We consider a general balanced urn. For all real $x, u$ and $v$, we set, provided it exists, $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ Let $\rho > 0$. We set $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2} = \{(x,u,v) \in \mathbb{R}^{3} ; |x| < \rho, 0 < u < 2, 0 < v < 2\}$.
Justify that, for $\rho$ sufficiently small, the function $H$ is well defined on $D_{\rho}$.

We consider a general balanced urn. For all real $x, u$ and $v$, we set $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ defined on $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$ for $\rho$ sufficiently small.
Justify that $H$ admits a first-order partial derivative with respect to $x$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $x$ of the expression for $H$.

In the general model of a Pólya urn ($b = c = 0$, $a = d$), the function $G$ is defined on $U$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ and admits the expansion $G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$ on $D_{\rho}$.
Justify that $G$ admits a first-order partial derivative with respect to $x$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $x$ of the expression for $G$.

In the general model of a Pólya urn ($b = c = 0$, $a = d$), the function $G$ is defined on $U$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ and admits the expansion $G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$ on $D_{\rho}$.
Prove that $G$ admits a first-order partial derivative with respect to $u$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $u$ of the expression for $G$.

Show that the series of functions $\sum u_k$ where for all $k \in \mathbb{N}^*$, the function $u_k$ is defined on $[0, +\infty[$ by $u_k : x \mapsto (1 + kx)^{-k}(1/2)^k$ is normally convergent on $[0, +\infty[$.

We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence, where $g_k(x) = \sqrt{2}\sin(k\pi x)$. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2\pi^2} \langle f, g_k \rangle g_k(x)$$ Show that $\Phi$ is continuous.

We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence, where $g_k(x) = \sqrt{2}\sin(k\pi x)$. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2\pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Show that $$\lim_{N \rightarrow +\infty} \left\| T(f_N) - \Phi \right\| = 0$$

We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence, where $g_k(x) = \sqrt{2}\sin(k\pi x)$. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2\pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Deduce $T(f) = \Phi$.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2 \pi^2} \langle f, g_k \rangle g_k(x)$$ Show that $\Phi$ is continuous.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2 \pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Show that $$\lim_{N \rightarrow +\infty} \left\| T(f_N) - \Phi \right\| = 0$$

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2 \pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Deduce $T(f) = \Phi$.

We assume that $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$. Let $k \in \mathbb{N}$. Show that, if $g$ is of class $\mathcal{C}^k$ and if the functions $g^{(j)}$ are bounded for $j \in \llbracket 0, k \rrbracket$, then $f * g$ is of class $\mathcal{C}^k$ and $(f * g)^{(k)} = f * (g^{(k)})$.

Assume that $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$. Let $k \in \mathbb{N}$. Show that, if $g$ is of class $\mathcal{C}^k$ and if the functions $g^{(j)}$ are bounded for $j \in \llbracket 0, k \rrbracket$, then $f * g$ is of class $\mathcal{C}^k$ and $(f*g)^{(k)} = f * (g^{(k)})$.

Given three real numbers $a, b$ and $c$ with $c \in D$, the Gauss hypergeometric function is defined by $$F_{a,b,c}(x) = \sum_{n=0}^{+\infty} \frac{[a]_n [b]_n}{[c]_n} \frac{x^n}{n!}$$ Justify that $F_{a,b,c}$ is of class $\mathcal{C}^1$ on $]-1,1[$. Calculate its derivative and express it using a Gauss hypergeometric function.

Given three real numbers $a, b$ and $c$ with $c \in D$, the Gauss hypergeometric function is defined by $$F_{a,b,c}(x) = \sum_{n=0}^{+\infty} \frac{[a]_n [b]_n}{[c]_n} \frac{x^n}{n!}$$ Justify that $F_{a,b,c}$ is of class $\mathcal{C}^\infty$ on $]-1,1[$ and express its $n$-th derivative using a Gauss hypergeometric function.

Deduce that the power series $\sum _ { n \geqslant 0 } C _ { n } t ^ { n }$ converges uniformly on the interval $I = \left[ - \frac { 1 } { 4 } , \frac { 1 } { 4 } \right]$.

Let $K \in \mathbb{N}^{\star}$, consider distinct real numbers $x_1 < \cdots < x_K$ in an interval $[a,b]$ (with $a < b$), and a sequence of functions $(f_n)$ of class $\mathcal{C}^K$ on $[a,b]$ with real values satisfying: (H1) the function series $\sum f_n^{(K)}$ converges normally on $[a,b]$; (H2) for all $\ell \in \llbracket 1, K \rrbracket$ the numerical series $\sum f_n(x_\ell)$ is absolutely convergent.
In the special case $[a,b] = [0,1]$, justify that the series $\sum f_n^{(k)}$ converges normally on $[a,b]$ for all $k \in \llbracket 0, K-1 \rrbracket$.

Let $K \in \mathbb{N}^{\star}$, consider distinct real numbers $x_1 < \cdots < x_K$ in an interval $[a,b]$ (with $a < b$), and a sequence of functions $(f_n)$ of class $\mathcal{C}^K$ on $[a,b]$ with real values satisfying: (H1) the function series $\sum f_n^{(K)}$ converges normally on $[a,b]$; (H2) for all $\ell \in \llbracket 1, K \rrbracket$ the numerical series $\sum f_n(x_\ell)$ is absolutely convergent.
Treat the question of showing that $\sum f_n^{(k)}$ converges normally on $[a,b]$ for all $k \in \llbracket 0, K-1 \rrbracket$ in the general case of a segment $[a,b]$ with $a < b$. One may examine $f_n \circ \sigma$ where $\sigma : [0,1] \rightarrow [a,b]$ is defined by $\sigma(t) = (1-t)a + tb$ for all $t \in [0,1]$.

Let $K \in \mathbb{N}^{\star}$, consider distinct real numbers $x_1 < \cdots < x_K$ in an interval $[a,b]$ (with $a < b$), and a sequence of functions $(f_n)$ of class $\mathcal{C}^K$ on $[a,b]$ satisfying hypotheses (H1) and (H2). According to the result of the previous question, we set $F_k(x) = \sum_{n=0}^{+\infty} f_n^{(k)}(x)$ for all $x \in [a,b]$. Prove that $F_0$ is of class $\mathcal{C}^K$ on $[a,b]$ and that $F_0^{(k)} = F_k$ for all $k \in \llbracket 1, K \rrbracket$.

Let $K \in \mathbb{N}^\star$, consider distinct real numbers $x_1 < \cdots < x_K$ in an interval $[a,b]$ (with $a < b$), and a sequence of functions $(f_n)$ of class $\mathcal{C}^K$ on $[a,b]$ with real values satisfying: (H1) the function series $\sum f_n^{(K)}$ converges normally on $[a,b]$; (H2) for all $\ell \in \llbracket 1, K \rrbracket$ the numerical series $\sum f_n(x_\ell)$ is absolutely convergent.
In the particular case $[a,b] = [0,1]$, justify that the series $\sum f_n^{(k)}$ converges normally on $[a,b]$ for all $k \in \llbracket 0, K-1 \rrbracket$.

Let $K \in \mathbb{N}^\star$, consider distinct real numbers $x_1 < \cdots < x_K$ in an interval $[a,b]$ (with $a < b$), and a sequence of functions $(f_n)$ of class $\mathcal{C}^K$ on $[a,b]$ with real values satisfying: (H1) the function series $\sum f_n^{(K)}$ converges normally on $[a,b]$; (H2) for all $\ell \in \llbracket 1, K \rrbracket$ the numerical series $\sum f_n(x_\ell)$ is absolutely convergent.
Treat the previous question in the general case of a segment $[a,b]$ with $a < b$. One may examine $f_n \circ \sigma$ where $\sigma : [0,1] \rightarrow [a,b]$ is defined by $\sigma(t) = (1-t)a + tb$ for all $t \in [0,1]$.

Let $K \in \mathbb{N}^\star$, consider distinct real numbers $x_1 < \cdots < x_K$ in an interval $[a,b]$ (with $a < b$), and a sequence of functions $(f_n)$ of class $\mathcal{C}^K$ on $[a,b]$ with real values satisfying: (H1) the function series $\sum f_n^{(K)}$ converges normally on $[a,b]$; (H2) for all $\ell \in \llbracket 1, K \rrbracket$ the numerical series $\sum f_n(x_\ell)$ is absolutely convergent.
According to the result of the previous question, we can set $F_k(x) = \sum_{n=0}^{+\infty} f_n^{(k)}(x)$ for all $x \in [a,b]$. Prove that $F_0$ is of class $\mathcal{C}^K$ on $[a,b]$ and that $F_0^{(k)} = F_k$ for all $k \in \llbracket 1, K \rrbracket$.

Let $(c_{j})_{j \in \mathbf{N}}$ be a sequence of complex numbers such that the series $\sum c_{j}$ converges absolutely. We set $$\forall t \in \mathbf{R}, \quad u(t) = \sum_{j=0}^{+\infty} c_{j} e^{-i(j+1)t}.$$
Justify the existence and continuity of the function $u$. For $k \in \mathbf{N}$, show that $$\frac{1}{2\pi} \int_{-\pi}^{\pi} u(t) e^{i(k+1)t} \mathrm{d}t = c_{k}$$

Show that $S$ is continuous on $\mathbf{R}_{+}$, where $S(t) = \operatorname{Ent}_{\varphi}\left(P_{t}(f)\right)$.
Hint: You may first show that, if $x \in \mathbf{R}$, then $t \mapsto P_{t}(f)(x)$ is continuous on $\mathbf{R}_{+}$.

Show that $S$ is continuous on $\mathbf{R}_+$.
Hint: You may first show that, if $x \in \mathbf{R}$, $t \mapsto P_t(f)(x)$ is continuous on $\mathbf{R}_+$.