From Bessel’s Inequality to an Expansion
For an orthonormal system, Bessel’s inequality says that the sum of squared coefficients of a vector cannot exceed its squared norm. Completeness is the additional condition that makes this bound an equality: the coefficient expansion accounts for the whole vector, with convergence in norm. Neither statement by itself asserts pointwise convergence of representatives of \(L^2\) functions.
Let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness or sigma-finiteness assumption is needed. We work in the complex Hilbert space \(L^2(\mu)=L^2(X,\mathcal{M},\mu)\). Its elements are equivalence classes, with functions identified when they agree \(\mu\)-a.e. We use the inner product \(\langle f,g\rangle=\int_X f\overline{g}\,d\mu\), linear in the first argument, and the norm \(\|f\|_2=\langle f,f\rangle^{1/2}\). The same arguments apply to real \(L^2\), omitting complex conjugates.
We first record the finite calculation behind Bessel’s inequality. For a finite subset \(F\subset I\), put \(s_F=\sum_{i\in F}c_i e_i\). Then \(f-s_F\) is orthogonal to every \(e_i\) with \(i\in F\), because \(\langle f-s_F,e_i\rangle=c_i-c_i=0\). Thus the two terms in \(f=s_F+(f-s_F)\) are orthogonal. Expanding the squared norm gives \(\|f\|_2^2=\|f-s_F\|_2^2+\sum_{i\in F}|c_i|^2\), and hence \(\sum_{i\in F}|c_i|^2\le\|f\|_2^2\). This is the finite form of Bessel’s inequality.
Coefficient Expansions and Parseval’s Identity
Proof. The finite calculation above gives \(\sum_{i\in F}|c_i|^2\le\|f\|_2^2\) for every finite \(F\subset I\), which proves the stated bound for the supremum of finite subsums. For each positive integer \(k\), the set \(J_k=\{i\in I:|c_i|\ge 1/k\}\) is finite: if it had more than \(k^2\|f\|_2^2\) elements, a finite subset with that many elements would have squared coefficient sum greater than \(\|f\|_2^2\), contradicting the bound. Every nonzero coefficient belongs to some \(J_k\), so \(J=\{i\in I:c_i\ne0\}=\bigcup_{k\ge1}J_k\) is at most countable.
If \(J\) is infinite, enumerate it as \(i_1,i_2,\ldots\), and define \(s_N=\sum_{r=1}^N c_{i_r}e_{i_r}\). If \(J\) is finite, use its finite sum instead. For \(M>N\), orthonormality gives \(\|s_M-s_N\|_2^2=\sum_{r=N+1}^M|c_{i_r}|^2\). The series of squared coefficients has finite sum by Bessel’s bound, so its tails tend to zero; therefore \((s_N)\) is Cauchy. Completeness of \(L^2(\mu)\) gives a limit \(p\in L^2(\mu)\). Every \(s_N\) is in the span, so \(p\) is in its closure.
For each \(i\in J\), continuity of the inner product gives \(\langle f-p,e_i\rangle=\lim_N\langle f-s_N,e_i\rangle=0\), since the coefficient of \(e_i\) in \(s_N\) is eventually \(c_i\). For \(i\notin J\), \(\langle f,e_i\rangle=0\), and every \(s_N\) is orthogonal to \(e_i\), so \(\langle f-p,e_i\rangle=0\) there as well. By linearity, \(f-p\) is orthogonal to the span, and by continuity it is orthogonal to its closure. Since \(p\) belongs to that closure, \(p\) and \(f-p\) are orthogonal. Applying the Pythagorean identity gives \(\|f\|_2^2=\|p\|_2^2+\|f-p\|_2^2\). Taking limits in \(\|s_N\|_2^2=\sum_{r=1}^N|c_{i_r}|^2\) gives \(\|p\|_2^2=\sum_{i\in I}|c_i|^2\), proving the identity. If the family is complete, its closed span is the whole space. The residual \(f-p\) lies in that space and is orthogonal to it, so \(\|f-p\|_2^2=\langle f-p,f-p\rangle=0\). Thus \(p=f\) in \(L^2(\mu)\), proving the expansion and Parseval’s identity. \(\square\)
Worked Example: A Complete Orthonormal Basis in Two Dimensions
Take \(X=\{1,2\}\) with counting measure and \(\mathcal{M}\) its power set. Then \(L^2(\mu)\) is \(\mathbb{C}^2\) with squared norm \(\|f\|_2^2=|f(1)|^2+|f(2)|^2\). Let \(e_1=(1,0)\), \(e_2=(0,1)\), and \(f=(3,4)\). Each \(e_j\) has norm one and their inner product is zero. Their span is all of \(L^2(\mu)\), so the system is complete.
The coefficients are \(\langle f,e_1\rangle=3\) and \(\langle f,e_2\rangle=4\). Hence the expansion is \(3e_1+4e_2=(3,4)=f\). Its coefficient sum is \(3^2+4^2=9+16=25\), while \(\|f\|_2^2=|3|^2+|4|^2=9+16=25\). Check: substituting the expansion gives \(\|3e_1+4e_2\|_2^2=|3|^2+|4|^2=25\), agreeing with Parseval’s identity.
Worked Example: Exact Expansion of a Function in a Two-Dimensional Span
On \([0,1]\) with Lebesgue measure, consider the two functions \(e_1(x)=1\) and \(e_2(x)=\sqrt{3}(2x-1)\). We use them as an orthonormal family in \(L^2([0,1],m)\); they are not a basis for the whole space. Indeed, \(\int_0^1 e_1(x)^2\,dx=1\), \(\int_0^1(2x-1)\,dx=0\), and \(\int_0^1 e_2(x)^2\,dx=3\int_0^1(2x-1)^2\,dx=3(1/3)=1\).
For \(f(x)=x\), the coefficients are $$ c_1=\int_0^1x\,dx=\frac12,\qquad c_2=\sqrt{3}\int_0^1(2x^2-x)\,dx =\sqrt{3}\left(\frac23-\frac12\right)=\frac{\sqrt{3}}6. $$ Thus \(c_1e_1(x)+c_2e_2(x)=\frac12+\frac12(2x-1)=x\) for every \(x\in[0,1]\). Also, \(\|f\|_2^2=\int_0^1x^2\,dx=1/3\), while \(|c_1|^2+|c_2|^2=1/4+3/36=1/4+1/12=1/3\). Check: the displayed expansion makes the residual zero, so the residual term in the theorem’s norm identity is zero, as the computed equality requires.
What Changes When the System Is Incomplete?
For an incomplete orthonormal family, the coefficient series still converges in norm, but its limit is the component of \(f\) in the closed span of the family. The theorem proves this directly: the residual is orthogonal to that span. In particular, Parseval equality for a given \(f\) holds exactly when the residual is zero, or equivalently when \(f\) belongs to the closed span. Bessel’s inequality is strict when the residual is nonzero.
Worked Example: Strict Bessel Inequality in \(L^2([0,1])\)
Use Lebesgue measure on \([0,1]\), the single-element orthonormal family \(e_1(x)=1\), and \(f(x)=x\). This family is incomplete in \(L^2([0,1],m)\). Its one coefficient is \(c_1=\int_0^1x\,dx=1/2\), so the coefficient sum is \(|c_1|^2=1/4\). But \(\|f\|_2^2=\int_0^1x^2\,dx=1/3\), and \(1/4<1/3\).
The coefficient expansion is \(p(x)=1/2\). Its residual is \(f-p=x-1/2\), and $$ \|f-p\|_2^2=\int_0^1(x-\tfrac12)^2\,dx =\left[\frac{x^3}{3}-\frac{x^2}{2}+\frac{x}{4}\right]_0^1 =\frac13-\frac12+\frac14=\frac1{12}. $$ The identity in the theorem checks exactly: \(1/4+1/12=1/3=\|f\|_2^2\). The strict inequality comes from the nonzero residual.
Common Pitfall: Norm Convergence Is Not Pointwise Convergence
A coefficient expansion in this theorem converges in the \(L^2\) norm, which means that the norm of the difference between \(f\) and its partial sums tends to zero. That conclusion alone is not a claim that the partial sums converge at every point, or even \(\mu\)-a.e. A second common error is to apply Parseval’s equality to an incomplete system: in the preceding example, asserting \(1/4=1/3\) omits the squared norm \(1/12\) of the orthogonal residual. The correct general identity includes that residual; equality with the coefficient sum follows when the system is complete, or for a particular \(f\) when \(f\) lies in its closed span.
Check Your Understanding
For each question, distinguish the coefficient sum from the squared norm of any residual.
- In \(\mathbb{C}^2\) with counting measure, compute the coefficients and verify Parseval’s identity for \(f=(1,2)\) and the standard coordinate basis.
- On \([0,1]\) with Lebesgue measure, use the family \(e(x)=1\) to find the coefficient expansion of \(f(x)=x\) and the squared norm of its residual.
- State precisely what completeness adds to Bessel’s inequality for an orthonormal family in \(L^2(\mu)\).
- For an incomplete orthonormal family, what condition on a particular \(f\) makes the coefficient sum equal \(\|f\|_2^2\)?
- Does convergence of the coefficient expansion in \(L^2\) assert pointwise convergence of its partial sums? Explain the distinction.