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 $s \in [0,1]$, the function $k_s$ is defined by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ We set, for all $f \in E_1$, $$U(f)(s) = \int_0^1 k_s'(t) f'(t) \, \mathrm{d}t$$ Show that $U$ is the identity map on $E_1$.
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 $s \in [0,1]$, the function $k_s$ is defined by,
$$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$
We set, for all $f \in E_1$,
$$U(f)(s) = \int_0^1 k_s'(t) f'(t) \, \mathrm{d}t$$
Show that $U$ is the identity map on $E_1$.