Why L2 Is A Hilbert Space

What It Means for L2 to Be a Hilbert Space

An inner product gives more structure than a norm: it records angles and makes orthogonality meaningful. For \(L^2\), the natural candidate is obtained by integrating the product of one function with the complex conjugate of another. To call \(L^2\) a Hilbert space, however, we must verify two separate facts: this inner product induces the usual \(L^2\) norm, and the resulting normed space is complete.

Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it is not assumed finite, sigma-finite, or complete. Functions are complex-valued and measurable. Two such functions are identified when they agree \(\mu\)-a.e. The space \(L^2(\mu)\) consists of the resulting equivalence classes \([f]\) for which \(\int_X |f|^2\,d\mu<\infty\). We write \(f\) for its equivalence class when no confusion can arise. Define

The integral defining the inner product is finite: the Cauchy–Schwarz inequality for integrals gives \(\int_X|f\overline g|\,d\mu\le \|f\|_2\|g\|_2\). This inequality applies to \(f,g\in L^2(\mu)\), without requiring \(\mu(X)<\infty\). If representatives are changed on null sets, the integrands change only on a null set, so the inner product and norm do not depend on the chosen representatives. We use the convention that the inner product is linear in its first argument and conjugate-linear in its second.

The Inner Product Induces the L2 Norm

We first verify the norm properties. The integral Cauchy–Schwarz inequality used below follows directly from nonnegativity: if \(\|g\|_2>0\), expand \(\|f-cg\|_2^2\ge0\) with \(c=\langle f,g\rangle/\|g\|_2^2\). The expansion gives \(\|f\|_2^2-|\langle f,g\rangle|^2/\|g\|_2^2\ge0\), hence \(|\langle f,g\rangle|\le\|f\|_2\|g\|_2\). If \(\|g\|_2=0\), then \(g=0\) \(\mu\)-a.e., and the same inequality holds because \(\langle f,g\rangle=0\).

Proof. Linearity in the first argument and conjugate symmetry follow from the corresponding properties of the integral: \(\langle af+bh,g\rangle=a\langle f,g\rangle+b\langle h,g\rangle\) and \(\langle f,g\rangle=\overline{\langle g,f\rangle}\), for \(a,b\in\mathbb{C}\). Also, \(\langle f,f\rangle=\int_X|f|^2\,d\mu\ge0\). This integral is zero exactly when \(f=0\) \(\mu\)-a.e.: for the nontrivial direction, each set \(E_n=\{|f|>1/n\}\) satisfies \(\mu(E_n)/n^2\le\int_X|f|^2\,d\mu=0\), so \(\mu(E_n)=0\); since \(\{|f|>0\}=\bigcup_{n\ge1}E_n\), that set is null. Thus the inner product is positive definite on equivalence classes. Finally, its induced norm is \((\langle f,f\rangle)^{1/2}=(\int_X|f|^2\,d\mu)^{1/2}\), as claimed. \(\square\)

For completeness, the norm axioms follow from this identity. Nonnegativity and definiteness have just been shown, and \(\|af\|_2=|a|\|f\|_2\). For the triangle inequality, expand the square and apply Cauchy–Schwarz:

Taking nonnegative square roots proves \(\|f+g\|_2\le\|f\|_2+\|g\|_2\). Thus the inner product induces a norm, not merely a nonnegative quadratic expression.

The Parallelogram Identity

Proof. Expand both squared norms using the inner product. Conjugate symmetry gives \(\langle g,f\rangle=\overline{\langle f,g\rangle}\), and so \(\langle f,g\rangle+\langle g,f\rangle=2\operatorname{Re}\langle f,g\rangle\). Hence

Adding the equalities cancels the two real-part terms and leaves \(2\|f\|_2^2+2\|g\|_2^2\). The expansions are valid because \(f+g\) and \(f-g\) belong to \(L^2(\mu)\), by the triangle inequality already proved. \(\square\)

Completeness: Every L2-Cauchy Sequence Has a Limit

The key point in the completeness proof is to extract from a Cauchy sequence a subsequence whose successive differences have summable \(L^2\) norms. The pointwise sum of their absolute values is then finite almost everywhere. We use the Monotone Convergence Theorem and Fatou's lemma for nonnegative measurable functions on \((X,\mathcal{M},\mu)\); neither result requires finite or sigma-finite measure. We also use the \(L^2\) triangle inequality proved above.

Proof. Let \((f_n)\) be a Cauchy sequence in \(L^2(\mu)\). Choose a subsequence \((f_{n_j})\) such that \(\|f_{n_{j+1}}-f_{n_j}\|_2\le 2^{-j}\) for every \(j\ge1\). This choice is possible by the Cauchy property: for each \(j\), all sufficiently late pairs have distance at most \(2^{-j}\), and the indices can be selected successively beyond those thresholds. Choose measurable representatives and put \(u_j=f_{n_{j+1}}-f_{n_j}\) and \(S_N=\sum_{j=1}^N|u_j|\). The triangle inequality gives \(\|S_N\|_2\le\sum_{j=1}^N\|u_j\|_2\le\sum_{j=1}^N2^{-j}\le1\).

The nonnegative functions \(S_N^2\) increase pointwise to \(S^2\), where \(S=\sum_{j=1}^{\infty}|u_j|\), with \(+\infty\) allowed. By the Monotone Convergence Theorem, \(\int_X S^2\,d\mu=\lim_N\int_X S_N^2\,d\mu\le1\). Therefore \(S<\infty\) \(\mu\)-a.e.; otherwise its square would have infinite integral on the measurable set where it is infinite, unless that set were null. On the measurable set \(S<\infty\), the series of successive differences \(\sum_j u_j(x)\) converges absolutely. Thus \(f_{n_j}(x)\) converges there to a finite limit. Define \(f\) to be that limit on \(\{S<\infty\}\) and zero on \(\{S=\infty\}\). It is measurable, since it is defined by a pointwise limit on a measurable set and by a constant on its measurable complement.

For a fixed \(k\), as \(r\to\infty\), \(f_{n_r}-f_{n_k}\to f-f_{n_k}\) \(\mu\)-a.e. For every \(r>k\), the triangle inequality gives \(\|f_{n_r}-f_{n_k}\|_2\le\sum_{j=k}^{r-1}2^{-j}\le 2^{1-k}\). Apply Fatou's lemma to the nonnegative functions \(|f_{n_r}-f_{n_k}|^2\); their a.e. limit is \(|f-f_{n_k}|^2\). It follows that \(\int_X|f-f_{n_k}|^2\,d\mu\le\liminf_{r\to\infty}\|f_{n_r}-f_{n_k}\|_2^2\le 2^{2-2k}\). In particular, \(f-f_{n_1}\in L^2(\mu)\), and \(f=f_{n_1}+(f-f_{n_1})\in L^2(\mu)\) by the triangle inequality. The same bound shows \(\|f-f_{n_k}\|_2\to0\).

Finally, let \(\varepsilon>0\). Since \((f_n)\) is Cauchy, there is \(N\) such that \(\|f_n-f_m\|_2<\varepsilon/2\) whenever \(n,m\ge N\). Choose \(k\) large enough that \(n_k\ge N\) and \(\|f_{n_k}-f\|_2<\varepsilon/2\). For every \(n\ge N\), the triangle inequality now gives \(\|f_n-f\|_2\le\|f_n-f_{n_k}\|_2+\|f_{n_k}-f\|_2<\varepsilon\). Hence \(f_n\to f\) in \(L^2(\mu)\), proving completeness. Combined with the inner product and norm properties, this proves that \(L^2(\mu)\) is a Hilbert space. \(\square\)

A Common Mistake: Dropping the Cross Term

It is incorrect to calculate \(\|f+g\|_2^2\) as \(\|f\|_2^2+\|g\|_2^2\) in general. For a concrete check on \([0,1]\) with Lebesgue measure, take \(f=g=1\). Then \(\|f\|_2^2=\|g\|_2^2=1\), but \(\|f+g\|_2^2=\int_0^1 2^2\,dx=4\), not \(1+1=2\). The correction is the cross term: \(\|f+g\|_2^2=\|f\|_2^2+\|g\|_2^2+2\operatorname{Re}\langle f,g\rangle=1+1+2=4\). The same care with almost-everywhere equivalence is essential: functions that differ only on a null set represent the same element of \(L^2(\mu)\), and positive definiteness holds for these equivalence classes, not for pointwise representatives.