We consider the Euclidean space $E = \mathscr{M}_{N,1}(\mathbf{R})$ equipped with the inner product $\langle X, Y \rangle = \sum_{i=1}^{N} X[i] Y[i] \pi[i]$, the endomorphism $u : X \mapsto (I_N - K)X$, the orthogonal projection $p : E \rightarrow E$ onto $\ker(u)$, and for $X \in E$, the function $\varphi_X : t \mapsto \|H_t X\|^2$. We set $Y = X - p(X)$. We denote by $\lambda$ the smallest nonzero eigenvalue of $u$. Show that for all real $t \in \mathbf{R}_+$, $\varphi_Y'(t) \leq -2\lambda \varphi_Y(t)$. Deduce that $\forall t \in \mathbf{R}_+, \|H_t X - p(X)\|^2 \leq e^{-2\lambda t} \|X - p(X)\|^2$.

We consider the Euclidean space $E = \mathscr{M}_{N,1}(\mathbf{R})$ equipped with the inner product $\langle X, Y \rangle = \sum_{i=1}^{N} X[i] Y[i] \pi[i]$, the endomorphism $u : X \mapsto (I_N - K)X$, the orthogonal projection $p : E \rightarrow E$ onto $\ker(u) = \operatorname{Vect}(U)$, and $\lambda$ the smallest nonzero eigenvalue of $u$. We have established that $\forall t \in \mathbf{R}_+, \|H_t X - p(X)\|^2 \leq e^{-2\lambda t} \|X - p(X)\|^2$. Let $i \in \llbracket 1;N \rrbracket$ and $t \in \mathbf{R}_+$. Show that $\|H_t E_i - \pi[i] U\| \leq e^{-\lambda t} \sqrt{\pi[i]}$.

We consider the matrix $H_t$ defined by $\forall (i,j) \in \llbracket 1;N \rrbracket^2, H_t[i,j] = e^{-t} \sum_{n=0}^{+\infty} \frac{t^n K^n[i,j]}{n!}$, and $\pi$ the stationary probability. Show that for all $(i,j) \in \llbracket 1;N \rrbracket^2$ and all $t \in \mathbf{R}_+$, $$H_t[i,j] - \pi[j] = \sum_{k=1}^{N} \left(H_{t/2}[i,k] - \pi[k]\right)\left(H_{t/2}[k,j] - \pi[j]\right)$$ One may use question 5.

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a continuous function such that for all $x \in \mathbb { R }$ and for all $t \geq 0$, $$f ( x ) = f \left( e ^ { t } x \right)$$ Show that $f$ is a constant function.