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 $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We denote by $G = \operatorname{Vect}\left((g_k)_{k \in \mathbb{N}^*}\right)$ and $H = G^\perp$. We admit that, $$H \neq \{0\} \Longrightarrow \exists f \in H \text{ such that } \left\{ \begin{array}{l} \|f\| = 1 \\ \langle T(f), f \rangle = \sup_{h \in H, \|h\|=1} \langle T(h), h \rangle \end{array} \right.$$ Deduce that $H = \{0\}$.
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 $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We denote by $G = \operatorname{Vect}\left((g_k)_{k \in \mathbb{N}^*}\right)$ and $H = G^\perp$.\\
We admit that,
$$H \neq \{0\} \Longrightarrow \exists f \in H \text{ such that } \left\{ \begin{array}{l} \|f\| = 1 \\ \langle T(f), f \rangle = \sup_{h \in H, \|h\|=1} \langle T(h), h \rangle \end{array} \right.$$
Deduce that $H = \{0\}$.