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 $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence. For all $f \in E$, we set, $$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2 \pi^2} \langle f, g_k \rangle g_k(x)$$ For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$. Deduce $T(f) = \Phi$.
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 $k \in \mathbb{N}^*$, we set $g_k(x) = \sqrt{2} \sin(k\pi x)$. We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence.\\
For all $f \in E$, we set,
$$\forall x \in [0,1], \quad \Phi(x) = \sum_{k=1}^{+\infty} \frac{1}{k^2 \pi^2} \langle f, g_k \rangle g_k(x)$$
For all $N \in \mathbb{N}$, we set $f_N = \sum_{k=1}^N \langle f, g_k \rangle g_k$.\\
Deduce $T(f) = \Phi$.