We say that the sequence of distributions $(T_n)_{n \in \mathbb{N}}$ converges to the distribution $T$ if $$\forall \varphi \in \mathcal{D}, \lim_{n \rightarrow \infty} T_n(\varphi) = T(\varphi)$$ For $n$ a non-zero natural number, we consider the function $U_n$ zero on the negative reals, affine on the interval $[0, 1/n]$, equal to 1 for reals greater than $1/n$ and continuous on $\mathbb{R}$. a) Show that the sequence of regular distributions $(T_{U_n})_{n \in \mathbb{N}}$ converges to $T_U$. b) Show that $$\forall \varphi \in \mathcal{D} \quad T_{U_n}'(\varphi) = \int_0^{1/n} n\varphi(t) \mathrm{d}t$$ c) Deduce that the distribution $T_{U_n}'$ is regular and give a function $V_n$ such that $T_{V_n} = T_{U_n}'$. d) Sketch $V_n$ for $n = 1, 2, 4$. e) Show that if the sequence of distributions $(T_n)_{n \in \mathbb{N}}$ converges to the distribution $T$, then $(T_n')_{n \in \mathbb{N}}$ converges to $T'$. f) What is the limit of $T_{U_n}'$ as $n$ tends to infinity?
We say that the sequence of distributions $(T_n)_{n \in \mathbb{N}}$ converges to the distribution $T$ if
$$\forall \varphi \in \mathcal{D}, \lim_{n \rightarrow \infty} T_n(\varphi) = T(\varphi)$$
For $n$ a non-zero natural number, we consider the function $U_n$ zero on the negative reals, affine on the interval $[0, 1/n]$, equal to 1 for reals greater than $1/n$ and continuous on $\mathbb{R}$.\\
a) Show that the sequence of regular distributions $(T_{U_n})_{n \in \mathbb{N}}$ converges to $T_U$.\\
b) Show that
$$\forall \varphi \in \mathcal{D} \quad T_{U_n}'(\varphi) = \int_0^{1/n} n\varphi(t) \mathrm{d}t$$
c) Deduce that the distribution $T_{U_n}'$ is regular and give a function $V_n$ such that $T_{V_n} = T_{U_n}'$.\\
d) Sketch $V_n$ for $n = 1, 2, 4$.\\
e) Show that if the sequence of distributions $(T_n)_{n \in \mathbb{N}}$ converges to the distribution $T$, then $(T_n')_{n \in \mathbb{N}}$ converges to $T'$.\\
f) What is the limit of $T_{U_n}'$ as $n$ tends to infinity?