In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
Deduce from V.D the expression of $(Lf)(x)$ for $x \in E$.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
What is the value of $Lf(0)$?

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$.
Let $f$ be fixed such that $E$ is non-empty, $x \in E$ and $a > 0$. We set $h(t) = \displaystyle\int_0^t e^{-xu} f(u)\,du$ for all $t \geqslant 0$.
VI.B.1) Show that $Lf(x+a) = a\displaystyle\int_0^{+\infty} e^{-at} h(t)\,dt$.
VI.B.2) We assume that for all $n \in \mathbb{N}$, we have $Lf(x + na) = 0$.
Show that, for all $n \in \mathbb{N}$, the integral $\displaystyle\int_0^1 u^n h\!\left(-\frac{\ln u}{a}\right)du$ converges and that it is zero.
VI.B.3) What do we deduce for the function $h$?

We assume that $f$ is positive and that $E$ is neither empty nor equal to $\mathbb{R}$. We denote by $\alpha$ its infimum.
VII.A.1) Show that if $Lf$ is bounded on $E$, then $\alpha \in E$.
VII.A.2) If $\alpha \notin E$, what can we say about $Lf(x)$ when $x$ tends to $\alpha^+$?

In this question, $f(t) = \cos t$ and $\lambda(t) = \ln(1+t)$.
VII.B.1) Determine $E$.
VII.B.2) Determine $E^{\prime}$.
VII.B.3) Show that $Lf$ admits a limit at $\alpha$, the infimum of $E$, and determine it.

For $n \in \mathbb { Z }$, we denote by $\mathcal { E } _ { n }$ the space of functions $f$ in $\mathcal { C } ^ { 2 } \left( \mathbb { R } _ { + } ^ { * } , \mathbb { C } \right)$ such that $$\forall t \in \mathbb { R } _ { + } ^ { * } , \quad t ^ { 2 } f ^ { \prime \prime } ( t ) + t f ^ { \prime } ( t ) - n ^ { 2 } f ( t ) = 0$$ For $\alpha \in \mathbb { R }$, let $\varphi _ { \alpha }$ be the function from $\mathbb { R } _ { + } ^ { * }$ to $\mathbb { R }$ defined by $$\forall t \in \mathbb { R } _ { + } ^ { * } , \quad \varphi _ { \alpha } ( t ) = t ^ { \alpha }$$
For all $n \in \mathbb { Z } ^ { * }$, determine the real numbers $\alpha$ such that $\varphi _ { \alpha }$ belongs to $\mathcal { E } _ { n }$.

We denote $\alpha > -1/2$, $F_n$ the vector subspace of $E$ of polynomial functions of degree less than or equal to $n$ (where $n \in \mathbb{N}$), and $$\varphi_\alpha(y) : t \mapsto \left(1-t^2\right)y''(t) - (2\alpha+1)t\,y'(t)$$ Show that any eigenvector of $\varphi_\alpha$ of degree greater than or equal to 1 vanishes at least once in the interval $]-1,1[$.

We assume $\alpha = 1$. We denote $\|\cdot\|$ the norm associated with $S_1$, and $$\varphi_1(y) : t \mapsto \left(1-t^2\right)y''(t) - 3t\,y'(t)$$ Justify that, for all $k \in \mathbb{N}$, there exists a unique polynomial eigenvector of $\varphi_1$ of degree $k$, of norm 1 and with positive leading coefficient. We denote it $T_k$.

We assume $\alpha = 1$ and use the notation $V_n(z) = U_{n+1}(z,-1)$, and $$\varphi_1(y) : t \mapsto \left(1-t^2\right)y''(t) - 3t\,y'(t)$$ By differentiating twice the function $t \mapsto (\sin t)\,V_n(\cos t) - \sin((n+1)t)$, show that for all $n \in \mathbb{N}$, $V_n$ is an eigenvector of $\varphi_1$.

We assume $\alpha = 1$. We denote $\|\cdot\|$ the norm associated with $S_1$, $T_k$ the unique polynomial eigenvector of $\varphi_1$ of degree $k$, of norm 1 and with positive leading coefficient, and $V_n(z) = U_{n+1}(z,-1)$. Deduce that, for all $n \in \mathbb{N}$, $V_n$ and $T_n$ are proportional. Explicitly state the proportionality coefficient.

We keep all the notations from Part I and we assume that hypotheses (H1), (H2), (H3), (H4) and (H5) are all satisfied. Let $(w_1, w_2)$ be a characterizing pair of $G$ (satisfying properties (A), (B) and (C) of question 12). For any $\lambda \in \mathbb{R}$, we consider the following problem, denoted $\mathcal{P}_\lambda$: $$\text{Find } u \in G \text{ such that: } \forall v \in G, (u \mid v) - \lambda (T(u) \mid T(v)) = 0$$ and we denote by $H_\lambda$ the set of solutions $u$ of this problem.
(a) Show that if $(\mathcal{P}_\lambda)$ admits a solution $u \neq 0_E$, then necessarily $\lambda > 0$.
(b) Let $u \in G$. Show that $u$ is a solution of $(\mathcal{P}_\lambda)$ if and only if $$\left(\operatorname{Id}_E + \lambda T^2\right)(u) \in G^\perp$$ Deduce that there exist two real numbers $\alpha$ and $\beta$ such that: $$u = \alpha \left(\operatorname{Id}_E + \lambda T^2\right)^{-1} T(w_1) + \beta \left(\operatorname{Id}_E + \lambda T^2\right)^{-1} T(w_2)$$ (c) Show that the problem $(\mathcal{P}_\lambda)$ admits a non-zero solution if and only if $$Q_1(\lambda) \cdot Q_2(\lambda) = 0$$ where for $i \in \{1,2\}$, $Q_i$ is the polynomial $$Q_i(X) = \sum_{k=0}^{m-1} (-1)^k \left(T^{2k+1}(w_i) \mid T(w_i)\right) X^k$$ (d) Suppose that $\lambda$ is a root of the product polynomial $Q_1 Q_2$. Show that $\operatorname{dim}(H_\lambda) = 2$ if $\lambda$ is a common root of $Q_1$ and $Q_2$, and $\operatorname{dim}(H_\lambda) = 1$ otherwise.
(e) Show that $$\forall i \in \{1,2\}, Q_i(X) = \sum_{k=0}^{m-1} (-1)^k S\left(w_i, T^{2k+1}(w_i)\right) X^k$$

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ The endomorphisms $T(P) = P'$ and $M(P) = P^*$ are defined as before. We set $$\mathbb{R}_{2m-1}^0[X] = \{P \in \mathbb{R}_{2m-1}[X] \mid P(-1) = 0 \text{ and } P(1) = 0\}$$ The polynomials $L_n$ are defined by $L_n(X) = \frac{1}{2^n n!} R_n^{(n)}(X)$ where $R_n(X) = (X^2-1)^n$.
Let $\lambda \in \mathbb{R}$. We consider the problem: find $P \in \mathbb{R}_{2m-1}^0[X]$ such that $$\forall Q \in \mathbb{R}_{2m-1}^0[X], \int_{-1}^{1} P(t)Q(t)\,dt - \lambda \int_{-1}^{1} P'(t)Q'(t)\,dt = 0$$ Show that this problem admits a non-identically zero solution $P$ if and only if $\lambda$ is a root of the polynomial $$K(X) = \left(\sum_{k=0}^{m-1} (-1)^k L_{2m-1}^{(2k+1)}(1) X^k\right) \cdot \left(\sum_{k=0}^{m-1} (-1)^k L_{2m}^{(2k+1)}(1) X^k\right)$$

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

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 $pU(P) - P$ satisfies $y'' - y' = 0$ on $\mathbb{R}_+^*$. Show that $P$ is an eigenvector of $U$ for the eigenvalue $1/p$.