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 $s \in [0,1]$, we define the function $k_s$ by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t) \, \mathrm{d}t$$ For all $f \in E$, show that $T(f)$ is of class $\mathcal{C}^2$ then that $T(f)'' = -f$.
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 $s \in [0,1]$, we define the function $k_s$ by,
$$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$
For all $f \in E$, we set,
$$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t) \, \mathrm{d}t$$
For all $f \in E$, show that $T(f)$ is of class $\mathcal{C}^2$ then that $T(f)'' = -f$.