We define the sequence of polynomials $\left(P_n\right)_{n \in \mathbb{N}}$ by: $$\left\{\begin{array}{l} P_0 = 1 \\ \forall n \in \mathbb{N}^*, \quad P_n = [X(X-1)]^n \end{array}\right.$$ We define the sequence of polynomials $\left(L_n\right)_{n \in \mathbb{N}}$ by: $$\left\{\begin{array}{l} L_0 = 1 \\ \forall n \in \mathbb{N}^*, \quad L_n = \frac{1}{P_n^{(n)}(1)} P_n^{(n)} \end{array}\right.$$
Let $n \in \mathbb{N}^*$. Show that, for all $Q \in \mathbb{R}_{n-1}[X]$, $\langle Q, L_n \rangle = 0$.
Hint: you may integrate by parts.

$\mathcal{P}$ denotes the vector space of polynomial functions with complex coefficients, with inner product $\langle P, Q \rangle = \displaystyle\int_0^{+\infty} e^{-t}\bar{P}(t)Q(t)\,dt$, and $U$ is the endomorphism of $\mathcal{P}$ defined by $U(P)(t) = e^t D\left(te^{-t}P^{\prime}(t)\right)$.
Show that for all $P$ and $Q$ in $\mathcal{P}$, we have $$\langle U(P), Q \rangle = \langle P, U(Q) \rangle.$$

Let $E_1$ denote the vector space of functions $f:[0,1] \rightarrow \mathbb{R}$ continuous, of class $\mathcal{C}^1$ piecewise, and satisfying $f(0) = f(1) = 0$, equipped with the inner product $$\forall (f,g) \in (E_1)^2, \quad (f \mid g) = \int_0^1 f'(t) g'(t)\,\mathrm{d}t$$ Prove that the pre-Hilbert space $(E_1, (\cdot \mid \cdot))$ is a reproducing kernel Hilbert space and that its reproducing kernel is the application $K$ defined by $K(s,t) = k_s(t)$ where $$k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$

We consider the space $E$ of continuous functions from $[0,1]$ to $\mathbb{R}$, equipped with the inner product defined by $$\langle f, g \rangle = \int_0^1 f(t) g(t)\,\mathrm{d}t$$ Show that $(E, \langle \cdot, \cdot \rangle)$ is not a reproducing kernel Hilbert space.

In this part, $E_1$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, of class $\mathcal{C}^1$ piecewise, and satisfying $f(0) = f(1) = 0$. We denote by $N$ the norm associated with the inner product $(f \mid g) = \int_0^1 f'(t) g'(t) \, \mathrm{d}t$. For all $(s,t) \in [0,1]^2$, $K(s,t) = k_s(t)$ where $$k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ Prove that the pre-Hilbert space $(E_1, (\cdot \mid \cdot))$ is a reproducing kernel Hilbert space and that its reproducing kernel is the application $K$ defined in the previous part.

We consider the space $E$ of continuous functions from $[0,1]$ to $\mathbb{R}$, equipped with the inner product defined by $$\langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ Show that $(E, \langle \cdot, \cdot \rangle)$ is not a reproducing kernel Hilbert space.

We fix two functions $f$ and $g$ in $E$. For $x > 0$, we set $F ( x ) = - U ( f ) ^ { \prime } ( x ) \mathrm { e } ^ { - x }$, which is an antiderivative of $x \mapsto f ( x ) \frac { \mathrm { e } ^ { - x } } { x }$. The limits of $t \mapsto F(t)U(g)(t)$ at $0$ and $+\infty$ are both $0$. Show that $$\langle f \mid U ( g ) \rangle = \int _ { 0 } ^ { + \infty } U ( f ) ^ { \prime } ( t ) U ( g ) ^ { \prime } ( t ) \mathrm { e } ^ { - t } \mathrm {~d} t.$$

We fix two functions $f$ and $g$ in $E$. Using the result of Question 28, deduce that $\langle f \mid U ( g ) \rangle = \langle U ( f ) \mid g \rangle$.