Real Analysis Essentials

Real analysis supplies the precise language for limits, continuity, differentiability, integrability, and function spaces. Quantum mechanics needs this language because wavefunctions are functions only after a representation is chosen, and differential operators are sensitive to more than square-integrability.

The practical rule is:

Normalizability is a Hilbert-space condition; differentiability, boundary values, and finite moments are additional analytic conditions.

This page is a physics-facing checklist, not a replacement for a real-analysis course.

Continuity

A function $f:\mathbb R\to\mathbb C$ is continuous at $x_0$ if

$$x\to x_0 \quad\Longrightarrow\quad f(x)\to f(x_0).$$

Equivalently, for every tolerance $\varepsilon>0$, there is a $\delta>0$ such that

$$\lvert x-x_0\rvert<\delta \quad\Longrightarrow\quad \lvert f(x)-f(x_0)\rvert<\varepsilon.$$

Continuity is a pointwise property. It concerns values near a point. By contrast, the $L^2$ norm concerns an integral over the whole domain.

This distinction matters because an element of $L^2(\mathbb R)$ is an equivalence class of functions that agree almost everywhere. Changing a representative at one point does not change the $L^2$ vector. Therefore pointwise statements about an arbitrary $L^2$ wavefunction require care.

Almost Everywhere

A property holds almost everywhere if it fails only on a set of measure zero. For ordinary one-dimensional wavefunctions, changing values at finitely many points changes no integral:

$$\int_{-\infty}^{\infty} \lvert\psi(x)\rvert^2\,dx$$

is unaffected by altering $\psi(0)$.

This is why $\psi(x_0)$ is not a well-defined property of a bare $L^2$ equivalence class. In many physics problems one works with a smooth or continuous representative, but the mathematical distinction prevents mistakes when discussing delta functions, boundary values, and point interactions.

For the Hilbert-space setting, see L2 Spaces.

Differentiability

A function is differentiable at $x_0$ if the limit

$$f'(x_0) = \lim_{h\to0} \frac{f(x_0+h)-f(x_0)}{h}$$

exists.

Differentiability is stronger than continuity. If $f$ is differentiable at $x_0$, then it is continuous at $x_0$, but the converse is false. The function

$$f(x)=\lvert x\rvert$$

is continuous at $0$ but not differentiable there.

Momentum and kinetic-energy operators involve derivatives. Thus a square-integrable function need not lie in the domain of the momentum operator, and a function with one derivative need not lie in the domain of the Hamiltonian. The domain issue is developed in Domains of Operators.

Classical and Weak Derivatives

The derivative used in introductory calculus is a classical derivative. Quantum mechanics also encounters weak or distributional derivatives. These are defined by how a function acts under integration against test functions.

For example, a step function is square-integrable on a finite interval, but its distributional derivative contains a delta distribution at the jump. That derivative is not an ordinary $L^2$ function. Consequently, the step function is not in the usual momentum-operator domain on that interval. The definition and jump rules are collected in Distributional Derivatives.

Distributional language is not optional decoration; it is the clean way to discuss delta potentials, Green functions, and generalized eigenstates. The general test-function language is introduced in Distributions, and the delta function itself is introduced in Delta Function.

Integrability

Integrability controls whether an expression has a finite integral. The most common examples are:

$$\int \lvert f(x)\rvert\,dx < \infty \qquad \text{and} \qquad \int \lvert f(x)\rvert^2\,dx < \infty.$$

The first condition says $f\in L^1$. The second says $f\in L^2$.

For a wavefunction, normalizability requires $L^2$ integrability:

$$\int \lvert\psi(x)\rvert^2\,dx = 1.$$

This does not automatically imply that every expectation value is finite. For example, finite position variance requires

$$\int x^2\lvert\psi(x)\rvert^2\,dx < \infty.$$

Finite kinetic energy requires derivative information, not only decay of $\psi$.

Function Spaces

Function spaces package analytic conditions into named homes. Common examples include:

• $C^0$: continuous functions;
• $C^1$: functions with a continuous first derivative;
• $C^k$: functions with continuous derivatives through order $k$;
• $C^\infty$: smooth functions;
• $C_0^\infty$: smooth functions with compact support;
• $L^p$: functions whose $p$th power is integrable, up to almost-everywhere equality;
• Sobolev spaces: functions with weak derivatives in specified $L^p$ spaces.

Quantum mechanics uses several of these at once. A wavefunction may live in $L^2$, while a Hamiltonian domain may require additional differentiability and boundary behavior. Test functions for distributions often live in $C_0^\infty$ or in the Schwartz space.

The same symbol $\psi(x)$ may therefore be used at different regularity levels. The surrounding sentence should make clear whether one is discussing an abstract state, a smooth trial function, an operator-domain vector, or a distributional idealization.

Boundary Values

Boundary conditions are pointwise or trace-like statements imposed at endpoints or surfaces:

$$\psi(0)=0, \qquad \psi'(L)=0, \qquad \psi(0)=\psi(L).$$

Such statements require enough regularity for the boundary values to make sense. A generic $L^2$ equivalence class has no well-defined value at an endpoint. In elementary problems this is often hidden because one works directly with smooth solutions of a differential equation.

The physics-facing boundary-condition page is Boundary Conditions. The operator-domain version is Domains of Operators.

Worked Example: Normalizable but Not in the Position Domain

Consider

$$\psi(x) = \frac{C}{1+\lvert x\rvert}$$

on the real line. It is square-integrable because

$$\int_{-\infty}^{\infty} \frac{dx}{(1+\lvert x\rvert)^2} < \infty.$$

Thus one can choose $C$ so that $\psi$ is normalized.

But

$$x\psi(x) = \frac{Cx}{1+\lvert x\rvert}$$

does not lie in $L^2(\mathbb R)$, because its magnitude approaches $\lvert C\rvert$ as $\lvert x\rvert\to\infty$. Therefore this normalized wavefunction is not in the domain of the position operator $X$ as an operator requiring $X\psi\in L^2$.

The lesson is that normalizability alone does not guarantee finite moments or membership in every observable’s domain.

Worked Example: A Cusp

The function

$$f(x)=e^{-\lvert x\rvert}$$

is continuous and square-integrable. Its derivative exists away from $x=0$:

$$f'(x) = \begin{cases} e^x, & x<0,\\ -e^{-x}, & x>0. \end{cases}$$

The left and right derivatives at $0$ are $1$ and $-1$, so the classical derivative does not exist at $0$.

This kind of cusp can be physically meaningful in problems with singular interactions, but it must be matched to the correct operator domain or boundary condition. Smoothness assumptions should not be silently imported from a nicer problem.

Practical Warnings

Several common manipulations require hypotheses:

• differentiating an infinite series term by term;
• moving a limit inside an integral;
• integrating by parts and dropping boundary terms;
• assigning values to an $L^2$ function at a point;
• applying an unbounded operator to a norm limit;
• treating a delta distribution as an ordinary function.

The convergence page Sequences, Series, and Convergence explains the first two warnings. The operator-domain pages explain the unbounded-operator warnings.

Common Mistakes

• Assuming every normalizable wavefunction is continuous.
• Treating a point value of an $L^2$ equivalence class as intrinsically meaningful.
• Assuming square-integrability implies finite expectation values of all observables.
• Applying a differential operator without checking differentiability and boundary conditions.
• Confusing a weak derivative with an ordinary derivative.
• Dropping boundary terms in integration by parts without checking decay or boundary conditions.
• Treating all function spaces as if they had the same convergence notion.

Cross-Links

• Sequences, Series, and Convergence
• Norms and Metrics
• L2 Spaces
• Domains of Operators
• Boundary Conditions
• Ordinary Differential Equations
• Delta Function
• Distributions
• Distributional Derivatives
• Fourier Transform
• Wavefunctions and Probability Density
• Born Rule for Continuous Spectra

References

• W. Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976.
• G. B. Folland, Real Analysis, 2nd ed., Wiley, 1999.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.

Exercises

1. Explain why changing the value of a square-integrable wavefunction at one point does not change the represented $L^2$ vector.

Solution

Changing a function at one point changes it only on a set of measure zero. The $L^2$ norm of the difference is

$$\int \lvert\psi(x)-\tilde\psi(x)\rvert^2\,dx = 0$$

because the integrand is nonzero only on a measure-zero set. Thus the two functions represent the same $L^2$ equivalence class.

2. Show that $\psi(x)=C/(1+\lvert x\rvert)$ can be normalized on the real line but need not have $x\psi(x)\in L^2(\mathbb R)$.

Solution

The normalization integral is finite:

$$\int_{-\infty}^{\infty} \frac{dx}{(1+\lvert x\rvert)^2} = 2.$$

So $C=1/\sqrt2$ normalizes the function. But

$$\left\lvert \frac{x}{1+\lvert x\rvert} \right\rvert \to 1$$

as $\lvert x\rvert\to\infty$, so

$$\int_{-\infty}^{\infty} \left\lvert \frac{Cx}{1+\lvert x\rvert} \right\rvert^2 dx = \infty.$$

Thus $x\psi(x)\notin L^2(\mathbb R)$.

3. Is $f(x)=\lvert x\rvert$ differentiable at $0$?

Solution

No. The right derivative is

$$\lim_{h\downarrow0} \frac{\lvert h\rvert-0}{h} = 1,$$

while the left derivative is

$$\lim_{h\uparrow0} \frac{\lvert h\rvert-0}{h} = -1.$$

Since the one-sided derivatives differ, the derivative at $0$ does not exist.

4. Why is a step function on a finite interval square-integrable but not in the usual momentum-operator domain?

Solution

A bounded step function on a finite interval has finite $L^2$ norm, so it is square-integrable. But it has a jump discontinuity. Its distributional derivative contains a delta distribution at the jump, not an ordinary square-integrable function. The usual momentum operator requires an $L^2$ derivative together with appropriate boundary/domain conditions, so the step function is not in that domain.