L2 Spaces

An $L^2$ space is a Hilbert space built from square-integrable functions, with functions identified when they agree almost everywhere. The measure is part of the definition:

$$L^2(X,\mu).$$

This space is the natural analytic home for normalizable wavefunction representations, Fourier expansions, and many differential-operator problems. The mathematical object is an equivalence class of functions, not an arbitrary pointwise formula.

The physical interpretation of a wavefunction belongs to Wavefunctions as Representations. This page develops the function-space mathematics.

Definition on a Measure Space

Let $(X,\Sigma,\mu)$ be a measure space. Begin with the measurable complex functions $f:X\to\mathbb C$ satisfying

$$\int_X\lvert f(x)\rvert^2\,d\mu(x)<\infty.$$

Two such functions are equivalent when they agree almost everywhere:

$$\begin{aligned} f\sim g &\Longleftrightarrow f(x)=g(x) \\ &\quad \text{for }\mu\text{-almost every }x. \end{aligned}$$

The space $L^2(X,\mu)$ is the set of equivalence classes under this relation. Its inner product and norm are

$$\begin{aligned} \langle f\vert g\rangle &= \int_X f(x)^*g(x)\,d\mu(x),\\ \lVert f\rVert_2 &= \left( \int_X\lvert f(x)\rvert^2\,d\mu(x) \right)^{1/2}. \end{aligned}$$

The subscript on $\lVert\cdot\rVert_2$ distinguishes this norm from pointwise, uniform, or derivative-sensitive norms.

Why Equivalence Classes Are Necessary

On raw functions, the integral expression is only a seminorm. A nonzero function can vanish almost everywhere. For example, on $\mathbb R$ with Lebesgue measure,

$$f(x) = \begin{cases} 1,&x=0,\\ 0,&x\ne0 \end{cases}$$

has

$$\lVert f\rVert_2=0.$$

A genuine norm must satisfy $\lVert f\rVert=0$ only for the zero vector. Passing to equivalence classes identifies every almost-everywhere-zero function with zero and restores positive definiteness.

This has practical consequences:

• changing a representative at finitely many points does not change the $L^2$ vector;
• a bare $L^2$ element has no well-defined value at a specified point;
• continuity and boundary values require extra regularity or a chosen representative;
• the Dirac delta is not an $L^2$ function.

Physicists often write a representative and call it the wavefunction. That is harmless when the almost-everywhere equivalence is understood.

The Inner Product Is Well Defined

If $f$ and $g$ are square integrable, their product is integrable by the Cauchy–Schwarz inequality:

$$\begin{aligned} \left\lvert \int_X f^*g\,d\mu \right\rvert &\le \int_X\lvert f\rvert\lvert g\rvert\,d\mu\\ &\le \lVert f\rVert_2\lVert g\rVert_2. \end{aligned}$$

Changing either function on a null set changes neither the integral nor the equivalence class. Thus the inner product is independent of the chosen representatives.

The induced distance is

$$d_2(f,g)=\lVert f-g\rVert_2.$$

Two representatives can differ substantially at isolated points and still have zero distance because they represent the same vector.

Completeness

The Riesz–Fischer theorem states that $L^2(X,\mu)$ is complete: every $L^2$-Cauchy sequence converges in the $L^2$ norm to an equivalence class in $L^2(X,\mu)$. Therefore $L^2$ is a Hilbert space.

The proof is analytic rather than pointwise. One standard route selects a subsequence $(f_{n_k})$ whose successive differences satisfy

$$\sum_{k=1}^{\infty} \lVert f_{n_{k+1}}-f_{n_k}\rVert_2 <\infty.$$

The sum of the absolute successive differences is then an $L^2$ function and is finite almost everywhere. It defines an almost-everywhere limit for the subsequence, and the original Cauchy property gives convergence of the full sequence in $L^2$.

The important conclusion is not that every sequence converges. It is that an $L^2$-Cauchy sequence cannot converge to a missing object outside the space.

Standard Models

The real line.

$$L^2(\mathbb R,dx) = \left\{ f: \int_{-\infty}^{\infty} \lvert f(x)\rvert^2\,dx<\infty \right\}$$

modulo equality almost everywhere.

Euclidean configuration space.

$$L^2(\mathbb R^d,d^dx)$$

uses the $d$-dimensional Lebesgue measure.

A finite interval.

$$L^2([a,b],dx)$$

contains constant functions because the interval has finite measure.

Counting measure. On $\mathbb N$ with counting measure,

$$L^2(\mathbb N) \cong \ell^2(\mathbb N),$$

because integration becomes summation.

Finite sets. On a set of $d$ points with counting measure,

$$L^2(X)\cong\mathbb C^d.$$

Thus finite coordinate spaces, square-summable sequences, and square-integrable functions are versions of one construction with different measure spaces.

The Measure Is Part of the Space

The notation $L^2(X)$ suppresses information. Different measures on the same underlying set generally give different norms and can give different equivalence classes.

In spherical coordinates on $\mathbb R^3$,

$$d^3x =r^2\sin\theta\, dr\,d\theta\,d\varphi.$$

The radial measure is therefore $r^2dr$, not merely $dr$. A radial function $R(r)$ has norm contribution

$$\int_0^\infty \lvert R(r)\rvert^2r^2\,dr.$$

If one instead defines $u(r)=rR(r)$, then

$$\int_0^\infty \lvert R(r)\rvert^2r^2\,dr = \int_0^\infty \lvert u(r)\rvert^2\,dr.$$

The two formulas use different representatives and measures for equivalent radial information.

More generally, under a coordinate change $x=x(q)$,

$$d^dx = \left\lvert \det\frac{\partial x}{\partial q} \right\rvert d^dq.$$

Dropping the Jacobian changes the inner product.

Square-Integrability Tests

Membership is determined by local singularities and large-distance tails. For a power law on $\mathbb R^d$,

$$\lvert f(x)\rvert \sim \lVert x\rVert^{-\alpha},$$

the radial contribution at large $r$ behaves as

$$\int^\infty r^{d-1-2\alpha}\,dr.$$

It converges at infinity precisely when

$$\alpha>\frac d2.$$

Near the origin, the same power law is locally square integrable precisely when

$$\alpha<\frac d2.$$

The direction of the inequality reverses because the endpoint changes.

On the line,

$$f_p(x) = \frac{1}{(1+\lvert x\rvert)^p}$$

belongs to $L^2(\mathbb R)$ exactly when $p>1/2$. It is bounded near the origin, so only the tail matters.

Common Function Examples

• Every Gaussian $e^{-a x^2}$ with $\operatorname{Re}a>0$ belongs to $L^2(\mathbb R)$.
• The exponential $e^{-a\lvert x\rvert}$ belongs to $L^2(\mathbb R)$ when $\operatorname{Re}a>0$.
• A nonzero constant belongs to $L^2([a,b])$ but not to $L^2(\mathbb R)$.
• A plane wave $e^{ikx}$ is not in $L^2(\mathbb R)$ because its modulus is one.
• The Dirac delta is a distribution, not a square-integrable function.
• A function can belong to $L^2$ without being continuous or differentiable.

Plane waves and delta functions remain useful as generalized spectral objects. Their canonical treatment is Generalized Eigenvectors.

Worked Example: Normalizing a Gaussian

Let

$$\psi(x)=C e^{-\alpha x^2/2}, \qquad \alpha>0.$$

Its squared norm is

$$\begin{aligned} \lVert\psi\rVert_2^2 &= \lvert C\rvert^2 \int_{-\infty}^{\infty} e^{-\alpha x^2}\,dx\\ &= \lvert C\rvert^2 \sqrt{\frac{\pi}{\alpha}}. \end{aligned}$$

Unit norm requires

$$\lvert C\rvert = \left( \frac{\alpha}{\pi} \right)^{1/4}.$$

The norm determines only the magnitude of $C$. Multiplication by a constant phase $e^{i\gamma}$ leaves the norm unchanged.

The probability interpretation of unit norm is a quantum postulate, not a consequence of the $L^2$ definition; see Normalization.

Norm Convergence versus Pointwise Convergence

Convergence in $L^2$ means

$$\lVert f_n-f\rVert_2 \longrightarrow0.$$

It does not mean $f_n(x)\to f(x)$ at every point. Conversely, pointwise convergence does not imply $L^2$ convergence.

For example, on $\mathbb R$ let

$$f_n=\mathbf 1_{[n,n+1]}.$$

For every fixed $x$, eventually $x\notin[n,n+1]$, so $f_n(x)\to0$. But

$$\lVert f_n\rVert_2^2=1$$

for every $n$, so the sequence does not converge to zero in $L^2$.

If $f_n\to f$ in $L^2$, one can extract a subsequence converging to $f$ almost everywhere, but the entire sequence need not converge pointwise. Additional hypotheses such as domination can connect pointwise and norm convergence.

This distinction is central to basis expansions: convergence of a Fourier series in $L^2$ does not automatically imply pointwise convergence at every location.

Dense Subspaces and Test Functions

For Lebesgue measure on $\mathbb R^d$, the smooth compactly supported functions

$$C_c^\infty(\mathbb R^d)$$

are dense in $L^2(\mathbb R^d)$. Every $L^2$ vector can be approximated in norm by smooth functions with bounded support.

Density does not mean every $L^2$ function is smooth. It means that for each $f\in L^2$ and every $\epsilon>0$, there is $\varphi\in C_c^\infty$ such that

$$\lVert f-\varphi\rVert_2<\epsilon.$$

Dense test spaces are useful initial domains for differential operators and distribution theory. Their completion in the $L^2$ norm forgets derivatives and pointwise boundary data; retaining that information requires stronger spaces, such as Sobolev spaces.

Orthonormal Bases and Fourier Maps

If $\{e_n\}$ is a complete orthonormal basis of a separable $L^2$ space, then

$$\begin{aligned} f &= \sum_{n=1}^{\infty} \langle e_n\vert f\rangle e_n,\\ \lVert f\rVert_2^2 &= \sum_{n=1}^{\infty} \left\lvert \langle e_n\vert f\rangle \right\rvert^2. \end{aligned}$$

These statements concern $L^2$ equivalence classes and norm convergence. Trigonometric functions on a finite interval and Hermite functions on $\mathbb R$ are standard examples.

With a consistent convention, the Fourier transform extends from nice test functions to a unitary map on $L^2(\mathbb R^d)$. Plancherel’s theorem gives

$$\lVert\widehat f\rVert_2=\lVert f\rVert_2.$$

See Fourier Transform for normalization conventions and the extension.

Multiplication Operators and Domains

Let $g:X\to\mathbb C$ be measurable. Multiplication by $g$ is the operator

$$(M_gf)(x)=g(x)f(x).$$

If $g$ is essentially bounded, then $M_g$ is bounded on all of $L^2$ and

$$\lVert M_g\rVert_{\mathrm{op}} = \operatorname*{ess\,sup}_{x\in X} \lvert g(x)\rvert.$$

If $g$ is unbounded, the natural domain is

$$\mathcal D(M_g) = \left\{ f\in L^2: gf\in L^2 \right\}.$$

For example, multiplication by $x$ on $L^2(\mathbb R)$ is not defined on every square-integrable function. Square integrability of $f$ does not imply square integrability of $xf$.

Differential operators require still more regularity. An arbitrary $L^2$ equivalence class need not possess a classical derivative or pointwise boundary values. See Domains of Operators and Real Analysis Essentials.

Product Spaces and Multiple Components

Under standard hypotheses on the measures,

$$\begin{gathered} L^2(X,\mu)\otimes L^2(Y,\nu) \\ \cong L^2(X\times Y,\mu\otimes\nu). \end{gathered}$$

A simple tensor corresponds to a product function:

$$(f\otimes g)(x,y)=f(x)g(y).$$

Finite linear combinations of such product functions are dense in the product-space $L^2$. General elements need not factor.

For a $k$-component wavefunction, use the vector-valued space

$$L^2(X,\mu;\mathbb C^k)$$

with norm

$$\lVert\psi\rVert_2^2 = \int_X \sum_{a=1}^{k} \lvert\psi_a(x)\rvert^2\,d\mu(x).$$

Spinor and coupled-channel wavefunctions fit this pattern.

Boundary with Probability

If a wavefunction representative is normalized,

$$\int_X\lvert\psi(x)\rvert^2\,d\mu(x)=1,$$

the continuous Born rule assigns a region $R\subseteq X$ the probability

$$\int_R\lvert\psi(x)\rvert^2\,d\mu(x).$$

The integral depends only on the $L^2$ equivalence class. But the assignment of this integral as a probability is physical structure beyond the definition of $L^2$. See The Continuous Born Rule.

The measure also controls units. If $x$ has dimensions of length and $d\mu=dx$, a normalized one-dimensional wavefunction has units $\text{length}^{-1/2}$ so that $\lvert\psi\rvert^2dx$ is dimensionless.

Numerical Practice

A grid and quadrature rule replace the integral by a weighted sum:

$$\langle f\vert g\rangle \approx \sum_{j=1}^{N} w_j f(x_j)^*g(x_j).$$

The weights are part of the discrete inner product. Treating sampled values with the ordinary Euclidean dot product is correct only when the quadrature and basis convention justify equal weights.

Useful checks include:

• verify all weights are nonnegative for a norm quadrature;
• include coordinate Jacobians and cell volumes;
• normalize with the same weights used by operators;
• refine the grid and domain independently;
• monitor tail probability outside the computational window;
• distinguish interpolation error from $L^2$ error;
• do not infer pointwise convergence from a stable discrete norm alone.

For a generalized eigenproblem with a nonorthogonal discretization, the mass or overlap matrix represents the discrete $L^2$ inner product.

Common Mistakes

• Defining $L^2$ as a set of pointwise functions rather than equivalence classes.
• Treating a single point value as a property of an arbitrary $L^2$ vector.
• Forgetting that the measure and coordinate Jacobian are part of the norm.
• Assuming square integrability implies continuity or differentiability.
• Treating a plane wave or Dirac delta as a normalizable $L^2$ vector.
• Confusing pointwise, uniform, and $L^2$ convergence.
• Assuming every multiplication or differential operator acts on all of $L^2$.
• Dropping component sums for spinor-valued functions.
• Normalizing a grid vector with weights different from those used in the Hamiltonian discretization.
• Inferring the Born rule from the function-space definition alone.

Exercises

1. On $[0,1]$ with Lebesgue measure, let $f(x)=0$ for all $x\ne1/2$ and $f(1/2)=7$. Compute $\lVert f\rVert_2$ and identify its $L^2$ equivalence class.

Solution

The function is nonzero only on the one-point set $\{1/2\}$, which has Lebesgue measure zero. Therefore

$$\lVert f\rVert_2^2 = \int_0^1\lvert f(x)\rvert^2\,dx =0.$$

It represents the zero equivalence class in $L^2([0,1])$.

2. For

   $$f_p(x) = \frac{1}{(1+\lvert x\rvert)^p},$$

   determine all real $p$ for which $f_p\in L^2(\mathbb R)$.

Solution

The function is bounded near $x=0$, so only the tail matters. For large $\lvert x\rvert$,

$$\lvert f_p(x)\rvert^2 \sim \lvert x\rvert^{-2p}.$$

The integral

$$\int_1^\infty x^{-2p}\,dx$$

converges exactly when $2p>1$. Hence

$$f_p\in L^2(\mathbb R) \quad\Longleftrightarrow\quad p>\frac12.$$

3. Normalize

   $$\psi(x)=C e^{-\alpha x^2/2}, \qquad \alpha>0,$$

   and state the remaining freedom in $C$.

Solution

Using the Gaussian integral,

$$\begin{aligned} 1 &= \lVert\psi\rVert_2^2\\ &= \lvert C\rvert^2 \int_{-\infty}^{\infty} e^{-\alpha x^2}\,dx\\ &= \lvert C\rvert^2 \sqrt{\frac{\pi}{\alpha}}. \end{aligned}$$

Therefore

$$\lvert C\rvert = \left( \frac{\alpha}{\pi} \right)^{1/4}.$$

The most general choice is

$$C = e^{i\gamma} \left( \frac{\alpha}{\pi} \right)^{1/4},$$

where $\gamma$ is any real constant phase.

4. On $L^2(\mathbb R)$, let $M_x$ multiply by $x$. Use normalized indicator functions supported on $[n,n+1]$ to prove that $M_x$ is unbounded.

Solution

Let

$$f_n=\mathbf 1_{[n,n+1]}.$$

The interval has length one, so $\lVert f_n\rVert_2=1$. But

$$\begin{aligned} \lVert M_xf_n\rVert_2^2 &= \int_n^{n+1}x^2\,dx\\ &\ge n^2. \end{aligned}$$

Hence

$$\lVert M_xf_n\rVert_2\ge n.$$

No finite constant $C$ can satisfy $\lVert M_xf\rVert_2\le C\lVert f\rVert_2$ for every $f$ in the domain. Thus $M_x$ is unbounded. Its natural domain is

$$\mathcal D(M_x) = \left\{ f\in L^2(\mathbb R): xf\in L^2(\mathbb R) \right\}.$$

References

• G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Wiley, 1999.
• E. M. Stein and R. Shakarchi, Real Analysis: Measure Theory, Integration, and Hilbert Spaces, Princeton University Press, 2005.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.