Assume $\mathcal{H}$ is a separable Hilbert space with orthonormal basis function $\{e_i\}_{i=1}^{\infty}$ that are countably infinite. We know there always exists a Schauder basis $\{ y_i \}_{i=1}^{\infty}$ for $\mathcal{H}$ such that
$$ \forall x \in \mathcal{H}, \;\; \exists ! \alpha_{(\cdot)} \in \mathbb{R}^{\mathbb{N}}, \;\;x = \sum_{i=1}^{\infty} \alpha_i y _i. $$
I have a question as follow: If $\{f_i\}_{i=1}^{d} \subset \mathcal{H}$ are linearly independent with $d \in \mathbb{N}$, can we always find functions $\{f_i\}_{i=d+1}^{\infty} \subset \mathcal{H}$ such that $\{f_i\}_{i=1}^{\infty}$ is a Schauder basis for $\mathcal{H}$?
Let $\{f_i\}_{i=d+1}^\infty$ be an orthonormal basis in $\mathcal{H}_0:=\{f_1,f_2,\ldots, f_d\}^\perp.$ Then $\{f_i\}_{i=1}^\infty $ is a Schauder basis for $\mathcal{H}.$ Indeed, let $P$ denote the orthogonal projection onto $\mathcal{H}_0.$ Then $I-P$ is the orthogonal projection onto $\mathcal{H}_0^\perp={\rm span}\,\{f_1,f_2,\ldots, f_d\}.$ For $x\in \mathcal{H}$ we have $$\displaylines{x=Px+(I-P)x=\sum_{i=d+1}^\infty \langle Px,f_i\rangle f_i+ \sum_{i=1}^d\alpha_i(x)f_i\\ = \sum_{i=d+1}^\infty \langle x,f_i\rangle f_i+ \sum_{i=1}^d\alpha_i(x)f_i}$$ The decomposition is unique. Indeed, assume $$0=\sum_{i=d+1}^\infty \alpha_if_i+\sum_{i=1}^d \alpha_if_i$$ The sums are orthogonal to each other hence $$\sum_{i=d+1}^\infty \alpha_if_i=0,\quad \sum_{i=1}^d \alpha_if_i=0$$ Thus $\alpha_i=0$ for any $i\ge d+1$ by orthogonality, and for $i=1,2,\ldots, d$ by linear independence.