Inner Products And Norms In L2

From Square-Integrable Functions to an Inner Product

On a Euclidean space, the dot product turns vectors into a geometry: it gives lengths, angles, and orthogonality. The same idea works for square-integrable functions, but there are two points to address. First, the product of two \(L^2\) functions must be integrable before its integral can define an inner product. Second, elements of \(L^2\) are equivalence classes: functions that agree almost everywhere represent the same element, so the inner product must not depend on the chosen representatives.

Let \((X,\mathcal{M},\mu)\) be any measure space. No assumption that \(\mu\) is finite, sigma-finite, or complete is needed here. For \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), take measurable, finite-valued functions \(f:X\to\mathbb{F}\) such that \(\int |f|^2\,d\mu<\infty\). Identify two such functions when they agree \(\mu\)-almost everywhere. The resulting space of equivalence classes is \(L^2(\mu;\mathbb{F})\); we write \(f\) for either a representative or its class when the meaning is clear. The usual \(L^2\) norm is \(\|f\|_2=(\int |f|^2\,d\mu)^{1/2}\).

The complex conjugate is essential: with this convention, \(\langle f,f\rangle=\int |f|^2\,d\mu\), a nonnegative real number. We first establish that the integral defining the inner product exists. For real numbers \(a,b\), and also for complex numbers by applying the inequality to their absolute values, \(2|ab|\le |a|^2+|b|^2\). Thus, for \(f,g\in L^2(\mu;\mathbb{F})\),

For real-valued functions, the same estimate applies to \(|fg|\). Therefore each proposed inner product is an integral of an integrable function. This argument also shows that no finite-measure assumption on \(X\) is required.

Independence of Representatives and Inner Product Properties

Proposition. For either choice \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), the formula above defines an inner product on \(L^2(\mu;\mathbb{F})\). In particular, it is independent of representatives, linear in its first argument, conjugate-symmetric, and positive definite.

Proof. Suppose \(f=f'\) \(\mu\)-a.e. and \(g=g'\) \(\mu\)-a.e. There are measurable null sets outside which the respective equalities hold. Outside their union, \(f\overline g=f'\overline{g'}\). The two products are integrable by the estimate above, and integrable functions equal almost everywhere have equal integrals. Thus \(\int f\overline g\,d\mu=\int f'\overline{g'}\,d\mu\). In the real case this is the same argument with \(fg\) in place of \(f\overline g\); the inner product is well defined on equivalence classes.

Linearity of the Lebesgue integral gives \(\langle af_1+bf_2,g\rangle=a\langle f_1,g\rangle+b\langle f_2,g\rangle\) for scalars \(a,b\in\mathbb{F}\). In the complex case, conjugating the integral gives \(\langle f,g\rangle=\overline{\langle g,f\rangle}\); in the real case this says \(\langle f,g\rangle=\langle g,f\rangle\). Finally, \(\langle f,f\rangle=\int |f|^2\,d\mu\ge0\). This integral is zero exactly when \(f=0\) \(\mu\)-a.e.: indeed, if it is zero, then for each positive integer \(n\), the set \(E_n=\{|f|>1/n\}\) satisfies \(\mu(E_n)/n^2\le\int |f|^2\,d\mu=0\). Thus each \(E_n\) is null, and \(\{|f|>0\}=\bigcup_{n=1}^{\infty}E_n\) is null. Conversely, if \(f=0\) a.e., then \(\int |f|^2\,d\mu=0\). This proves positive definiteness on equivalence classes. \(\square\)

Cauchy–Schwarz and the Induced Norm

Proof. If \(\|g\|_2=0\), positive definiteness shows that \(g=0\) a.e., so \(\langle f,g\rangle=0\) and the inequality holds. Now suppose \(\|g\|_2>0\), and set \(c=\langle f,g\rangle/\|g\|_2^2\). Nonnegativity of the squared norm gives \(0\le\|f-cg\|_2^2\). Expanding by linearity in the first argument and conjugate-linearity in the second gives

For the last equality, \(\langle g,f\rangle=\overline{\langle f,g\rangle}\) and \(c=\langle f,g\rangle/\|g\|_2^2\), so each of the two subtracted terms is \(|\langle f,g\rangle|^2/\|g\|_2^2\), while the final term is one copy of that quantity. Multiplying by the positive number \(\|g\|_2^2\) gives \(|\langle f,g\rangle|^2\le\|f\|_2^2\|g\|_2^2\). Taking nonnegative square roots proves the result, including the case \(g=0\). \(\square\)

Proof. The identity follows from \(\langle f,f\rangle=\int|f|^2\,d\mu\). Nonnegativity and definiteness were proved above. For a scalar \(a\), \(\|af\|_2^2=\int|a|^2|f|^2\,d\mu=|a|^2\|f\|_2^2\), so \(\|af\|_2=|a|\|f\|_2\). For the triangle inequality, expand the square and use conjugate symmetry and Cauchy–Schwarz:

Both sides are nonnegative, so taking square roots yields \(\|f+g\|_2\le\|f\|_2+\|g\|_2\). Hence the induced quantity is a norm and agrees with the usual \(L^2\) norm. \(\square\)

A Complex-Valued Pitfall

For complex functions, omitting the conjugate is not a harmless change of notation. On the one-point space \(X=\{a\}\) with counting measure, take \(f(a)=1\) and \(g(a)=i\). The prescribed inner product is \(\langle f,g\rangle=1\cdot\overline{i}=-i\), whereas \(\langle g,f\rangle=i\cdot\overline{1}=i\), so \(\langle f,g\rangle=\overline{\langle g,f\rangle}\). A wrong calculation using \(\int fg\,d\mu\) gives \(i\) for both orders, which violates conjugate symmetry because \(i\ne\overline{i}=-i\). The correction is to conjugate the second factor: \(\langle f,g\rangle=\int f\overline g\,d\mu\). Then \(\|f\|_2^2=\|g\|_2^2=1\), and the inner product has the required properties.

The inner product therefore gives \(L^2\) its familiar geometric structure while respecting the fact that its elements are equivalence classes, not individual pointwise representatives. Cauchy–Schwarz is the key estimate: it controls pairings by the \(L^2\) norms and supplies the triangle inequality for the induced norm. The completeness of this normed space is a separate property; identifying that additional property is the next step in understanding why \(L^2\) is a Hilbert space.