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] The best-approximation property has a clean geometric interpretation. The Fourier partial sum $S_n(x)$ is the orthogonal projection of $x$ onto the finite-dimensional subspace $V_n := \operatorname{span}\{e_1, \dots, e_n\}$. This follows from the [Projection Theorem](/theorems/240): $V_n$ is a closed subspace (it is finite-dimensional), and the characterisation of the projection as the unique element $y \in V_n$ satisfying $(x - y, v)_H = 0$ for all $v \in V_n$ is exactly the condition that the coefficients of $y$ equal the Fourier coefficients. The minimal error formula $\|x - S_n(x)\|^2 = \|x\|^2 - \sum_{k=1}^n c_k^2$ is therefore the Pythagorean theorem applied to the decomposition $x = S_n(x) + (x - S_n(x))$. Bessel's inequality $\sum c_k^2 \le \|x\|^2$ is a consequence of this projection interpretation: the energy in the projection cannot exceed the total energy. The difference $\|x\|^2 - \sum c_k^2$ measures how much of $x$ "escapes" the span of the orthonormal system. The question of when this gap closes — when does $\sum c_k^2 = \|x\|^2$? — is precisely the question of completeness. ### Bessel's Inequality is Sharp in General The inequality in [Bessel's Inequality](/theorems/540) can be strict. Consider the orthonormal system consisting of a single vector $e_1 = (1, 0, 0, \dots) \in \ell^2(\mathbb{N})$. For $x = (0, 1, 0, 0, \dots)$, the Fourier coefficient is $c_1 = (x, e_1) = 0$, so $\sum c_k^2 = 0 < 1 = \|x\|^2$. The entire "energy" of $x$ is invisible to this one-element orthonormal system. More generally, for any incomplete orthonormal system, there exists $x \in H$ with $\sum c_k^2 < \|x\|^2$. The deficit is $\|x - P_M x\|^2$ where $M = \overline{\operatorname{span}\{e_k\}}$ and $P_M$ is the orthogonal projection — this is zero for all $x$ if and only if $M = H$, i.e. the system is complete. ## Complete Orthonormal Systems and Parseval's Identity The central question of the theory is: when does an orthonormal system capture *all* of $H$? An orthonormal system in $\mathbb{R}^n$ is a basis if and only if it has exactly $n$ elements — the dimension count suffices. In infinite dimensions, there is no such count, and a more structural condition is needed. The right condition is *completeness*: the orthonormal system should have trivial orthogonal complement. Equivalently, the system should be "maximal" in the sense that no nonzero vector can be added while preserving orthogonality. What makes this condition powerful is that it is equivalent to several seemingly stronger conditions — Parseval's identity, the convergence of every Fourier expansion, and the density of the linear span — each of which captures a different aspect of the same phenomenon. [definition:Complete Orthonormal System] An orthonormal system $\{e_k\}_{k=1}^N$ in a Hilbert space $H$ is **complete** (or is an **orthonormal basis** of $H$) if the only element of $H$ orthogonal to every $e_k$ is $0$: \begin{align*} (x, e_k)_H = 0 \text{ for all } k \implies x = 0. \end{align*} Equivalently, $\{e_k\}^\perp = \{0\}$. [/definition] The term "orthonormal basis" is standard but potentially misleading: it does *not* mean that every element of $H$ is a *finite* linear combination of the $e_k$ (that would be an algebraic basis, or Hamel basis). Rather, it means that every element is an *infinite* series $\sum c_k e_k$ converging in the norm of $H$. In a separable infinite-dimensional Hilbert space, a Hamel basis is necessarily uncountable (by the [Baire category theorem](/theorems/630)), while a complete orthonormal system is countable. The two notions are fundamentally different. The following theorem collects the equivalent characterisations of completeness. The equivalence of these five conditions is the core result of the theory. [quotetheorem:541] The power of this result is that it allows one to verify completeness by whichever characterisation is most convenient. In practice, condition (2) — density of the span — is often the easiest to check, since it can leverage known approximation theorems (e.g. density of polynomials in $L^2$ via the Weierstrass theorem, or density of trigonometric polynomials via the Fejér kernel). The analytic payoff is condition (3), which guarantees that *every* element of $H$ has a convergent Fourier expansion, and condition (4), Parseval's identity, which provides an exact norm computation via the Fourier coefficients. The implication $(1) \Rightarrow (2)$ is where the [Orthogonal Decomposition Theorem](/theorems/241) enters: it converts the "no nonzero element is orthogonal to every $e_k$" condition into $\overline{\operatorname{span}\{e_k\}} = H$. The implication $(2) \Rightarrow (3)$ uses the best-approximation property of [Bessel's Inequality](/theorems/540): if the span is dense, the Fourier partial sums — which are the *best* approximations in the span — must converge. The chain $(3) \Rightarrow (4) \Rightarrow (5) \Rightarrow (1)$ is a sequence of algebraic manipulations: norm convergence gives Parseval, Parseval polarises to the generalised identity, and the generalised identity with $y = x$ immediately gives completeness. [example:Removing A Basis Element] Let $H = \ell^2(\mathbb{N})$ and let $\{e_k\}_{k=1}^\infty$ be the standard basis, which is complete. Now remove $e_1$ and consider the subsystem $\{e_k\}_{k=2}^\infty$. This subsystem is *not* complete: $e_1$ is orthogonal to every $e_k$ with $k \ge 2$, and $e_1 \ne 0$. Consequently, Parseval's identity fails: for $x = e_1$, the Fourier coefficients with respect to $\{e_k\}_{k=2}^\infty$ are all zero, so $\sum_{k=2}^\infty c_k(x)^2 = 0 < 1 = \|x\|^2$. The closed span of $\{e_k\}_{k=2}^\infty$ is the subspace $M = \{x \in \ell^2 : x_1 = 0\}$, which is a proper closed subspace of $\ell^2$. The Fourier expansion with respect to $\{e_k\}_{k=2}^\infty$ converges — but only to the projection $P_M(x) = (0, x_2, x_3, \dots)$, not to $x$ itself. This example illustrates the precise relationship between completeness and Parseval's identity: the gap $\|x\|^2 - \sum c_k^2 = \|x - P_M x\|^2$ measures the component of $x$ in the orthogonal complement $M^\perp = \operatorname{span}\{e_1\}$. [/example] ### Completeness is Not Redundancy-Free It is worth noting that a complete orthonormal system is "just barely enough" to span $H$: removing any single element breaks completeness (as the previous example shows), but the remaining elements still form a dense set in a codimension-one subspace. This is a consequence of the [Orthogonal Decomposition Theorem](/theorems/241): if $\{e_k\}$ is complete and we remove $e_j$, then $\overline{\operatorname{span}\{e_k : k \ne j\}} = \{e_j\}^\perp$, which has codimension one but is not all of $H$. ## The Gram-Schmidt Process The existence of orthonormal systems is not obvious — it requires a constructive procedure. In $\mathbb{R}^n$, one can extract an orthonormal basis from any spanning set by a linear algebra argument. The Gram-Schmidt process extends this to infinite dimensions: given any linearly independent sequence, it produces an orthonormal system with the same closed span. The significance of Gram-Schmidt goes beyond mere construction. It is the mechanism by which the abstract guarantee of [separability](/page/Separable) (the existence of a countable dense subset) is converted into a concrete orthonormal basis. Without it, the passage from "countable dense set" to "orthonormal basis" would require an appeal to Zorn's lemma, which gives existence without any constructive information. [quotetheorem:542] The key property is **span preservation**: the orthonormal system $\{e_1, \dots, e_n\}$ spans exactly the same subspace as $\{v_1, \dots, v_n\}$, for every $n$. This is crucial because it means Gram-Schmidt does not change the approximation properties of the original collection — any function approximable by $\{v_k\}$ is equally well approximable by $\{e_k\}$. In the infinite case, the closure of the span is preserved, so if the original collection is dense in $H$, the resulting orthonormal system is complete. The construction is numerically straightforward but can be unstable in floating-point arithmetic: the subtraction step $\tilde{e}_n = v_n - \sum (v_n, e_k) e_k$ involves cancellation, and for nearly linearly dependent vectors the result $\tilde{e}_n$ can have large relative error. In computational practice, the *modified* Gram-Schmidt algorithm reorders the subtractions to improve stability, and for large-scale problems, Householder reflections are preferred. These numerical considerations do not affect the mathematical theory, but they are worth noting because Gram-Schmidt is often the first encounter with the gap between mathematical elegance and computational reliability. [example:Legendre Polynomials] Apply Gram-Schmidt to the monomials $\{1, t, t^2, t^3, \dots\}$ in $L^2(-1, 1)$ with the inner product $(f, g) = \int_{-1}^1 f(t)\, g(t)\, d\mathcal{L}^1(t)$. **$e_0$:** Start with $v_0(t) = 1$. Then $\|v_0\|^2 = \int_{-1}^1 1\, d\mathcal{L}^1 = 2$, so $e_0(t) = 1/\sqrt{2}$. **$e_1$:** Take $v_1(t) = t$. The projection onto $e_0$ is $(v_1, e_0)\, e_0 = \left(\int_{-1}^1 t \cdot \frac{1}{\sqrt{2}}\, d\mathcal{L}^1\right) \frac{1}{\sqrt{2}} = 0$ (by symmetry, since $t$ is odd on $[-1,1]$). So $\tilde{e}_1(t) = t$, and $\|\tilde{e}_1\|^2 = \int_{-1}^1 t^2\, d\mathcal{L}^1 = \frac{2}{3}$, giving $e_1(t) = t\sqrt{3/2}$. **$e_2$:** Take $v_2(t) = t^2$. The projection onto $\{e_0, e_1\}$ is: \begin{align*} (v_2, e_0)\, e_0 &= \left(\int_{-1}^1 t^2 \cdot \frac{1}{\sqrt{2}}\, d\mathcal{L}^1\right) \frac{1}{\sqrt{2}} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}, \\ (v_2, e_1)\, e_1 &= \left(\int_{-1}^1 t^2 \cdot t\sqrt{\tfrac{3}{2}}\, d\mathcal{L}^1\right) t\sqrt{\tfrac{3}{2}} = 0 \quad \text{(odd integrand)}. \end{align*} So $\tilde{e}_2(t) = t^2 - \frac{1}{3}$, and $\|\tilde{e}_2\|^2 = \int_{-1}^1 (t^2 - \frac{1}{3})^2\, d\mathcal{L}^1 = \int_{-1}^1 \left(t^4 - \frac{2}{3}t^2 + \frac{1}{9}\right) d\mathcal{L}^1 = \frac{2}{5} - \frac{4}{9} + \frac{2}{9} = \frac{8}{45}$, giving $e_2(t) = (t^2 - \frac{1}{3})\sqrt{45/8}$. The unnormalised polynomials $\tilde{e}_n$ are scalar multiples of the **Legendre polynomials** $P_n(t)$, which are the standard orthogonal polynomials on $[-1, 1]$ with weight $w \equiv 1$. The [Weierstrass approximation theorem](/theorems/480) guarantees that polynomials are dense in $L^2(-1, 1)$, so by the span-preservation property of Gram-Schmidt, $\{e_n\}_{n=0}^\infty$ is a complete orthonormal system. [/example] ## Existence of Orthonormal Bases The examples above — $\ell^2$, the trigonometric system, the Legendre polynomials — each exhibit a complete orthonormal system in a particular space. The natural question is whether every Hilbert space possesses one. In full generality (non-separable spaces), the existence requires Zorn's lemma. For separable Hilbert spaces — the setting of virtually all applications — the existence is constructive and follows from Gram-Schmidt. The hypothesis of separability is not merely a technical convenience: it is equivalent to the existence of a *countable* complete orthonormal system. A non-separable Hilbert space has uncountable orthonormal bases, and the theory becomes measure-theoretically more delicate. In practice, the Hilbert spaces of analysis ($L^2$, $H^1$, $\ell^2$) are all separable. [quotetheorem:543] The proof strategy is clean: take a countable dense subset (guaranteed by separability), extract a linearly independent subcollection that is still dense, and apply [Gram-Schmidt Orthonormalisation](/theorems/542). The completeness of the resulting orthonormal system follows from the density of its span, via condition (2) of the [Characterisation of Complete Orthonormal Systems](/theorems/541). The countability statement in part (2) has an appealing geometric proof: distinct elements of an orthonormal system are at distance $\sqrt{2}$ from each other (by the Pythagorean theorem), so the balls of radius $\frac{1}{2}$ centred at them are disjoint. In a separable space, any collection of pairwise disjoint [open sets](/page/Open%20Set) is at most countable (each must contain a point of the countable dense subset). ## Isomorphism with $\ell^2$ The preceding results combine into a remarkable structural conclusion: every separable infinite-dimensional Hilbert space is, up to isometric isomorphism, the same space — namely, $\ell^2(\mathbb{N})$. The isomorphism is nothing other than the Fourier coefficient map. This is a striking contrast with [Banach spaces](/page/Banach%20Space), where the isometric classification is extremely rich: $L^p$ spaces for different values of $p$ are non-isomorphic, and even isomorphically, the classification of separable Banach spaces is a deep and incomplete subject. In the Hilbert space category, separability and infinite-dimensionality determine the space uniquely. [quotetheorem:544] The isometric isomorphism $\Phi: H \to \ell^2$ sends $x$ to its sequence of Fourier coefficients, and the inverse $\Phi^{-1}$ reconstructs $x$ from its Fourier expansion. Parseval's identity is precisely the statement that $\Phi$ is an isometry, and the surjectivity of $\Phi$ is the statement that every square-summable sequence of coefficients gives rise to a convergent series in $H$. ### What the Isomorphism Does and Does Not Preserve The isomorphism $\Phi$ preserves the Hilbert space structure: inner products, norms, convergence, orthogonality, projections, and all the tools developed in the [Hilbert Space](/page/Hilbert%20Space) page. Any theorem proved in $\ell^2$ about inner products and norms transfers immediately to any separable Hilbert space. However, $\Phi$ does *not* preserve additional structure that the space may carry. For instance, $L^2(0, 1)$ has a pointwise product $(fg)(x) = f(x)g(x)$, but $\ell^2$ does not inherit a natural product from this operation — the Fourier coefficients of $fg$ are not simply the termwise products of the Fourier coefficients of $f$ and $g$. (They are given by *convolution* of the coefficient sequences.) Similarly, the notion of "support" of a function in $L^2$ has no natural counterpart in $\ell^2$. The isomorphism also depends on the *choice* of orthonormal basis: different bases give different isomorphisms, and there is no canonical choice. The classification theorem should therefore be understood as a statement about *Hilbert space structure*, not about the richer structures (algebraic, order-theoretic, measure-theoretic) that specific Hilbert spaces may possess. [example:Isomorphism Of L Two With Ell Two] Let $H = L^2(-1, 1)$ and let $\{e_n\}_{n=0}^\infty$ be the normalised Legendre polynomials constructed in the Gram-Schmidt example above. The isomorphism $\Phi: L^2(-1, 1) \to \ell^2$ sends each function $f$ to its sequence of Legendre-Fourier coefficients: \begin{align*} \Phi(f) = \left(\int_{-1}^1 f(t)\, e_0(t)\, d\mathcal{L}^1(t),\; \int_{-1}^1 f(t)\, e_1(t)\, d\mathcal{L}^1(t),\; \dots\right). \end{align*} Parseval's identity becomes the concrete statement that for every $f \in L^2(-1, 1)$: \begin{align*} \int_{-1}^1 |f(t)|^2\, d\mathcal{L}^1(t) = \sum_{n=0}^\infty \left(\int_{-1}^1 f(t)\, e_n(t)\, d\mathcal{L}^1(t)\right)^2. \end{align*} This is a non-trivial identity: the left side is a single integral, while the right side is an infinite series of squared [integrals](/page/Integral). The equality asserts that no $L^2$ energy is lost in the passage from the function to its Legendre coefficients. [/example] ## Problems [problem] Let $f: [-1, 1] \to \mathbb{R}$ be defined by $f(t) = t$. Compute the Fourier-Legendre coefficients of $f$ with respect to the normalised Legendre polynomials $\{e_n\}_{n=0}^\infty$ in $L^2(-1, 1)$, and verify Parseval's identity directly. [/problem] [solution] **Step 1: Identify the relevant orthonormal polynomials.** From the Gram-Schmidt construction: \begin{align*} e_0(t) = \frac{1}{\sqrt{2}}, \quad e_1(t) = t\sqrt{\frac{3}{2}}, \quad e_2(t) = \left(t^2 - \frac{1}{3}\right)\sqrt{\frac{45}{8}}. \end{align*} Since $f(t) = t$ is a polynomial of degree $1$, it lies in $\operatorname{span}\{e_0, e_1\}$, so all Fourier coefficients $c_n$ with $n \ge 2$ vanish. (This also follows from the span-preservation property of Gram-Schmidt: $\operatorname{span}\{e_0, e_1\} = \operatorname{span}\{1, t\}$.) **Step 2: Compute the Fourier coefficients.** The zeroth coefficient is: \begin{align*} c_0 = (f, e_0)_{L^2} = \int_{-1}^1 t \cdot \frac{1}{\sqrt{2}}\, d\mathcal{L}^1(t) = \frac{1}{\sqrt{2}} \cdot 0 = 0, \end{align*} since $t$ is an odd function on the symmetric interval $[-1, 1]$. The first coefficient is: \begin{align*} c_1 = (f, e_1)_{L^2} = \int_{-1}^1 t \cdot t\sqrt{\frac{3}{2}}\, d\mathcal{L}^1(t) = \sqrt{\frac{3}{2}} \int_{-1}^1 t^2\, d\mathcal{L}^1(t) = \sqrt{\frac{3}{2}} \cdot \frac{2}{3} = \frac{2}{3}\sqrt{\frac{3}{2}} = \sqrt{\frac{2}{3}}. \end{align*} **Step 3: Verify the Fourier expansion.** The expansion gives: \begin{align*} \sum_{n=0}^\infty c_n\, e_n(t) = 0 \cdot e_0(t) + \sqrt{\frac{2}{3}} \cdot t\sqrt{\frac{3}{2}} = t = f(t). \end{align*} The series terminates and reproduces $f$ exactly, as expected for a polynomial of degree $1$. **Step 4: Verify Parseval's identity.** The left side of Parseval's identity is: \begin{align*} \|f\|_{L^2}^2 = \int_{-1}^1 t^2\, d\mathcal{L}^1(t) = \frac{2}{3}. \end{align*} The right side is: \begin{align*} \sum_{n=0}^\infty c_n^2 = 0^2 + \left(\sqrt{\frac{2}{3}}\right)^2 + 0 + 0 + \cdots = \frac{2}{3}. \end{align*} Both sides equal $\frac{2}{3}$, confirming Parseval's identity. [/solution] [problem] Let $H$ be a separable Hilbert space, and let $\{e_k\}_{k=1}^\infty$ be a complete orthonormal system. Let $\{f_k\}_{k=1}^\infty$ be a sequence in $H$ satisfying \begin{align*} \sum_{k=1}^\infty \|e_k - f_k\|_H^2 < 1. \end{align*} Prove that $\{f_k\}$ is also a complete system in $H$ (not necessarily orthonormal, but its closed linear span is $H$). [/problem] [solution] **Step 1: Define the perturbation operator.** Define the linear operator $T: H \to H$ by \begin{align*} Tx := \sum_{k=1}^\infty (x, e_k)_H\, (f_k - e_k). \end{align*} This series converges for every $x \in H$: by the Cauchy-Schwarz inequality in $\ell^2$, \begin{align*} \|Tx\|_H^2 &= \left\|\sum_{k=1}^\infty c_k(x)(f_k - e_k)\right\|_H^2. \end{align*} To bound this, apply the generalised triangle inequality. For any finite partial sum: \begin{align*} \left\|\sum_{k=1}^n c_k(x)(f_k - e_k)\right\|_H \le \sum_{k=1}^n |c_k(x)| \cdot \|f_k - e_k\|_H \le \left(\sum_{k=1}^n c_k(x)^2\right)^{1/2} \left(\sum_{k=1}^n \|f_k - e_k\|_H^2\right)^{1/2}, \end{align*} by the Cauchy-Schwarz inequality for finite sums. By [Bessel's Inequality](/theorems/540) and the hypothesis: \begin{align*} \|Tx\|_H \le \|x\|_H \cdot \left(\sum_{k=1}^\infty \|e_k - f_k\|_H^2\right)^{1/2} < \|x\|_H. \end{align*} Therefore $T$ is a bounded linear operator with $\|T\| < 1$. **Step 2: Invertibility of $I + T$.** Since $\|T\| < 1$, the operator $I + T$ is invertible with bounded inverse given by the Neumann series $(I + T)^{-1} = \sum_{n=0}^\infty (-T)^n$. In particular, $I + T$ is surjective. **Step 3: Relate $I + T$ to the systems.** For any $x \in H$, using the Fourier expansion $x = \sum_{k=1}^\infty (x, e_k)_H\, e_k$ (which converges by completeness of $\{e_k\}$): \begin{align*} (I + T)x = x + Tx = \sum_{k=1}^\infty (x, e_k)_H\, e_k + \sum_{k=1}^\infty (x, e_k)_H\, (f_k - e_k) = \sum_{k=1}^\infty (x, e_k)_H\, f_k. \end{align*} Therefore $\operatorname{Range}(I + T) \subseteq \overline{\operatorname{span}\{f_k\}}$. **Step 4: Conclude completeness.** Since $I + T$ is surjective, $\operatorname{Range}(I + T) = H$. Combined with the inclusion from Step 3: \begin{align*} H = \operatorname{Range}(I + T) \subseteq \overline{\operatorname{span}\{f_k\}} \subseteq H, \end{align*} so $\overline{\operatorname{span}\{f_k\}} = H$, proving that $\{f_k\}$ is a complete system. [/solution] ## References 1. H. Brezis, *Functional Analysis, [Sobolev Spaces](/page/Sobolev%20Space) and Partial Differential Equations* (2010). 2. W. Rudin, *Functional Analysis* (1991). 3. M. Reed and B. Simon, *Methods of Modern Mathematical Physics I: Functional Analysis* (1980). 4. N. Young, *An Introduction to Hilbert Space* (1988).