. Show that for all $(x_1, x_2) \in \mathbb{R}^d \times \mathbb{R}^d$, we have $$\left(\operatorname{proj}_C(x_1) - \operatorname{proj}_C(x_2)\right) \cdot \left(x_1 - x_2\right) \geqslant \left\|\operatorname{proj}_C(x_1) - \operatorname{proj}_C(x_2)\right\|^2,$$ and deduce that $\operatorname{proj}_C$ is continuous.

Let $M \in \mathcal{M}_{k \times d}(\mathbb{R}), b = (b_1, \ldots, b_k) \in \mathbb{R}^k$ and $p \in \mathbb{R}^d$. Set $$\alpha := \inf\left\{p \cdot x : x \in \mathbb{R}^d, x \geq 0, Mx \leqslant b\right\}$$ and $$\beta := \sup\left\{b \cdot q : q \in \mathbb{R}^k, q \leqslant 0, M^T q \leqslant p\right\}$$ (adopting the convention: $\inf \emptyset = +\infty$ and $\sup \emptyset = -\infty$).
Show that $\alpha \geqslant \beta$.

Show the interpolation inequality $$\forall f \in \mathcal{C}^{1}([0,1]), \quad \|f\|_{\infty} \leqslant \left\|f^{\prime}\right\|_{\infty} + C\left|f\left(x_{1}\right)\right|$$ with $C = 1$.

We fix two distinct real numbers $x_1 < x_2$ in $[0,1]$. For all $x \in [0,1]$ and $f \in \mathcal{C}^{2}([0,1])$, prove the inequality $$\left|f^{\prime}(x) - \frac{f\left(x_{2}\right) - f\left(x_{1}\right)}{x_{2} - x_{1}}\right| \leqslant \left\|f^{\prime\prime}\right\|_{\infty}.$$

We fix two distinct real numbers $x_1 < x_2$ in $[0,1]$. Conclude the case $K = 2$ by showing the interpolation inequality $$\forall f \in \mathcal{C}^{2}([0,1]), \quad \max\left(\|f\|_{\infty}, \left\|f^{\prime}\right\|_{\infty}\right) \leqslant \left\|f^{\prime\prime}\right\|_{\infty} + C\left(\left|f\left(x_{1}\right)\right| + \left|f\left(x_{2}\right)\right|\right)$$ with $C = 1 + \frac{1}{x_{2} - x_{1}}$.

Show the interpolation inequality $$\forall f \in \mathcal{C}^1([0,1]), \quad \|f\|_\infty \leqslant \left\|f^\prime\right\|_\infty + C\left|f\left(x_1\right)\right|$$ with $C = 1$.

We fix two distinct real numbers $x_1 < x_2$ in $[0,1]$. For all $x \in [0,1]$ and $f \in \mathcal{C}^2([0,1])$, prove the inequality $$\left|f^\prime(x) - \frac{f\left(x_2\right) - f\left(x_1\right)}{x_2 - x_1}\right| \leqslant \left\|f^{\prime\prime}\right\|_\infty.$$

We fix two distinct real numbers $x_1 < x_2$ in $[0,1]$. Conclude the case $K = 2$ by showing the interpolation inequality $$\forall f \in \mathcal{C}^2([0,1]), \quad \max\left(\|f\|_\infty, \left\|f^\prime\right\|_\infty\right) \leqslant \left\|f^{\prime\prime}\right\|_\infty + C\left(\left|f\left(x_1\right)\right| + \left|f\left(x_2\right)\right|\right)$$ with $C = 1 + \frac{1}{x_2 - x_1}$.

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 $k \in \llbracket 0 , n \rrbracket$, $z _ { k } = \cos \left( \frac { k \pi } { n } \right)$. We now assume that $\sup _ { x \in [ - 1,1 ] } | W ( x ) | = \frac { 1 } { 2 ^ { n - 1 } }$. Show that, for all $k \in \llbracket 0 , n \rrbracket$, $$\frac { Q \left( z _ { k } \right) } { \prod _ { \substack { j = 0 \\ j \neq k } } ^ { n } \left( z _ { k } - z _ { j } \right) } \geqslant 0.$$

We consider the vector $$w_3 = \begin{pmatrix}-1\\0\\-1\end{pmatrix}.$$ Show that if $B(v,w_3)<0$, then $d(v_0, s_{w_3}(v)) < d(v_0,v)$.

Show the inequality $$\forall (A, B) \in S_n^{++}(\mathrm{R})^2, \quad \operatorname{det}^{1/n}(A + B) \geq \operatorname{det}^{1/n}(A) + \operatorname{det}^{1/n}(B)$$

Show the inequality
$$\forall ( A , B ) \in S _ { n } ^ { + + } ( \mathrm { R } ) ^ { 2 } , \quad \operatorname { det } ^ { 1 / n } ( A + B ) \geq \operatorname { det } ^ { 1 / n } ( A ) + \operatorname { det } ^ { 1 / n } ( B )$$

We consider $\alpha = (\alpha_i)_{i \in I} \in (\mathbb{R}_+^*)^I$ and $\beta = (\beta_j)_{j \in J} \in (\mathbb{R}_+^*)^J$ such that $\sum_{i \in I} \alpha_i = \sum_{j \in J} \beta_j = 1$. We denote by $\boldsymbol{p}$ the element of $F(\alpha, \beta)$ defined by $p_{ij} = \alpha_i \beta_j > 0$ for all $(i,j) \in I \times J$. Let $C = (C_{ij})_{(i,j) \in I \times J} \in \mathbb{R}_+^{I \times J}$ and $\epsilon > 0$. We consider $J_\epsilon : Q \rightarrow \mathbb{R}$ defined by $$J_\epsilon(\boldsymbol{q}) = \sum_{ij} q_{ij} C_{ij} + \epsilon \operatorname{KL}(\boldsymbol{q}, \boldsymbol{p})$$ where $\mathrm{KL}(\boldsymbol{q}, \boldsymbol{p})$ is defined by taking $X = I \times J$. Show that $J_\epsilon$ is strictly convex on $Q$.

Let $(f^0, g^0) \in \mathbb{R}^{I \times J}$. For all $k \geq 0$, we consider $$g^{k+1} = g_*(f^k) \text{ and } f^{k+1} = f_*(g^{k+1})$$ Show that the sequence $(G(f^k, g^k))_{k \geq 0}$ is increasing.