Let $E$ be the set of continuous functions $f$ from $I$ to $\mathbb{R}$ such that $f^2 w$ is integrable on $I$. For all functions $f$ and $g$ in $E$, we set $$\langle f, g \rangle = \int_I f(x) g(x) w(x)\,\mathrm{d}x.$$
Show that we thus define an inner product on $E$.

Let $P \in \mathbb { R } [ X ]$ and $Q \in \mathbb { R } [ X ]$. Show that the function $x \mapsto P ( 4 x ) Q ( 4 x ) \frac { \sqrt { 1 - x } } { \sqrt { x } }$ is integrable on $\left. ] 0,1 \right]$.

Let $a , b , c , d$ be four real numbers such that $a \leqslant b$ and $c \leqslant d$. Let $U$ be an open set of $\mathbb { R } ^ { 2 }$ containing $[ a , b ] \times [ c , d ]$. Let $h : U \rightarrow \mathbb { R }$ be a function of class $\mathcal { C } ^ { 2 }$.
(a) Show the identity $$h ( b , d ) - h ( a , d ) - h ( b , c ) + h ( a , c ) = \int _ { a } ^ { b } \hat { h } \left( s _ { 1 } \right) d s _ { 1 }$$ where $\hat { h }$ is defined by $$\hat { h } \left( s _ { 1 } \right) = \int _ { c } ^ { d } \frac { \partial ^ { 2 } h } { \partial s _ { 1 } \partial s _ { 2 } } \left( s _ { 1 } , s _ { 2 } \right) d s _ { 2 }$$ (b) Deduce that there exists a point $\left( \bar { s } _ { 1 } , \bar { s } _ { 2 } \right)$ of $[ a , b ] \times [ c , d ]$ such that we have the two equalities $$h ( b , d ) - h ( a , d ) - h ( b , c ) + h ( a , c ) = ( b - a ) \hat { h } \left( \bar { s } _ { 1 } \right) = ( b - a ) ( d - c ) \frac { \partial ^ { 2 } h } { \partial s _ { 1 } \partial s _ { 2 } } \left( \bar { s } _ { 1 } , \bar { s } _ { 2 } \right)$$

We keep, until the end of this third part, the hypotheses and notation of question 3.2. For $x , y \in I$ such that $y \neq x$, we set $$H _ { f } ( x , y ) = \frac { x f ( y ) - y f ( x ) } { f ( y ) - f ( x ) }$$ (a) Show that for all $x , y \in I$ such that $y \neq x$ we have $$H _ { f } ( x , y ) = x - f ( x ) \int _ { 0 } ^ { 1 } g ^ { \prime } ( \lambda f ( x ) + ( 1 - \lambda ) f ( y ) ) d \lambda$$ (b) Deduce that $H _ { f }$ admits a unique continuous extension to $I \times I$ as a whole. We still denote this extension by $H _ { f } : I \times I \rightarrow \mathbb { R }$.
(c) Show that $H _ { f }$ is of class $\mathcal { C } ^ { 2 }$ on $I \times I$.
(d) Compute $H _ { f } ( x , x )$.

We keep the hypotheses and notation of question 3.2. We now assume $0 \in f ( I )$ and we denote $x ^ { * } = g ( 0 )$. For $x \in I$ we denote by $I _ { x }$ the closed interval with endpoints $x$ and $x ^ { * }$.
(a) Let $x , y \in I$. Show that there exists $( \bar { x } , \bar { y } ) \in I _ { x } \times I _ { y }$, such that $$H _ { f } ( x , y ) - x ^ { * } = \left( x - x ^ { * } \right) \left( y - x ^ { * } \right) \frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } ( \bar { x } , \bar { y } )$$ (b) Compute $$\frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } \left( x ^ { * } , x ^ { * } \right)$$ as a function of the derivatives of $f$.

Let $a , b , c , d$ be four real numbers such that $a \leqslant b$ and $c \leqslant d$. Let $U$ be an open set of $\mathbb { R } ^ { 2 }$ containing $[ a , b ] \times [ c , d ]$. Let $h : U \rightarrow \mathbb { R }$ be a function of class $\mathcal { C } ^ { 2 }$.
(a) Show the identity $$h ( b , d ) - h ( a , d ) - h ( b , c ) + h ( a , c ) = \int _ { a } ^ { b } \hat { h } \left( s _ { 1 } \right) d s _ { 1 }$$ where $\hat { h }$ is defined by $$\hat { h } \left( s _ { 1 } \right) = \int _ { c } ^ { d } \frac { \partial ^ { 2 } h } { \partial s _ { 1 } \partial s _ { 2 } } \left( s _ { 1 } , s _ { 2 } \right) d s _ { 2 }$$
(b) Deduce that there exists a point $\left( \bar { s } _ { 1 } , \bar { s } _ { 2 } \right)$ of $[ a , b ] \times [ c , d ]$ such that we have the two equalities $$h ( b , d ) - h ( a , d ) - h ( b , c ) + h ( a , c ) = ( b - a ) \hat { h } \left( \bar { s } _ { 1 } \right) = ( b - a ) ( d - c ) \frac { \partial ^ { 2 } h } { \partial s _ { 1 } \partial s _ { 2 } } \left( \bar { s } _ { 1 } , \bar { s } _ { 2 } \right)$$

We keep, until the end of this third part, the hypotheses and notation of the previous question. For $x , y \in I$ such that $y \neq x$, we set $$H _ { f } ( x , y ) = \frac { x f ( y ) - y f ( x ) } { f ( y ) - f ( x ) }$$
(a) Show that for all $x , y \in I$ such that $y \neq x$ we have $$H _ { f } ( x , y ) = x - f ( x ) \int _ { 0 } ^ { 1 } g ^ { \prime } ( \lambda f ( x ) + ( 1 - \lambda ) f ( y ) ) d \lambda$$
(b) Deduce that $H _ { f }$ admits a unique continuous extension to $I \times I$ as a whole. We still denote this extension by $H _ { f } : I \times I \rightarrow \mathbb { R }$.
(c) Show that $H _ { f }$ is of class $\mathcal { C } ^ { 2 }$ on $I \times I$.
(d) Compute $H _ { f } ( x , x )$.

We keep the hypotheses and notation of questions 3.2 and 3.3. We now assume $0 \in f ( I )$ and we denote $x ^ { * } = g ( 0 )$. For $x \in I$ we denote by $I _ { x }$ the closed interval with endpoints $x$ and $x ^ { * }$.
(a) Let $x , y \in I$. Show that there exists $( \bar { x } , \bar { y } ) \in I _ { x } \times I _ { y }$, such that $$H _ { f } ( x , y ) - x ^ { * } = \left( x - x ^ { * } \right) \left( y - x ^ { * } \right) \frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } ( \bar { x } , \bar { y } )$$
(b) Compute $$\frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } \left( x ^ { * } , x ^ { * } \right)$$ as a function of the derivatives of $f$.

The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$.
Show that for all integer $n > 1$,
$$\int _ { 1 } ^ { n } \frac { q ( u ) } { u } \mathrm {~d} u = \ln ( n ! ) + ( n - 1 ) - n \ln ( n ) - \frac { 1 } { 2 } \ln ( n ) = \ln \left( \frac { n ! e ^ { n } } { n ^ { n } \sqrt { n } } \right) - 1$$

Show that for all integer $n > 1$, $$\int_{1}^{n} \frac{q(u)}{u} \mathrm{~d}u = \ln(n!) + (n-1) - n\ln(n) - \frac{1}{2}\ln(n) = \ln\left(\frac{n! e^n}{n^n \sqrt{n}}\right) - 1.$$

We assume that $f$ is a function from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ of class $\mathcal { C } ^ { 1 }$ satisfying $$\left\{ \begin{array} { l } \lim _ { x \rightarrow 0 } f ( x ) = 0 \\ \exists C > 0 ; \forall x > 0 , \quad \left| f ^ { \prime } ( x ) \right| \leqslant C \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } } \end{array} \right.$$ Using the result of Question 7, deduce that $f \in E$, where $E$ is the set of continuous functions $f$ from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ such that the integral $\int _ { 0 } ^ { + \infty } f ^ { 2 } ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$ converges.

The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$.
Show that $\int _ { \lfloor x \rfloor } ^ { x } \frac { q ( u ) } { u } \mathrm {~d} u$ tends to 0 as $x$ tends to $+ \infty$, and deduce the convergence of the integral $\int _ { 1 } ^ { + \infty } \frac { q ( u ) } { u } \mathrm {~d} u$, as well as the equality
$$\int _ { 1 } ^ { + \infty } \frac { q ( u ) } { u } \mathrm {~d} u = \frac { \ln ( 2 \pi ) } { 2 } - 1$$

Show that $\int_{\lfloor x \rfloor}^{x} \frac{q(u)}{u} \mathrm{~d}u$ tends to 0 when $x$ tends to $+\infty$, and deduce the convergence of the integral $\int_{1}^{+\infty} \frac{q(u)}{u} \mathrm{~d}u$, as well as the equality $$\int_{1}^{+\infty} \frac{q(u)}{u} \mathrm{~d}u = \frac{\ln(2\pi)}{2} - 1$$

By observing that the function $|\cos|$ is $2\pi$-periodic, calculate, for $\omega \in \mathbf{R}$, the integral $$\int_{-\pi}^{\pi} |\cos(u - \omega)| \mathrm{d}u$$ Deduce that, if $(a,b) \in \mathbf{R}^{2}$, $$\int_{-\pi}^{\pi} |a\cos(u) + b\sin(u)| \mathrm{d}u = 4\sqrt{a^{2} + b^{2}}$$

To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Is the endomorphism $U$ surjective?

We fix two functions $f$ and $g$ in $E$. For $x > 0$, we set $$F ( x ) = - U ( f ) ^ { \prime } ( x ) \mathrm { e } ^ { - x }$$ where $U(f)^\prime(x) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Verify that $F$ is an antiderivative of $x \mapsto f ( x ) \frac { \mathrm { e } ^ { - x } } { x }$ on the interval $\mathbb { R } _ { + } ^ { * }$.

For any real $\alpha > 0$, consider the function $h _ { \alpha } : t \mapsto \ln \left( \frac { 1 - t ^ { 2 } } { \alpha ^ { 2 } + t ^ { 2 } } \right)$. Show that $h _ { \alpha }$ is a continuous decreasing integrable function on $[ 0,1 [$.

For any real $\alpha > 0$, consider the function $h _ { \alpha } : t \mapsto \ln \left( \frac { 1 - t ^ { 2 } } { \alpha ^ { 2 } + t ^ { 2 } } \right)$ and set $J _ { \alpha } = \int _ { 0 } ^ { 1 } h _ { \alpha } ( t ) \, \mathrm { d } t$. Justify that $$J _ { \alpha } = \int _ { 0 } ^ { 1 } \ln ( 1 - t ) \, \mathrm { d } t + \int _ { 0 } ^ { 1 } \ln ( 1 + t ) \, \mathrm { d } t - \int _ { 0 } ^ { 1 } \ln \left( \alpha ^ { 2 } + t ^ { 2 } \right) \mathrm { d } t = \int _ { 0 } ^ { 2 } \ln ( u ) \, \mathrm { d } u - \int _ { 0 } ^ { 1 } \ln \left( \alpha ^ { 2 } + t ^ { 2 } \right) \mathrm { d } t.$$

For any real $\alpha > 0$, consider the function $h _ { \alpha } : t \mapsto \ln \left( \frac { 1 - t ^ { 2 } } { \alpha ^ { 2 } + t ^ { 2 } } \right)$ and set $J _ { \alpha } = \int _ { 0 } ^ { 1 } h _ { \alpha } ( t ) \, \mathrm { d } t$. Deduce that $$J _ { \alpha } = 2 \ln ( 2 ) - \ln \left( 1 + \alpha ^ { 2 } \right) - 2 \alpha \arctan \left( \frac { 1 } { \alpha } \right).$$

For any real $\alpha > 0$, consider $J _ { \alpha } = 2 \ln ( 2 ) - \ln \left( 1 + \alpha ^ { 2 } \right) - 2 \alpha \arctan \left( \frac { 1 } { \alpha } \right)$. Show that there exists $\gamma > 0$ such that, for all $\alpha \in ] 0 , \gamma [$, $J _ { \alpha } > 0$.

For all $n \in \mathbb { N } ^ { * }$, consider in $]0,1[$ the points $a _ { k , n }$ given, for $k \in \llbracket 0 , n - 1 \rrbracket$, by $a _ { k , n } = \frac { 2 k + 1 } { 2 n }$ and $$S _ { n } \left( h _ { \alpha } \right) = \frac { 1 } { n } \sum _ { k = 0 } ^ { n - 1 } h _ { \alpha } \left( a _ { k , n } \right).$$ For all $n \in \mathbb { N } ^ { * }$, show that $$\int _ { 1 / 2 n } ^ { ( 2 n - 1 ) / 2 n } h _ { \alpha } ( t ) \, \mathrm { d } t + \frac { 1 } { n } h _ { \alpha } \left( \frac { 2 n - 1 } { 2 n } \right) \leqslant S _ { n } \left( h _ { \alpha } \right) \leqslant \frac { 1 } { n } h _ { \alpha } \left( \frac { 1 } { 2 n } \right) + \int _ { 1 / 2 n } ^ { ( 2 n - 1 ) / 2 n } h _ { \alpha } ( t ) \, \mathrm { d } t.$$

For all $n \in \mathbb { N } ^ { * }$, consider in $]0,1[$ the points $a _ { k , n } = \frac { 2 k + 1 } { 2 n }$ for $k \in \llbracket 0 , n - 1 \rrbracket$, and $S _ { n } \left( h _ { \alpha } \right) = \frac { 1 } { n } \sum _ { k = 0 } ^ { n - 1 } h _ { \alpha } \left( a _ { k , n } \right)$, where $J_\alpha = \int_0^1 h_\alpha(t)\,\mathrm{d}t$. Deduce that the sequence $\left( S _ { n } \left( h _ { \alpha } \right) \right) _ { n \in \mathbb { N } ^ { * } }$ converges to $J _ { \alpha }$.

Show that $$I _ { n } \geqslant \frac { 1 } { 2 ^ { n } }.$$ where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.

Justify the existence of $K _ { n } = \int _ { 0 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$ and give the exact value of $K _ { 1 }$.

For $\mu > 0$ and $\varphi \in \mathcal{C}_{c}(\mathbb{R})$, we define $T_{\mu} : \varphi \mapsto T_{\mu}\varphi$, where for all $x \in \mathbb{R}$,
$$T_{\mu}\varphi(x) = \frac{1}{2\mu} \int_{x-\mu}^{x+\mu} \varphi(t)\, dt$$
Show that for all $k \geqslant 0$, if $\varphi \in \mathcal{C}_{c}^{k}(\mathbb{R})$ then $T_{\mu}\varphi \in \mathcal{C}_{c}^{k+1}(\mathbb{R})$. Also show that
$$\left\|(T_{\mu}\varphi)^{(k)}\right\|_{\infty} \leqslant \left\|\varphi^{(k)}\right\|_{\infty}$$