From Square Integrability to \(L^2\)
The \(L^1\) space measures a function by the integral of its absolute value. The corresponding \(L^2\) space measures its squared size: a measurable function belongs to \(L^2\) when the integral of its square is finite. Taking the square root of that integral gives the \(L^2\) norm. As in \(L^1\), functions equal almost everywhere represent the same element, because changing a function on a null set does not change its integral.
The choice of a square is more than a different way to measure size. It lets us define an inner product by integrating a product of two functions. Hölder’s inequality with exponent \(2\), established earlier in the course, ensures that this product is integrable for any two \(L^2\) functions. The inner product gives \(L^2\) a useful geometric structure, while a summable-differences argument will show that the space is complete.
The definition does not depend on the chosen representative. If \(f=g\) almost everywhere, then \(|f|^2=|g|^2\) almost everywhere, so their integrals agree. The zero-norm criterion for nonnegative integrals also shows that \(\|f\|_2=0\) exactly when \(f=0\) almost everywhere. Thus identifying functions equal almost everywhere is necessary for the norm to be definite on the resulting space.
Worked Example: The \(L^2\) Norm of an Indicator
Let \(A=[1,4)\) in \(\mathbb{R}\) with Lebesgue measure, and let \(f=\mathbf{1}_A\). Since \(|\mathbf{1}_A|^2=\mathbf{1}_A\), the Integral of an Indicator theorem gives $$ \|f\|_2^2 =\int_{\mathbb{R}}\mathbf{1}_A\,d\lambda =\lambda(A) =4-1=3. $$ Therefore \(\|f\|_2=\sqrt{3}\), and \(f\in L^2(\mathbb{R},\lambda)\). More generally, an indicator belongs to \(L^2\) exactly when its set has finite measure, and its \(L^2\) norm is the square root of that measure.
Worked Example: A Singularity That Is Square-Integrable
On \((0,1)\) with Lebesgue measure, consider \(f(x)=x^{-1/4}\). Its square is \(x^{-1/2}\). For \(0<a<1\), $$ \int_a^1 |f(x)|^2\,dx =\int_a^1 x^{-1/2}\,dx =2(1-\sqrt{a}). $$ As \(a\) decreases to zero, these integrals increase to \(2\). Hence \(\|f\|_2^2=2\), so \(f\in L^2(0,1)\), even though \(f\) is unbounded near zero. In contrast, \(g(x)=x^{-1/2}\) has \(|g(x)|^2=x^{-1}\), and $$ \int_a^1 x^{-1}\,dx=-\log a\longrightarrow\infty \quad\text{as }a\downarrow0. $$ Thus \(g\notin L^2(0,1)\). The relevant test is the integral of the square, not boundedness alone.
The Inner Product and the Norm
For real-valued functions, the natural inner product is the integral of their product. The product need not be nonnegative, but it is integrable: Hölder’s inequality with conjugate exponents \(2\) and \(2\) gives $$ \int_X |fg|\,d\mu\leq \|f\|_2\|g\|_2<\infty. $$ This also ensures that the inner product is unchanged if either representative is changed on a null set.
Bilinearity follows from linearity of the Lebesgue integral, and symmetry follows from \(fg=gf\). Also, \(\langle f,f\rangle=\int_X|f|^2\,d\mu\). The zero-integral criterion applied to \(|f|^2\) shows that this quantity is zero exactly when \(f=0\) almost everywhere. The \(L^2\) norm is therefore the norm induced by this inner product.
Worked Example: Orthogonal Functions on an Interval
On \((0,2\pi)\) with Lebesgue measure, let \(f(x)=\sin x\) and \(g(x)=\cos x\). Both functions are bounded on an interval of finite measure, so their squares have finite integrals. Direct calculation gives $$ \langle f,g\rangle =\int_0^{2\pi}\sin x\cos x\,dx =\left[\frac{\sin^2 x}{2}\right]_0^{2\pi}=0. $$ Thus \(f\) and \(g\) are orthogonal in the inner-product sense. Their norms satisfy $$ \|f\|_2^2=\int_0^{2\pi}\sin^2x\,dx=\pi, \qquad \|g\|_2^2=\int_0^{2\pi}\cos^2x\,dx=\pi. $$ For example, the first integral follows from \(\sin^2x=(1-\cos(2x))/2\), whose integral from \(0\) to \(2\pi\) is \(\pi\); the second follows from \(\cos^2x=(1+\cos(2x))/2\), also giving \(\pi\). Hence both norms are \(\sqrt{\pi}\).
Proof. Since \(f+g\) and \(f-g\) belong to \(L^2\) by the Linear Closure of \(L^p\) theorem, all four norms are finite. Expanding the squares inside the integrals gives $$ \|f+g\|_2^2 =\int_X(f^2+2fg+g^2)\,d\mu $$ and $$ \|f-g\|_2^2 =\int_X(f^2-2fg+g^2)\,d\mu. $$ Adding these equalities cancels the product terms and leaves $$ 2\int_X f^2\,d\mu+2\int_X g^2\,d\mu =2\|f\|_2^2+2\|g\|_2^2, $$ which proves the identity.
The identity expresses a geometric feature of the norm: the sum of the squared lengths of \(f+g\) and \(f-g\) depends only on the separate squared lengths of \(f\) and \(g\). It is a useful test for inner-product geometry. In this course, it also makes explicit that the \(L^2\) norm is not merely one instance of an \(L^p\) formula; it is directly tied to a bilinear inner product.
Summable Differences in \(L^2\)
To prove completeness, we need a way to sum functions whose \(L^2\) norms have finite total. The corresponding \(L^1\) result controlled the pointwise sum of absolute values by its integral. In \(L^2\), the useful control is instead the \(L^2\) norm of finite sums of absolute values. Minkowski’s inequality bounds those norms by sums of the individual norms; the Monotone Convergence Theorem then passes the bound to the infinite sum.
Proof. Choose measurable representatives of the \(u_k\), and set $$ H_N(x)=\sum_{k=1}^N|u_k(x)|, \qquad H(x)=\sum_{k=1}^{\infty}|u_k(x)|. $$ Each \(H_N\) is measurable, and \(H_N\) increases pointwise to \(H\). By Minkowski’s inequality, $$ \|H_N\|_2\leq\sum_{k=1}^N\|u_k\|_2 \leq C, \qquad C=\sum_{k=1}^{\infty}\|u_k\|_2. $$ Applying the Monotone Convergence Theorem to \(H_N^2\) gives $$ \int_X H^2\,d\mu =\lim_{N\to\infty}\int_X H_N^2\,d\mu \leq C^2<\infty. $$ Consequently \(H\) is finite almost everywhere. On the measurable set \(E=\{x:H(x)<\infty\}\), the numerical series \(\sum_k u_k(x)\) converges absolutely. Define \(u\) to be that sum on \(E\), and set \(u=0\) on \(X\setminus E\). Measurability follows from pointwise convergence of the measurable partial sums on \(E\) and the measurable definition on its complement. Since \(|u|\leq H\) on \(E\) and \(u=0\) off \(E\), we have \(\int_X|u|^2\,d\mu\leq\int_XH^2\,d\mu<\infty\), so \(u\in L^2\).
Let \(s_N=\sum_{k=1}^N u_k\). For each \(N\), the tail function \(T_N=\sum_{k>N}|u_k|\) satisfies $$ |u-s_N|\leq T_N\quad\text{almost everywhere}. $$ Applying the same finite-sum Minkowski bound and Monotone Convergence argument to this tail gives $$ \|T_N\|_2\leq\sum_{k>N}\|u_k\|_2. $$ Therefore, by monotonicity of the \(L^2\) norm, $$ \|u-s_N\|_2\leq\sum_{k>N}\|u_k\|_2\longrightarrow0. $$ The last limit holds because the series of norms converges. Thus the partial sums converge to \(u\) in \(L^2\), as claimed.
Completeness of \(L^2\)
A normed space is complete when every Cauchy sequence in its norm converges to an element of the space. The series theorem supplies the main construction: from a Cauchy sequence, choose a subsequence whose successive differences have summable \(L^2\) norms. The sum of those differences produces a limit for the subsequence, and the Cauchy property then gives convergence of the full sequence.
Proof. Let \((F_n)\) be a Cauchy sequence in \(L^2\), where each \(F_n\) is an equivalence class. For each positive integer \(j\), choose \(N_j\) so that $$ \|F_r-F_s\|_2<2^{-j} \quad\text{whenever }r,s\geq N_j. $$ Choose strictly increasing indices \(n_j\) with \(n_j\geq N_j\). Then $$ \|F_{n_{j+1}}-F_{n_j}\|_2<2^{-j}. $$ Choose measurable representatives \(f_{n_j}\) and define \(u_j=f_{n_{j+1}}-f_{n_j}\). The series of their norms is bounded by $$ \sum_{j=1}^{\infty}\|u_j\|_2 \leq\sum_{j=1}^{\infty}2^{-j}=1. $$ By the Absolute Convergence of a Series in \(L^2\) theorem, \(\sum_j u_j\) converges in \(L^2\) to a function \(u\in L^2\).
Telescoping gives, for each \(m\geq1\), $$ f_{n_m}=f_{n_1}+\sum_{j=1}^{m-1}u_j. $$ The partial sums on the right converge in \(L^2\) to \(u\). Hence \(F_{n_m}\) converges in \(L^2\) to the class \(F\) represented by \(f_{n_1}+u\). This representative belongs to \(L^2\) by linear closure.
It remains to show that the entire sequence converges to \(F\). Given \(\varepsilon>0\), choose \(N\) such that $$ \|F_r-F_s\|_2<\frac{\varepsilon}{2} \quad\text{whenever }r,s\geq N. $$ Because \(F_{n_m}\to F\), choose \(m\) so large that \(n_m\geq N\) and \(\|F_{n_m}-F\|_2<\varepsilon/2\). For every \(n\geq N\), the triangle inequality gives $$ \|F_n-F\|_2 \leq\|F_n-F_{n_m}\|_2+\|F_{n_m}-F\|_2 <\varepsilon. $$ Thus \(F_n\to F\) in \(L^2\), proving completeness.
What the \(L^2\) Norm Measures
The \(L^2\) norm measures the integrated square of an error, not its largest pointwise value. A function can have a large value on a sufficiently small set and still have a small \(L^2\) norm. For example, on \((0,1)\), let $$ f_n(x)=n^{1/4}\mathbf{1}_{(0,1/n)}(x). $$ Then $$ \|f_n\|_2^2 =\int_0^1 n^{1/2}\mathbf{1}_{(0,1/n)}(x)\,dx =n^{1/2}\cdot\frac1n =\frac1{\sqrt n}, $$ so \(\|f_n\|_2=n^{-1/4}\to0\). Yet the height of \(f_n\) on its support is \(n^{1/4}\), which grows. This illustrates why norm convergence and uniform convergence are different kinds of control.
Completeness matters whenever functions are built by successive approximations. If the total \(L^2\) size of the successive corrections is finite, the series theorem constructs a limit in \(L^2\); more generally, completeness ensures that every \(L^2\)-Cauchy approximation sequence has a limit even when no formula for that limit is known in advance. The inner product adds a geometric language for comparing functions, including orthogonality and the parallelogram identity.
A common pitfall is to infer \(L^2\) membership from \(L^1\) membership, or the reverse, without additional hypotheses. The singular-function example shows the relevant tests can differ because \(L^2\) integrates a square. Another pitfall is to confuse a small \(L^2\) norm with a pointwise bound: the tall, narrow functions have norms tending to zero while their maximum values grow.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why does Hölder’s inequality ensure that \(\langle f,g\rangle\) is finite for \(f,g\in L^2\)?
- What condition on a measurable set \(A\) makes its indicator belong to \(L^2\), and what is the resulting norm?
- Why does the function \(x^{-1/4}\) belong to \(L^2(0,1)\), while \(x^{-1/2}\) does not?
- How does the Absolute Convergence of a Series in \(L^2\) theorem help establish completeness?
- What does the parallelogram identity say about the \(L^2\) norms of \(f+g\) and \(f-g\)?