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 $\forall n \in \mathbb{N},\, x_{n+1} := x_n - \tau f'(x_n)$.)

Show that $c$ is a constant solution of $(E)$, then that $(E)$ admits exactly two constant solutions denoted $c _ { 1 }$ and $c _ { 2 }$ such that $c _ { 1 } < 0 < c _ { 2 }$. Deduce the value of $c$ as a function of $c _ { 1 }$ and $c _ { 2 }$.
We admit that $y$ is decreasing on $\mathbf { R } _ { + }$ and $\lim _ { x \rightarrow + \infty } y ( x ) = c$, where $c \in \mathbf { R }$. The equation $(E)$ is: $$( E ) : \quad y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }.$$

Problem 1: calculation of an integral
For $x \geqslant 0$ we define $$f ( x ) = \int _ { 0 } ^ { \infty } \frac { e ^ { - t x } } { 1 + t ^ { 2 } } \mathrm {~d} t \quad \text { and } \quad g ( x ) = \int _ { 0 } ^ { \infty } \frac { \sin t } { t + x } \mathrm {~d} t$$
Show that on $] 0 , \infty [$ we have $f = g$. For this you may use a differential equation satisfied by $( f - g )$ and use the behavior of $f$ and $g$ at $+ \infty$.

In this question only, we set $f(x) := \frac{1}{2}Lx^2$ for all $x \in \mathbb{R}$, where $L > 0$ is fixed. a) Show that $x_{n+1} = (1 - \tau L)x_n$, then express directly $x_n$ as a function of $x_0$ and $n$. b) We suppose $x_0 \neq 0$. Justify that $x_n \rightarrow 0$ if and only if $0 < \tau < 2/L$.

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 $\alpha \leq L$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, and $f$ is $\alpha$-convex with $\alpha > 0$, that is $g(x) := f(x) - \frac{1}{2}\alpha x^2$ is a convex function on $\mathbb{R}$. Show that $f(x) \geq f(0) + f'(0)x + \alpha x^2/2$ for all $x \in \mathbb{R}$. Deduce that $f$ admits a minimizer on $\mathbb{R}$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$ such that $f'$ is $L$-Lipschitzian, with $L > 0$, $f$ is $\alpha$-convex with $\alpha > 0$, and $x_*$ denotes a minimizer of $f$. Show that for all $x, y \in \mathbb{R}$ $$\alpha|x-y|^2 \leq \left(f'(x) - f'(y)\right)(x-y)$$

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 $S _ { 0 } = 0$. Give the expression of the solution triplet $( S , I , R )$ of system $( F )$.
The system $(F)$ is: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$ with $S_0 + I_0 + R_0 = 1$ and $S_0, I_0, R_0 \in [0,1]$.

Show that if $S _ { 0 } > 0$ then the function $S$ of the solution triplet $( S , I , R )$ of $( F )$ never vanishes, and deduce that $S$ is strictly positive.
The system $(F)$ is: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$

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: $$( F ) : \left\{ \begin{array} { l } S ^ { \prime } ( x ) = - I ( x ) S ( x ) \\ I ^ { \prime } ( x ) = I ( x ) S ( x ) - I ( x ) \\ R ^ { \prime } ( x ) = I ( x ) \\ S ( 0 ) = S _ { 0 } , \quad I ( 0 ) = I _ { 0 } , \quad R ( 0 ) = R _ { 0 } \end{array} \right.$$

With $S _ { 0 } = 1 / 2$, $I _ { 0 } = 1 / 2$, $R _ { 0 } = 0$, and $h : \mathbf { R } _ { + } \rightarrow \mathbf { R }$ defined by
$$\forall x \in \mathbf { R } _ { + } \quad h ( x ) = \ln \left( \frac { S ( x ) } { S _ { 0 } } \right) = \ln ( 2 S ( x ) ),$$
show that $h$ is a solution of the Cauchy problem $(C)$:
$$( C ) : \left\{ \begin{array} { l } y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) } \\ y ( 0 ) = 0 \end{array} \right.$$
using the relation established in question 18.

Show that $S _ { N }$ converges uniformly to $S$ on $\mathbf { R } _ { + }$ when $N \rightarrow + \infty$ and that
$$\left\| S _ { N } - S \right\| _ { \infty , \mathbf { R } _ { + } } \leq \frac { M \mathrm { e } ^ { 2 M } } { 2 ^ { N + 1 } }$$
where $S _ { N } ( x ) = S _ { 0 } \mathrm { e } ^ { y _ { N } ( x ) } = \frac { 1 } { 2 } \exp \left( \sum _ { n = 0 } ^ { N } a _ { n } \mathrm { e } ^ { - \lambda _ { n } x } \right)$, $S(x) = S_0 e^{y(x)} = \frac{1}{2} e^{y(x)}$, and $\left\| y_N - y \right\|_{\infty, \mathbf{R}_+} \leq \frac{M}{2^N}$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. In this question only, we suppose that $f \in \mathcal{C}^1(\mathbb{R})$. Show that $x_1 := p_f(x_0)$ satisfies $$x_1 = x_0 - \tau f'(x_1)$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$, where $p_f(x_0)$ is the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. In this question only, we set $f(x) := |x|$ for all $x \in \mathbb{R}$. a) Show that $$p_f(x) = \begin{cases} x - \tau & \text{if } x \geq \tau \\ x + \tau & \text{if } x \leq -\tau \\ 0 & \text{otherwise} \end{cases}$$ b) Deduce that $x_n \rightarrow 0$ as $n \rightarrow \infty$, for all $x_0 \in \mathbb{R}$ and $\tau > 0$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. Show that $\frac{1}{2}|x_1 - x_0|^2 + \tau f(x_1) \leq \tau f(x_0)$. Deduce that for all integers $N > M \geq 0$ $$\frac{1}{2}\sum_{M < n \leq N}|x_n - x_{n-1}|^2 \leq \tau\left(f(x_M) - f(x_N)\right)$$ Deduce that $|x_{n+1} - x_n| \rightarrow 0$ as $n \rightarrow \infty$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x \in \mathbb{R}$ and $\tilde{x} := p_f(x)$. Show that for all $v \in \mathbb{R}$ and $t \in \mathbb{R}$ $$\tau f(\tilde{x}) + \frac{1}{2}|\tilde{x} - x|^2 \leq \tau f(\tilde{x} + tv) + \frac{1}{2}|\tilde{x} + tv - x|^2$$ Let also $y \in \mathbb{R}$, and $\tilde{y} := p_f(y)$. Deduce that $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Let $x, y \in \mathbb{R}$, $\tilde{x} := p_f(x)$, $\tilde{y} := p_f(y)$. Show that the right-hand side in inequality $$2\tau(f(\tilde{x}) + f(\tilde{y}) - f(\tilde{x} + tv) - f(\tilde{y} - tv)) \leq |\tilde{x} + tv - x|^2 + |\tilde{y} - tv - y|^2 - |\tilde{x} - x|^2 - |\tilde{y} - y|^2$$ admits the asymptotic expansion $2tv(\tilde{x} - x + y - \tilde{y}) + o(t)$ as $t \rightarrow 0$.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that $$\frac{1}{2y} \int_{x-y}^{x+y} f(t)\, dt = f(x), \quad \text{for all } x \in \mathbb{R},\ y > 0$$ Show that there exist $a, b \in \mathbb{R}$ such that $f(x) = ax + b$ for all $x \in \mathbb{R}$.

The function $y = e ^ { k x }$ satisfies
$$\left( \frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } \right) \left( \frac { d y } { d x } - y \right) = y \frac { d y } { d x }$$
for
(A) exactly one value of $k$.
(B) two distinct values of $k$.
(C) three distinct values of $k$.
(D) infinitely many values of $k$.

Suppose $f ( x )$ is a twice differentiable function on $[ a , b ]$ such that $$f ( a ) = 0 = f ( b )$$ and $$x ^ { 2 } \frac { d ^ { 2 } f ( x ) } { d x ^ { 2 } } + 4 x \frac { d f ( x ) } { d x } + 2 f ( x ) > 0 \text { for all } x \in ( a , b )$$ Then,
(A) $f$ is negative for all $x \in ( a , b )$.
(B) $f$ is positive for all $x \in ( a , b )$.
(C) $f ( x ) = 0$ for exactly one $x \in ( a , b )$.
(D) $f ( x ) = 0$ for at least two $x \in ( a , b )$.

Consider a differentiable function $u : [ 0,1 ] \rightarrow \mathbb { R }$. Assume the function $u$ satisfies $$u ( a ) = \frac { 1 } { 2 r } \int _ { a - r } ^ { a + r } u ( x ) d x , \quad \text{for all } a \in ( 0,1 ) \text{ and all } r < \min ( a , 1 - a ).$$ Which of the following four statements must be true?
(A) $u$ attains its maximum but not its minimum on the set $\{ 0,1 \}$.
(B) $u$ attains its minimum but not maximum on the set $\{ 0,1 \}$.
(C) If $u$ attains either its maximum or its minimum on the set $\{ 0,1 \}$, then it must be constant.
(D) $u$ attains both its maximum and its minimum on the set $\{ 0,1 \}$.

If $f : [ 0 , \infty ) \rightarrow \mathbb { R }$ is a continuous function such that $$f ( x ) + \ln 2 \int _ { 0 } ^ { x } f ( t ) d t = 1 , x \geq 0$$ then for all $x \geq 0$,
(A) $f ( x ) = e ^ { x } \ln 2$.
(B) $f ( x ) = e ^ { - x } \ln 2$.
(C) $f ( x ) = 2 ^ { x }$.
(D) $f ( x ) = \left( \frac { 1 } { 2 } \right) ^ { x }$.

49. The differential equation $\frac { d y } { d x } = \frac { \sqrt { 1 - y ^ { 2 } } } { y }$ determines a family of circles with
(A) variable radii and a fixed centre at $( 0,1 )$
(B) variable radii and a fixed centre at $( 0 , - 1 )$
(C) fixed radius 1 and variable centres along the $x$-axis
(D) fixed radius 1 and variable centres along the $y$-axis Answer O O O O
(A)
(B)
(C)
(D)