Let $\lambda > 0$ and let $f \in L^1(\mathbb{R}) \cap \mathcal{C}^1(\mathbb{R})$ such that $\hat{f} \in L^1(\mathbb{R})$ and such that $\hat{f}$ is zero outside the segment $[-\lambda, \lambda]$. We denote by $r_\lambda$ the function from $\mathbb{R}$ to $\mathbb{C}$ such that $r_\lambda(x) = r(\lambda x)$ for all real $x$, where $r(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi}\rho(\xi)\,\mathrm{d}\xi$ and $\rho$ is the function from question 30. We admit that $f * r_\lambda$ is integrable. Show that $f = \lambda f * r_\lambda$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ For any real number $x$, express $\Phi_n^{(n)}(x)$ and $\Phi_n^{(n+1)}(x)$ in terms of $L_n(x)$ and $L_n'(x)$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Use the equality $\Phi_{n+1}^{(n+1)}(x) = \frac{\mathrm{d}^{n+1} x\Phi_n(x)}{\mathrm{d}x^{n+1}}$, which we will justify, to prove the equality $$L_{n+1}(x) = \left(1 - \frac{x}{n+1}\right) L_n(x) + \frac{x}{n+1} L_n'(x)$$ valid for any real number $x$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Use the equality $\Phi_{n+1}^{(n+2)}(x) = \frac{\mathrm{d}^{n+1} \Phi_{n+1}^{(1)}}{\mathrm{d}x^{n+1}}(x)$ to prove the equality $$L_{n+1}'(x) = L_n'(x) - L_n(x)$$ valid for any real number $x$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ The confluent hypergeometric function $M_{a,c}$ is the solution of $$x y''(x) + (c-x) y'(x) - a y(x) = 0$$ satisfying $M_{a,c}(0) = 1$. Show that $L_n$ is a confluent hypergeometric function.

We assume that $I = [a,b]$ with $a < b$, $\forall x \in I, w(x) = 1$ (general weight $w$ in the formula for $e(f)$), and that $f$ is of class $\mathcal{C}^{m+1}$ on $I$, where $m$ is the order of the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$.
For every natural number $m$, consider the function $\varphi_m : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$\forall (x,t) \in \mathbb{R}^2, \quad \varphi_m(x,t) = \begin{cases} (x-t)^m & \text{if } x \geqslant t, \\ 0 & \text{if } x < t. \end{cases}$$
Using the Taylor formula with integral remainder, show that $e(f) = e(R_m)$, where $R_m$ is defined by $$\forall x \in [a,b], \quad R_m(x) = \frac{1}{m!} \int_a^b \varphi_m(x,t) f^{(m+1)}(t)\,\mathrm{d}t.$$

We assume that $I = [a,b]$ with $a < b$, and that $f$ is of class $\mathcal{C}^{m+1}$ on $I$, where $m \geqslant 1$ is the order of the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$.
For every natural number $m$, consider the function $\varphi_m : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$\forall (x,t) \in \mathbb{R}^2, \quad \varphi_m(x,t) = \begin{cases} (x-t)^m & \text{if } x \geqslant t, \\ 0 & \text{if } x < t. \end{cases}$$
Deduce that, if $m \geqslant 1$, $$e(f) = \frac{1}{m!} \int_a^b K_m(t) f^{(m+1)}(t)\,\mathrm{d}t$$ where the function $K_m : [a,b] \rightarrow \mathbb{R}$ is defined by $$\forall t \in [a,b], \quad K_m(t) = e\left(x \mapsto \varphi_m(x,t)\right) = \int_a^b \varphi_m(x,t) w(x)\,\mathrm{d}x - \sum_{j=0}^n \lambda_j \varphi_m(x_j, t).$$ You may use the following admitted result: for every continuous function $g : [a,b]^2 \rightarrow \mathbb{R}$, we have $$\int_a^b \left(\int_a^b g(x,t)\,\mathrm{d}t\right)\mathrm{d}x = \int_a^b \left(\int_a^b g(x,t)\,\mathrm{d}x\right)\mathrm{d}t.$$

We assume that $I = [0,1]$, $\forall x \in I, w(x) = 1$, and we consider the quadrature formula $$I_1(g) = \frac{g(0) + g(1)}{2},$$ which is of order $m = 1$.
Calculate the associated Peano kernel $t \mapsto K_1(t)$ and show that, for every function $g$ of class $\mathcal{C}^2$ from $[0,1]$ to $\mathbb{R}$, we have the following bound on the associated quadrature error: $$|e(g)| \leqslant \frac{1}{12} \sup_{x \in [0,1]} |g''(x)|.$$

For all $n \in \mathbb{N}^{\star}$, justify that there exists a unique function $f_n \in \mathcal{C}^{2}(]0, +\infty[)$ satisfying $f_n(1) = 0$, $f_n(2) = 0$ and $f_n^{\prime\prime}(x) = (-1)^n 2^{-nx^2}$ for all $x > 0$.

To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $$U ( f ) ( x ) = \int _ { 0 } ^ { x } \left( 1 - \mathrm { e } ^ { - t } \right) \frac { f ( t ) } { t } \mathrm {~d} t + \left( \mathrm { e } ^ { x } - 1 \right) \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ Let $f \in E$. Show that $U ( f )$ is of class $\mathcal { C } ^ { 1 }$ on $\mathbb { R } _ { + } ^ { * }$ and satisfies, for all $x > 0$, $$( U ( f ) ) ^ { \prime } ( x ) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$

To each function $f \in E$, we associate the endomorphism $U$ of $E$ defined for all $x > 0$ by $$U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ It has been shown that $U(f)$ satisfies $y'' - y' = -f(x)/x$ on $\mathbb{R}_+^*$. Let $\lambda \in \mathbb { R } ^ { * }$. We assume that $\lambda$ is an eigenvalue of $U$. Let $f$ be an eigenvector associated with it. Show that $f$ is a solution of the differential equation $(E_{1/\lambda}) : x(y'' - y') + \frac{1}{\lambda} y = 0$.

For $p \in \mathbb{N}^*$, let $P$ be a non-zero polynomial solution of $(E_p) : x(y'' - y') + py = 0$. It has been shown that $U(f)$ satisfies $y'' - y' = -f(x)/x$ and that $U$ is self-adjoint ($\langle f | U(g) \rangle = \langle U(f) | g \rangle$). Prove that the function $pU(P) - P$ satisfies on $\mathbb { R } _ { + } ^ { * }$ the differential equation $y ^ { \prime \prime } - y ^ { \prime } = 0$.

We consider $\mathcal{F} : \mathbb{R}^d \rightarrow \mathcal{P}_c(\mathbb{R}^d)$ taking values in the set $\mathcal{P}_c(\mathbb{R}^d)$ of compact subsets of $\mathbb{R}^d$, and the differential inclusion problem: $$\left\{\begin{array}{l} y'(t) \in \mathcal{F}(y(t)) \\ y(0) = y_{\text{init}} \end{array}\right.$$
Show that if for every compact $K \subset \mathbb{R}^d$, there exists $C_K > 0$ such that $\mathcal{F}$ satisfies: $$\forall x, y \in K, \forall v_x \in \mathcal{F}(x), \forall v_y \in \mathcal{F}(y), \quad \langle v_x - v_y, x - y \rangle \leqslant C_K \|x - y\|^2$$ then problem (2) admits at most one maximal solution. (Hint: You may look at $\|X(t) - Y(t)\|^2$ for $X$ and $Y$ two solutions.)

We consider the differential inclusion problem given by $d = 2$ and $\mathcal{F} : \mathbb{R}^2 \rightarrow \mathcal{P}_c(\mathbb{R}^2)$ defined for all $x = (x_1, x_2) \in \mathbb{R}^2$ by: $$\mathcal{F}(x) = \begin{cases} \{v^-\} & \text{if } x_1 < 0 \\ \{v^+\} & \text{if } x_1 > 0 \\ [v_1^+, v_1^-] \times [v_2^+, v_2^-] & \text{if } x_1 = 0 \end{cases}$$ where $v^- = (v_1^-, v_2^-) \in \mathbb{R}^2$ and $v^+ = (v_1^+, v_2^+) \in \mathbb{R}^2$ with $v_1^- \geqslant v_1^+$ and $v_2^- \geqslant v_2^+$.
We set $v^- = (1, 2)$ and $v^+ = (-1, 2)$.
(a) Show that $\mathcal{F}$ satisfies condition (3).
(b) We choose $y_{\text{init}} = (0, 0)$. Find all maximal solutions of problem (2).
(c) We choose $y_{\text{init}} = (1, 0)$. Find all maximal solutions of problem (2).

We consider the differential inclusion problem given by $d = 2$ and $\mathcal{F} : \mathbb{R}^2 \rightarrow \mathcal{P}_c(\mathbb{R}^2)$ defined for all $x = (x_1, x_2) \in \mathbb{R}^2$ by: $$\mathcal{F}(x) = \begin{cases} \{v^-\} & \text{if } x_1 < 0 \\ \{v^+\} & \text{if } x_1 > 0 \\ [v_1^+, v_1^-] \times [v_2^+, v_2^-] & \text{if } x_1 = 0 \end{cases}$$ where $v^- = (v_1^-, v_2^-) \in \mathbb{R}^2$ and $v^+ = (v_1^+, v_2^+) \in \mathbb{R}^2$ with $v_1^- \geqslant v_1^+$ and $v_2^- \geqslant v_2^+$.
We set $v^- = (0, 1)$ and $v^+ = (1, 1)$.
(a) Show that $\mathcal{F}$ does not satisfy condition (3).
(b) We choose $y_{\text{init}} = (1, 0)$. Find all maximal solutions of problem (2).
(c) We choose $y_{\text{init}} = (0, 0)$. Find all maximal solutions of problem (2).

Deduce that for $f \in C^{2}(\mathbf{R}) \cap CL(\mathbf{R})$ such that $f^{\prime}$ and $f^{\prime\prime}$ have slow growth, we have
$$\forall t \in \mathbf{R}_{+}^{*}, \forall x \in \mathbf{R}, \quad \frac{\partial P_{t}(f)(x)}{\partial t} = L\left(P_{t}(f)\right)(x)$$
where $\forall x \in \mathbf{R}, \quad L(f)(x) = f^{\prime\prime}(x) - x f^{\prime}(x).$

We admit that $S$ is of class $C^{1}$ on $\mathbf{R}_{+}^{*}$ and that
$$\forall t \in \mathbf{R}_{+}^{*}, \quad S^{\prime}(t) = \int_{-\infty}^{+\infty} \frac{\partial P_{t}(f)(x)}{\partial t} \left(1 + \ln\left(P_{t}(f)(x)\right)\right) \varphi(x) \mathrm{d}x$$
Show that
$$\forall t \in \mathbf{R}_{+}^{*}, \quad S^{\prime}(t) = \int_{-\infty}^{+\infty} L\left(P_{t}(f)\right)(x) \left(1 + \ln\left(P_{t}(f)(x)\right)\right) \varphi(x) \mathrm{d}x$$

Deduce that for $f \in C^2(\mathbf{R}) \cap CL(\mathbf{R})$ such that $f'$ and $f''$ have slow growth, we have $$\forall t \in \mathbf{R}_+^*, \forall x \in \mathbf{R}, \quad \frac{\partial P_t(f)(x)}{\partial t} = L\!\left(P_t(f)\right)(x),$$ where $L(f)(x) = f''(x) - xf'(x)$.

We admit that $S$ is of class $C^1$ on $\mathbf{R}_+^*$ and that $$\forall t \in \mathbf{R}_+^*, \quad S'(t) = \int_{-\infty}^{+\infty} \frac{\partial P_t(f)(x)}{\partial t}\left(1 + \ln\!\left(P_t(f)(x)\right)\right)\varphi(x)\,\mathrm{d}x.$$ Show that $$\forall t \in \mathbf{R}_+^*, \quad S'(t) = \int_{-\infty}^{+\infty} L\!\left(P_t(f)\right)(x)\left(1 + \ln\!\left(P_t(f)(x)\right)\right)\varphi(x)\,\mathrm{d}x.$$

Show that a power series $f(x) = \sum_{n=0}^{\infty} \frac{c_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ is a solution of a differential equation if and only if its Laplace transform $$\widehat{f}(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} c_n x^n$$ is a solution of a differential equation.

Using equation $(E)$ satisfied by $y$, calculate $b _ { 1 }$.
The equation $(E)$ is $y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }$, and $y$ is the sum of a Dirichlet series $y(x) = \sum_{n=0}^{+\infty} a_n e^{-\lambda_n x}$ with $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.

Let $k \in \mathbf { N } ^ { * }$. Using equation $(E)$ satisfied by $y$, exhibit a recurrence relation linking $b _ { k + 1 } , b _ { k }$ and $d _ { k }$.
The equation $(E)$ is $y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }$, with $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$ and $d_k$ as defined in question 11.

Suppose that $S _ { 0 } > 0$. Show that the function $S$ of the solution triplet $( S , I , R )$ of $(F)$ satisfies the relation
$$\left( - \frac { S ^ { \prime } } { S } \right) ^ { \prime } = - S ^ { \prime } + \frac { S ^ { \prime } } { S }$$
The system $(F)$ is: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$

We suppose in this question that $f \in \mathcal{C}^1(\mathbb{R})$ is convex, and that $f'$ is $L$-Lipschitzian, for some $L > 0$. a) Show that for all $x, y \in \mathbb{R}$ $$\left|f'(x) - f'(y)\right|^2 \leq L(x-y)\left(f'(x) - f'(y)\right)$$ b) Let $x, y \in \mathbb{R}$, and let $\tilde{x} := x - \tau f'(x)$ and $\tilde{y} := y - \tau f'(y)$. Show that $$|\tilde{x} - \tilde{y}|^2 \leq |x-y|^2 - \tau(2 - \tau L)(x-y)\left(f'(x) - f'(y)\right)$$ c) We further suppose that $f$ admits a minimizer $x_*$, and that $0 < \tau \leq 2/L$. Show that the sequence $\left(\left|x_n - x_*\right|\right)_{n \in \mathbb{N}}$ is decreasing. (Recall that $\left(x_n\right)_{n \in \mathbb{N}}$ satisfies the recurrence relation $\forall n \in \mathbb{N},\, x_{n+1} := x_n - \tau f'(x_n)$.)

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, and we fix $\tau$ such that $0 < \tau \leq 2/L$. We further suppose that $f$ is $\alpha$-convex, with $\alpha > 0$, that is $$g(x) := f(x) - \frac{1}{2}\alpha x^2 \quad \text{is a convex function on } \mathbb{R}$$ Justify that $f'(x) - \alpha x$ is an increasing function of $x \in \mathbb{R}$. Deduce that $\alpha \leq L$.