We denote by $K$ the function defined from $[0,1]^2$ to $\mathbb{R}$ by the following relation: $K(s,t) = (1-s)t$ if $0 \leq t \leq s \leq 1$ and $K(s,t) = (1-t)s$ otherwise. We denote by $T$ the application defined on $E = C([0,1], \mathbb{R})$, equipped with the norm $\|.\|_2 = \sqrt{\int_0^1 |f(x)|^2 \, dx}$, by the relation: $$\forall f \in E, \quad \forall s \in [0,1], \quad T(f)(s) = \int_0^1 K(s,t) f(t) \, dt$$ Let $f \in E$. By decomposing $T(f)$ into two integrals, show that $T(f)$ is a $C^2$ function and express $(T(f))'$ then $(T(f))''$.
We denote by $K$ the function defined from $[0,1]^2$ to $\mathbb{R}$ by the following relation:
$K(s,t) = (1-s)t$ if $0 \leq t \leq s \leq 1$ and $K(s,t) = (1-t)s$ otherwise.
We denote by $T$ the application defined on $E = C([0,1], \mathbb{R})$, equipped with the norm $\|.\|_2 = \sqrt{\int_0^1 |f(x)|^2 \, dx}$, by the relation:
$$\forall f \in E, \quad \forall s \in [0,1], \quad T(f)(s) = \int_0^1 K(s,t) f(t) \, dt$$
Let $f \in E$. By decomposing $T(f)$ into two integrals, show that $T(f)$ is a $C^2$ function and express $(T(f))'$ then $(T(f))''$.