[proofplan] We prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5) \Rightarrow (1)$. The key insight is that completeness ($\{e_k\}^\perp = \{0\}$) forces the closed span to be all of $H$ by the [Orthogonal Decomposition Theorem](/theorems/241). Once density is established, the Fourier expansion and Parseval's identity follow from the best-approximation property of [Bessel's Inequality](/theorems/540). The return from (5) to (1) is immediate by specialisation. [/proofplan] [step:$(1) \Rightarrow (2)$: Completeness implies density of span] [claim:Completeness Implies Density] If $(x, e_k)_H = 0$ for all $k$ implies $x = 0$, then $\overline{\operatorname{span}\{e_k\}} = H$. [/claim] [proof] Let $M = \overline{\operatorname{span}\{e_k\}}$. By the [Orthogonal Decomposition Theorem](/theorems/241), $H = M \oplus M^\perp$. If $x \in M^\perp$, then $(x, e_k)_H = 0$ for all $k$, so $x = 0$ by (1). Therefore $M^\perp = \{0\}$ and $H = M$. [/proof] [/step] [step:$(2) \Rightarrow (3)$: Density implies norm convergence of the Fourier series] [claim:Density Implies Fourier Convergence] If $\overline{\operatorname{span}\{e_k\}} = H$, then $x = \sum c_k(x)\, e_k$ for every $x \in H$. [/claim] [proof] Given $\varepsilon > 0$, density provides $\sum a_k e_k$ with $\|x - \sum a_k e_k\|_H < \varepsilon$. By the best-approximation property of [Bessel's Inequality](/theorems/540), $\|x - S_n\|_H \leq \|x - \sum a_k e_k\|_H < \varepsilon$ for $n$ large enough, and $\|x - S_m\|_H \leq \|x - S_n\|_H$ for $m \geq n$. [/proof] [/step] [step:$(3) \Rightarrow (4)$: Fourier expansion implies Parseval's identity] If $x = \sum c_k(x) e_k$ in norm, then \begin{align*} \|x\|_H^2 = \lim_{n \to \infty} \left\|\sum_{k=1}^n c_k(x)\, e_k\right\|_H^2 = \lim_{n \to \infty} \sum_{k=1}^n c_k(x)^2 = \sum_{k=1}^N c_k(x)^2. \end{align*} [/step] [step:$(4) \Rightarrow (5)$: Parseval implies the generalised Parseval identity via polarisation] [claim:Parseval Implies Generalised Parseval] If $\sum c_k(x)^2 = \|x\|_H^2$ for all $x$, then $\sum c_k(x) c_k(y) = (x, y)_H$ for all $x, y$. [/claim] [proof] Apply (4) to $x + y$ and $x - y$, subtract, and divide by $4$: \begin{align*} \sum_{k=1}^N c_k(x)\, c_k(y) = \frac{1}{4}\left(\|x+y\|_H^2 - \|x-y\|_H^2\right) = (x, y)_H \end{align*} by the polarisation identity. [/proof] [/step] [step:$(5) \Rightarrow (1)$: Generalised Parseval implies completeness] If $(x, e_k)_H = 0$ for all $k$, then all $c_k(x) = 0$, and (5) with $y = x$ gives $\|x\|_H^2 = 0$, so $x = 0$. [/step]