We admit that $(g_k)_{k \in \mathbb{N}^*}$ is a total sequence, where $g_k(x) = \sqrt{2}\sin(k\pi x)$.
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 $T(f) = \Phi$.