1\leq j\leq m}$ is defined by $$\begin{aligned}
||A||:=\sqrt{\sum_{j=1}^n\sum_{i=1}^n|a_{ij}|^2}.\end{aligned}$$

Frobenius norm and euclidean norm are the same for vectors.

\[ch1Minkowski\]**\[Minkowski inequality : Integral form\]**.

Let $1\leq p<+\infty$ and let $(X, \mathcal{A}, dx)$ and $(Y, \mathcal{B}, dy)$ be $\sigma$-finite measures spaces. Let $F$ be a measurable function on the product space $X\times Y$. Then $$\begin{aligned}
\left(\int_X\left|\int_YF(x,y)dy\right|^pdx\right)^{1/p}\leq \int_Y\left(\int_X|F(x,y)|^pdx\right)^{1/p}dy.\end{aligned}$$ The above inequality can be writen as $$\begin{aligned}
\left\|\int_YF(.,y)dy\right\|_{L^p(X,\mathcal{A}, dx)}\leq \int_Y||F(.,y)||_{L^p(X, \mathcal{A}, dx)}dy.\end{aligned}$$

**\[Gronwall inequality\] : Continous form**

Let $a(t)$ and $b(t)$ be two continuous and positives functions defines on $\mathbb{R}_+$ such that $$\begin{aligned}
a(t)\leq b(t)+c\int_0^ta(s)ds,  \hspace{0.5cm} \forall\, t\in \mathbb{R}_+,\end{aligned}$$ then $$\begin{aligned}
a(t)\leq b(t)+c\int_0^tb(s)e^{c(t-s)}ds, \hspace{0.5cm}\forall\, t\in \mathbb{R}_+.\end{aligned}$$ **\[Gronwall inequality\] : Discrete form**.

Let $\theta$ and $K$ be two constants and $(v_n)$ be a sequence satisfying : $$\begin{aligned}
v_{n+1}\leq (1+\theta)v_n+K,\end{aligned}$$ then $$\begin{aligned}
v_n\leq e^{n\theta}v_0+K\dfrac{e^{n\theta}-1}{e^{\theta}-1}.\end{aligned}$$

[@Gron].

**\[Borel Cantelli\]**

Let $(A_n)_{n\in \mathbb{N}}$ be a family of subset of $\Omega$.

1.  If $\sum\limits_{n\in\mathbb{N}}\mathbb{P}(A_n)<\infty$, then $\mathbb{P}\left[\limsup\limits_{n\longrightarrow\infty}A_n\right]=0$.

2.  If the events $(A_n)$ are independent and $\sum\limits_{n\in\mathbb{N}}\mathbb{P}(A_n)=0$, then $\mathbb{P}\left[\limsup\limits_{n\longrightarrow \infty}A_n\right]=1$.

[@Phi].

Stochastic processes
--------------------

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. A family $(\mathcal{F}_t)_{t\geq 0}$ of sub $\sigma$-algebra of $\mathcal{F}$ is called filtration if $\mathcal{F}_s\subset \mathcal{F}_t$, $\forall\hspace{0.2cm} 0\leq s\leq t$.

If $(\mathcal{F}_t)$ is such that $\mathcal{F}_t=\cap_{t>s}\mathcal{F}_s$, then $(\mathcal{F}_t)_{t\geq 0}$ is said 