The [Hilbert Space](/page/Hilbert%20Space) page developed the foundational geometry — projections, orthogonal complements, the decomposition $H = M \oplus M^\perp$ for a closed subspace $M$. That theory decomposes a Hilbert space into two complementary pieces. The present page asks: can we go further and decompose $H$ into *countably many* mutually orthogonal one-dimensional pieces, obtaining an infinite-dimensional coordinate system?

In $\mathbb{R}^n$, the answer is immediate: any orthogonal basis $\{e_1, \dots, e_n\}$ gives a unique expansion $x = \sum_{k=1}^n (x, e_k) e_k$. The coefficients are computed by inner products, the Pythagorean theorem controls norms, and the representation is exact. The question is whether — and in what sense — this finite-dimensional picture survives the passage to infinite dimensions. The answer is that it does, but with a crucial caveat: the expansion $x = \sum_{k=1}^\infty c_k e_k$ is now an *infinite series*, and convergence must be established. The theory of orthogonal systems is the systematic study of when and how such expansions work.

## Motivation

[motivation] ### Finite-Dimensional Success In $\mathbb{R}^n$ with the Euclidean inner product, every orthonormal basis $\{e_1, \dots, e_n\}$ provides a complete coordinate system. Given any vector $x \in \mathbb{R}^n$, the expansion \begin{align*} x = \sum_{k=1}^n (x, e_k)\, e_k \end{align*} holds as an algebraic identity, and the Pythagorean theorem gives $\|x\|^2 = \sum_{k=1}^n (x, e_k)^2$. Both facts are elementary consequences of linear independence and the dimension count: $n$ orthonormal vectors in an $n$-dimensional space must span. Convergence is not an issue because the sum is finite. ### The Infinite-Dimensional Attempt Now consider $H = L^2(0, 2\pi)$ and the orthonormal sequence $\{e_k\}_{k=1}^\infty$ where, say, $e_k(x) = \frac{1}{\sqrt{2\pi}} e^{ikx}$. Given $f \in L^2(0, 2\pi)$, we can compute the Fourier coefficients $c_k = (f, e_k)_{L^2}$ and form the partial sums $S_N(f) = \sum_{k=1}^N c_k\, e_k$. Three questions arise that had no counterpart in finite dimensions: 1. **Does the [series](/page/Series) converge?** The partial sums $S_N(f)$ are elements of $H$, but there is no a priori reason why $\|f - S_N(f)\|$ should tend to zero. Convergence is an analytic question, not an algebraic one. 2. **Is the energy accounted for?** Even if $\sum c_k^2$ converges, it could converge to a value strictly less than $\|f\|^2$, meaning that the orthonormal system "misses" part of $f$. The Pythagorean identity $\|f\|^2 = \sum c_k^2$ — which was automatic in $\mathbb{R}^n$ — becomes a substantive claim (Parseval's identity) that must be proved. 3. **Is the system large enough?** A finite orthonormal set in an infinite-dimensional space cannot span. We need a criterion that distinguishes an orthonormal system that captures the entire space from one that captures only a proper subspace. ### The Resolution: Completeness The correct condition is *completeness*: an orthonormal system $\{e_k\}$ is complete if the only element orthogonal to every $e_k$ is $0$. This is the exact analogue of the dimension count in $\mathbb{R}^n$: a complete orthonormal system has trivial orthogonal complement, so the [Orthogonal Decomposition Theorem](/theorems/241) forces $\overline{\operatorname{span}\{e_k\}} = H$. With completeness in hand, all three questions above are resolved: the [Fourier series](/page/Fourier%20Series) converges, Parseval's identity holds, and every element of $H$ is recovered by its Fourier expansion. The development below makes this precise. [/motivation]

## Orthogonal and Orthonormal Systems

The basic objects of the theory are systems of mutually perpendicular elements.

A natural setting for studying orthogonal expansions requires a space with an inner product, but the definitions make sense even without completeness. However, the deep results — convergence of Fourier series, Parseval's identity, existence of complete orthonormal systems — all rely on the completeness of $H$.

[definition:Orthogonal System] Let $H$ be an inner product space over $\mathbb{R}$. A collection $\{v_k\}_{k \in I}$ (where $I$ is any index set) is an **orthogonal system** in $H$ if: 1. $v_k \ne 0$ for every $k \in I$, 2. $(v_j, v_k)_H = 0$ for all $j \ne k$ in $I$. [/definition]

The exclusion of the zero vector is a convention that avoids degenerate situations: the zero vector is orthogonal to everything but carries no directional information. Without this convention, statements about linear independence and completeness would require constant qualification.

[definition:Orthonormal System] An orthogonal system $\{e_k\}_{k \in I}$ in $H$ is an **orthonormal system** if every element has unit norm: \begin{align*} (e_j, e_k)_H = \delta_{jk} := \begin{cases} 1 & \text{if } j = k, \\ 0 & \text{if } j \ne k, \end{cases} \end{align*} for all $j, k \in I$, where $\delta_{jk}$ is the Kronecker delta. [/definition]

Any orthogonal system can be normalised to an orthonormal one by replacing each $v_k$ with $v_k / \|v_k\|_H$, so the two notions are essentially interchangeable. The orthonormal convention simplifies all subsequent formulas — the Fourier coefficients, Bessel's inequality, and Parseval's identity all take their cleanest form with respect to orthonormal systems — so we work exclusively with orthonormal systems from this point on.

Every orthonormal system is linearly independent: if $\sum_{k=1}^n a_k e_k = 0$, then for each $j$, taking the inner product with $e_j$ gives $a_j = (\sum a_k e_k, e_j)_H = 0$. In finite dimensions, a linearly independent set of size $n$ in an $n$-dimensional space is automatically a basis. In infinite dimensions, linear independence alone guarantees nothing about the span — the orthonormal system could span a proper closed subspace.

[example:Standard Basis of $\ell^2$] The sequence space $\ell^2(\mathbb{N})$ consists of all real [sequences](/page/Sequence) $x = (x_1, x_2, \dots)$ with $\sum_{k=1}^\infty x_k^2 < \infty$, equipped with the inner product $(x, y)_{\ell^2} = \sum_{k=1}^\infty x_k y_k$. The **standard basis** $\{e_k\}_{k=1}^\infty$, where $e_k$ has a $1$ in position $k$ and $0$ elsewhere, is an orthonormal system. It is also complete: if $x \in \ell^2$ satisfies $(x, e_k)_{\ell^2} = 0$ for all $k$, then $x_k = 0$ for all $k$, so $x = 0$. The Fourier expansion recovers the identity \begin{align*} x = \sum_{k=1}^\infty x_k\, e_k, \end{align*} which is simply the statement that a sequence is the sum of its components — a tautology in $\ell^2$, but one that becomes a theorem once the convergence of the series is understood as convergence in the $\ell^2$ norm. [/example]

[example:Trigonometric System] The space $L^2(0, 2\pi)$ with the inner product $(f, g)_{L^2} = \int_0^{2\pi} f(t)\, g(t)\, d\mathcal{L}^1(t)$ carries the classical **trigonometric orthonormal system**: \begin{align*} e_0(t) = \frac{1}{\sqrt{2\pi}}, \quad e_k^{(c)}(t) = \frac{\cos(kt)}{\sqrt{\pi}}, \quad e_k^{(s)}(t) = \frac{\sin(kt)}{\sqrt{\pi}}, \quad k = 1, 2, 3, \dots \end{align*} Orthonormality is verified by direct integration. For instance, for $j \ne k$: \begin{align*} \int_0^{2\pi} \cos(jt)\cos(kt)\, d\mathcal{L}^1(t) = \frac{1}{2}\int_0^{2\pi} \left[\cos((j-k)t) + \cos((j+k)t)\right] d\mathcal{L}^1(t) = 0, \end{align*} since both $j - k$ and $j + k$ are nonzero integers, and $\int_0^{2\pi} \cos(nt)\, d\mathcal{L}^1(t) = 0$ for every nonzero integer $n$. The normalisation $\int_0^{2\pi} \cos^2(kt)\, d\mathcal{L}^1(t) = \pi$ gives the $1/\sqrt{\pi}$ factor. That this system is *complete* — i.e. the only $L^2$ [function](/page/Function) orthogonal to all sines and cosines is the zero function — is a deep fact equivalent to the density of trigonometric polynomials in $L^2(0, 2\pi)$. This density is established via [convolution](/page/Convolution) with the [Fejér kernel](/page/Fej%C3%A9r%20Kernel) and is one of the central results of classical Fourier analysis. [/example]

## Fourier Coefficients and Best Approximation

Given an orthonormal system $\{e_k\}$ and an element $x \in H$, the most natural way to approximate $x$ in $\operatorname{span}\{e_1, \dots, e_n\}$ is by its **Fourier partial sum** $S_n(x) := \sum_{k=1}^n (x, e_k)_H\, e_k$. Why are the coefficients $(x, e_k)_H$ the right choice? The answer is that they solve the best-approximation problem: among all linear combinations $\sum a_k e_k$, the Fourier partial sum is the unique closest element to $x$. Moreover, the error has a clean formula: the squared distance from $x$ to the span of the first $n$ basis vectors is $\|x\|^2$ minus the sum of squared Fourier coefficients.

[definition:Fourier Coefficient] Let $\{e_k\}_{k=1}^N$ (where $N \in \mathbb{N} \cup \{\infty\}$) be an orthonormal system in a Hilbert space $H$, and let $x \in H$. The **Fourier coefficient** of $x$ with respect to $e_k$ is \begin{align*} c_k(x) := (x, e_k)_H. \end{align*} The **Fourier partial sum** of order $n$ is \begin{align*} S_n(x) := \sum_{k=1}^n c_k(x)\, e_k. \end{align*} [/definition]

The notation $c_k(x)$ emphasises that the Fourier coefficient is a bounded linear functional of $x$: by the Cauchy-Schwarz inequality, $|c_k(x)| = |(x, e_k)_H| \le \|x\|_H \|e_k\|_H = \|x\|_H$. In the language of the [Riesz Representation Theorem](/theorems/221), the functional $x \mapsto c_k(x)$ is the element of $H^*$ represented by $e_k$, and its norm is $\|e_k\|_H = 1$.

The geometric content of the Fourier expansion is captured by the following result, which simultaneously characterises the best approximation and controls the approximation error.

[quotetheorem:540]