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Criteria For Completeness Of Orthonormal Systems

Completeness can be tested either by orthogonality or by approximation; the key is to use the closed span, not just the finite linear span.

Advanced 10 min read

What You'll Learn

  • Define completeness using the closure of an orthonormal system’s finite linear span.
  • Prove that completeness is equivalent to having zero orthogonal complement.
  • Test completeness by checking whether a vector orthogonal to every system member must vanish.
  • Use dense subsets and norm approximation to establish completeness.
  • Distinguish a dense algebraic span from a span equal to the entire space.

Completeness as Density in \(L^2\)

An orthonormal system need not span every vector by a finite linear combination. Its significance is instead determined by whether finite linear combinations approximate every vector in norm. This distinction matters in infinite-dimensional spaces: a subspace can be proper and still be dense. A second test turns approximation into an orthogonality question. If the only vector orthogonal to every member of the system is zero, then the system is complete.

Throughout, \((X,\mathcal{M},\mu)\) is a measure space, with no assumption that \(\mu\) is finite or sigma-finite. The space \(L^2(\mu)\) consists of equivalence classes, modulo equality \(\mu\)-a.e., of complex-valued measurable functions \(f\) for which \(\int_X |f|^2\,d\mu<\infty\). It is a Hilbert space with inner product and norm

$$ \langle f,g\rangle=\int_X f\overline{g}\,d\mu, \qquad \|f\|_2=\left(\int_X|f|^2\,d\mu\right)^{1/2}. $$

For real-valued \(L^2\), omit the complex conjugate. An orthonormal system is a family \((u_i)_{i\in I}\) in \(L^2(\mu)\) such that \(\langle u_i,u_j\rangle=0\) when \(i\ne j\), and \(\|u_i\|_2=1\) for every \(i\in I\). Its finite linear span is the set of all finite sums \(\sum_{j=1}^k a_j u_{i_j}\), with scalar coefficients \(a_j\). Let \(M\) be the closure of this span in the \(L^2\) norm.

Definition: The orthonormal system \((u_i)_{i\in I}\) is complete in \(L^2(\mu)\) if its finite linear span is dense in \(L^2(\mu)\); equivalently, if \(M=L^2(\mu)\). Its orthogonal complement is $$ M^\perp=\{h\in L^2(\mu):\langle h,g\rangle=0\text{ for every }g\in M\}. $$ Since \(M\) is the closed span of the system, \(h\in M^\perp\) exactly when \(\langle h,u_i\rangle=0\) for every \(i\in I\).

The last equivalence follows directly from continuity of the inner product: if \(h\) is orthogonal to each \(u_i\), it is orthogonal to every finite linear combination, and then to every norm limit of those combinations. The converse holds because every \(u_i\) belongs to \(M\).

The Orthogonal-Complement Criterion

Theorem: Let \((X,\mathcal{M},\mu)\) be any measure space, and let \((u_i)_{i\in I}\) be an orthonormal system in the real or complex Hilbert space \(L^2(\mu)\). If \(M\) is its closed linear span, then the system is complete if and only if \(M^\perp=\{0\}\). Equivalently, it is complete if and only if the only \(h\in L^2(\mu)\) satisfying \(\langle h,u_i\rangle=0\) for all \(i\in I\) is \(h=0\) as an \(L^2\) element (that is, \(h=0\) \(\mu\)-a.e.).

Proof. We use the orthogonal projection theorem for Hilbert spaces: if \(M\) is a closed subspace of a Hilbert space \(H\), then every \(f\in H\) has a unique decomposition \(f=p+r\), where \(p\in M\) and \(r\in M^\perp\). Here \(H=L^2(\mu)\), which is complete, and \(M\) is closed by definition, so the theorem applies.

If the system is complete, then \(M=H\). A vector \(h\in M^\perp\) is therefore orthogonal to every vector in \(H\), including itself. Thus \(\|h\|_2^2=\langle h,h\rangle=0\), so \(h=0\) as an \(L^2\) element.

Conversely, suppose \(M^\perp=\{0\}\), and take any \(f\in H\). Write \(f=p+r\) using the orthogonal projection theorem, with \(p\in M\) and \(r\in M^\perp\). The assumption forces \(r=0\), so \(f=p\in M\). Since this holds for every \(f\in H\), \(M=H\), and the system is complete. The equivalence with the coefficient test follows from the definition of \(M^\perp\) and the continuity argument above. This also covers the zero Hilbert space and the empty orthonormal system: in that case \(M=\{0\}=H\), so the empty system is complete. \(\square\)

This criterion replaces a potentially difficult construction of approximations to every \(f\) by a test for nonzero vectors satisfying a family of equations. Conversely, the density formulation is often easier to verify when a convenient dense set is available.

Approximation criterion: Let \(H=L^2(\mu)\), let \(M\) be the closed span of an orthonormal system, and let \(D\subseteq H\) be dense in \(H\). If every \(d\in D\) can be approximated arbitrarily closely in norm by finite linear combinations of the system, then the system is complete. Conversely, if the system is complete, every \(d\in D\) has such approximations.

Proof. Suppose first that each \(d\in D\) can be approximated arbitrarily closely by vectors in the finite linear span. Every such \(d\) then belongs to its closure \(M\). Because \(M\) is closed and contains the dense set \(D\), it contains the closure of \(D\), which is all of \(H\). Hence \(M=H\). In the other direction, if the system is complete, its finite linear span is dense in \(H\). Given \(d\in D\) and \(\varepsilon>0\), density supplies a finite linear combination \(s\) with \(\|d-s\|_2<\varepsilon\). \(\square\)

Worked Tests in Concrete Spaces

Worked Example: An Incomplete System in a Finite-Dimensional Space

Take \(X=\{1,2,3\}\), \(\mathcal{M}\) equal to the set of all subsets of \(X\), and \(\mu\) counting measure. Then \(L^2(\mu)=\mathbb{C}^3\) with inner product \(\langle x,y\rangle=\sum_{k=1}^3x_k\overline{y_k}\). Consider \(u_1=(1,0,0)\) and \(u_2=(0,1,0)\). Each has squared norm \(1\), and \(\langle u_1,u_2\rangle=1\cdot0+0\cdot1+0\cdot0=0\), so they form an orthonormal system.

Their span is \(\{(a,b,0):a,b\in\mathbb{C}\}\). A vector \(h=(x,y,z)\) belongs to its orthogonal complement exactly when \(\langle h,u_1\rangle=x=0\) and \(\langle h,u_2\rangle=y=0\). Thus \(M^\perp=\{(0,0,z):z\in\mathbb{C}\}\), which contains the nonzero vector \(u_3=(0,0,1)\). The criterion shows the system is not complete.

Check: \(\langle u_3,u_1\rangle=0\cdot1+0\cdot0+1\cdot0=0\), and \(\langle u_3,u_2\rangle=0\cdot0+0\cdot1+1\cdot0=0\), while \(\|u_3\|_2^2=0^2+0^2+1^2=1\). So the witness is indeed nonzero and orthogonal to both system members.

Worked Example: A Nonzero Function Orthogonal to Constants

Let \(X=[0,1]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra, and let \(\mu=m\), Lebesgue measure. The function \(u(x)=1\) has squared \(L^2\) norm \(\int_0^1 1^2\,dx=1\), so the one-element system \(\{u\}\) is orthonormal. Define \(h(x)=\sqrt{12}(x-\tfrac12)\). It is real-valued and belongs to \(L^2(m)\), since it is bounded on an interval of finite measure.

Its inner product with \(u\) is

$$ \langle h,u\rangle =\sqrt{12}\int_0^1\left(x-\frac12\right)\,dx =\sqrt{12}\left(\frac12-\frac12\right)=0. $$

Its squared norm is

$$ \|h\|_2^2 =12\int_0^1\left(x-\frac12\right)^2\,dx =12\left[\frac{(x-\frac12)^3}{3}\right]_0^1 =12\left(\frac1{24}-\left(-\frac1{24}\right)\right)=1. $$

Thus \(h\) is nonzero in \(L^2(m)\) and lies in the orthogonal complement of the constants. The system \(\{1\}\) is not complete. Check: the integral of \(h\) is zero, as required for orthogonality, while its norm is \(1\), so it cannot be the zero \(L^2\) element.

Worked Example: The Standard Orthonormal System in \(\ell^2\)

Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure. Then \(L^2(\mu)=\ell^2\), the space of complex sequences \(a=(a_n)\) with \(\sum_{n=1}^{\infty}|a_n|^2<\infty\). For each \(k\), let \(e_k\) have value \(1\) at index \(k\) and \(0\) at every other index. Directly, \(\langle e_j,e_k\rangle=0\) for \(j\ne k\), and \(\|e_k\|_2^2=1\).

For an arbitrary \(a\in\ell^2\), its truncation \(a^{(N)}=\sum_{k=1}^{N}a_ke_k\) is a finite linear combination of the system. The error is

$$ \|a-a^{(N)}\|_2^2 =\sum_{k=N+1}^{\infty}|a_k|^2 \longrightarrow 0 \quad\text{as }N\to\infty, $$

because the series \(\sum_{k=1}^{\infty}|a_k|^2\) converges. Therefore the finite linear span is dense and the system is complete. The orthogonality test confirms this: if \(a\) is orthogonal to every \(e_k\), then \(\langle a,e_k\rangle=a_k=0\) for every \(k\), so \(a=0\) in \(\ell^2\).

Check: for \(a=(1,\tfrac12,0,\ldots)\) and \(N=1\), \(a^{(1)}=(1,0,\ldots)\) and \(\|a-a^{(1)}\|_2^2=|\tfrac12|^2=\tfrac14\). For \(N=2\), \(a^{(2)}=a\) and the squared error is \(0\). These values agree with the tail-sum formula.

Common Pitfall: Confusing Span with Closed Span

In the preceding example, the finite linear span of the \(e_k\) consists exactly of the sequences with only finitely many nonzero entries. It is not all of \(\ell^2\): for example, \(a=(2^{-1},2^{-2},2^{-3},\ldots)\) has infinitely many nonzero entries. A mistaken argument would therefore say, “The span is proper, so the system is incomplete.” That conclusion uses the wrong definition. Completeness requires the closure of the span to be the whole space.

For this particular \(a\), the first \(N\) coordinates give a finite linear combination, and the squared approximation error is

$$ \sum_{k=N+1}^{\infty}4^{-k} =\frac{4^{-(N+1)}}{1-\frac14} =\frac{4^{-N}}{3} \longrightarrow 0. $$

For example, at \(N=1\) the squared error is \(4^{-1}/3=1/12\), and at \(N=2\) it is \(4^{-2}/3=1/48\). The terms are nonzero, so \(a\) is not in the finite span; nevertheless, its truncations converge to it in \(\ell^2\). The system is complete precisely because these approximations work for every \(\ell^2\) sequence.

The two tests are useful in different settings. Use the orthogonal-complement criterion when equations \(\langle h,u_i\rangle=0\) can be analyzed directly. Use a dense-set approximation argument when a familiar class of vectors can be approximated by finite linear combinations. In either case, the target is the same: the closed span must be the whole \(L^2\) space.

Check Your Understanding

For each question, distinguish the finite linear span from its closure and interpret zero in \(L^2\) as equality almost everywhere where appropriate.

  1. State the orthogonal-complement criterion for completeness and identify the role of the closed span.
  2. In \(\mathbb{C}^3\) with counting measure, find a nonzero vector orthogonal to \(u_1=(1,0,0)\) and \(u_2=(0,1,0)\).
  3. On \([0,1]\) with Lebesgue measure, verify that \(x-\tfrac12\) is orthogonal to the constant function \(1\), and explain why this alone shows that \(\{1\}\) is not complete.
  4. Why does the fact that finite sequences form a proper subspace of \(\ell^2\) not show that the standard orthonormal system is incomplete?
  5. Suppose \(D\) is dense in \(L^2(\mu)\). What approximation property for every \(d\in D\) is sufficient to establish completeness of an orthonormal system?

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