An Orthonormal Family $\{e_k\}_{k\in\mathbb{N}}$ is a basis if and only if
$$f=\sum^\infty_{n=1}\hat{f}(n)e_n \ \ \ \text{in} \ \mathcal{L}^2(\mathbb{R})$$
where $f\in\mathcal{L}^2(\mathbb{R})$ and $\hat{f}(k):=(f,e_k)=\int_\mathbb{R}f(x)\overline{e_k(x)}dx$
I am unsure how to prove this Lemma. It is apparently a Corollary from
$$\left\vert\left\vert f-\sum^n_{k=1}\hat{f}(k)e_k\right\vert\right\vert\leq\left\vert\left\vert f-\sum^n_{k=1}c_ke_k\right\vert\right\vert$$
Any hints?
Since orthonormal families are automatically linearly independent, the "only if" direction is trivial.
On the other hand, suppose the orthonormal family is a basis, and fix $\varepsilon > 0$. Then by definition of a (Schauder) basis, there exists some positive integer $n$ and $c_i \in \mathbb{C}$ for $i = 1, \ldots n$ such that $$\left\lVert f - \sum_{i=1}^n c_i e_i \right\rVert < \varepsilon.$$ By the lemma, this implies $$\left\lVert f - \sum_{i=1}^n \hat{f}(i) e_i \right\rVert < \varepsilon.$$ Now, fix $k \ge n$. Applying the lemma once more (with some of the $c_i = 0$), we have $$\left\lVert f - \sum_{i=1}^k \hat{f}(i) e_i \right\rVert \le \left\lVert f - \sum_{i=1}^n \hat{f}(i) e_i \right\rVert < \varepsilon.$$ Therefore, by definition of a limit, $$\sum_{i=1}^\infty \hat{f}(i) e_i = f,$$ and we are done.