The Cauchy Functional Equation

Why This Matters

The Cauchy functional equation is

$f(x + y) = f(x) + f(y) \qquad \forall x, y \in \mathbb{R}.$

A function satisfying this is called additive. The equation is the entry point to the entire field of functional equations, the model problem every olympiad student should know cold, and the gateway to many subsequent functional-equation techniques (Jensen's, multiplicative, Pexider's).

The result that drives most olympiad applications:

Under any one of these conditions: continuity at a single point, monotonicity on any interval, boundedness on a set of positive Lebesgue measure, or Lebesgue measurability — every additive function $f: \mathbb{R} \to \mathbb{R}$ is linear: $f(x) = cx$ for some constant $c$ .

Without any regularity, additive but non-linear (so-called pathological) solutions exist. Their construction (Hamel 1905) requires the axiom of choice; the graphs of these functions are dense in $\mathbb{R}^2$ .

For olympiad and Putnam problems, the canonical move is:

Derive Cauchy's equation (or a near-variant) from the problem's hypotheses.
Plug in $x = y = 0$ , $x = 0$ , $y = -x$ , etc., to extract structure.
Show $f(\text{rational multiple of } x) = (\text{rational}) \cdot f(x)$ .
Use a regularity condition (continuity, monotonicity, boundedness) given by the problem to extend from $\mathbb{Q}$ -linearity to $\mathbb{R}$ -linearity.
Conclude $f(x) = cx$ and verify by plugging back in.

This page treats the technique in the order an olympiad student would apply it.

Theorem: Solutions Over Q

Theorem

Q-Solutions Are Linear

Statement

For any additive $f: \mathbb{R} \to \mathbb{R}$ and any rational $r \in \mathbb{Q}$ and real $x$ :

$f(rx) = r f(x).$

In particular, $f$ restricted to $\mathbb{Q}$ is of the form $f(q) = q f(1)$ , i.e., $\mathbb{Q}$ -linear.

Intuition

Additivity over $\mathbb{R}$ already forces $f$ to be a $\mathbb{Q}$ -linear map. The proof is by induction on positive integers, then extension to negatives, then to rationals via $f(p/q) = (1/q) f(p) = (p/q) f(1)$ .

Proof Sketch

Step 1: $f(0) = 0$ . Set $x = y = 0$ in additivity: $f(0) = f(0) + f(0)$ , so $f(0) = 0$ .

Step 2: $f(-x) = -f(x)$ . Set $y = -x$ : $f(0) = f(x) + f(-x)$ , so $f(-x) = -f(x)$ .

Step 3: $f(nx) = n f(x)$ for $n \in \mathbb{N}$ . Induction: $f((n+1)x) = f(nx + x) = f(nx) + f(x) = n f(x) + f(x) = (n+1) f(x)$ .

Step 4: $f(nx) = n f(x)$ for $n \in \mathbb{Z}$ . Combine steps 2 and 3.

Step 5: $f(x/n) = f(x)/n$ . From step 3, $f(n \cdot (x/n)) = n f(x/n)$ , so $f(x) = n f(x/n)$ , giving $f(x/n) = f(x)/n$ .

Step 6: $f(p/q \cdot x) = (p/q) f(x)$ . Combine steps 4 and 5.

Why It Matters

Step 6 is the key intermediate result. Without any regularity, $f$ is already determined on the $\mathbb{Q}$ -span of any single point. To extend to $\mathbb{R}$ -linearity, you only need to control $f$ between rationals: and any of the standard regularity conditions does that.

Failure Mode

This theorem only gives $\mathbb{Q}$ -linearity, not $\mathbb{R}$ -linearity. The pathological non-linear solutions do satisfy step 6 (they are $\mathbb{Q}$ -linear), but they fail to be $\mathbb{R}$ -linear because they don't extend "continuously" between rationals. See the Hamel-basis construction below.

report a correction →

Theorem: Regularity Forces Linearity

Theorem

Regularity Implies Linearity

Statement

If $f$ is additive and any one of the following holds:

$f$ is continuous at some point $x_0 \in \mathbb{R}$ .
$f$ is monotonic on some interval $(a, b)$ .
$f$ is bounded above (or below) on some set of positive Lebesgue measure.
$f$ is Lebesgue measurable.
$f$ is bounded on some interval (hence on every interval by additivity).

Then $f(x) = cx$ for some constant $c = f(1)$ , i.e., $f$ is $\mathbb{R}$ -linear.

Intuition

Each regularity condition rules out the wild oscillations that pathological solutions exhibit. Continuity at one point is enough: additivity bootstraps it to continuity everywhere. Boundedness on a positive-measure set is the most striking: even very weak regularity is enough to collapse the solution space to the linear functions.

Proof Sketch

Continuity at one point implies continuity everywhere. Suppose $f$ is continuous at $x_0$ . For any $a$ and any sequence $x_n \to a$ , $x_n - a + x_0 \to x_0$ , so by continuity at $x_0$ : $f(x_n - a + x_0) \to f(x_0)$ . By additivity, $f(x_n) - f(a) + f(x_0) \to f(x_0)$ , so $f(x_n) \to f(a)$ . So $f$ is continuous at $a$ .

Continuity + $\mathbb{Q}$ -linearity = $\mathbb{R}$ -linearity. The rationals are dense in $\mathbb{R}$ , and on $\mathbb{Q}$ we have $f(q) = q f(1)$ . Continuity extends this: $f(x) = \lim_{q_n \to x, q_n \in \mathbb{Q}} f(q_n) = \lim q_n f(1) = x f(1)$ .

Monotonicity on $(a, b)$ . Sandwich any irrational $x$ between two rationals $q_1 < x < q_2$ in $(a, b)$ . Monotonicity gives $q_1 f(1) \le f(x) \le q_2 f(1)$ . Take $q_1, q_2 \to x$ .

Boundedness on a positive-measure set, measurability. The Steinhaus theorem (1920) says any positive-measure set $A \subseteq \mathbb{R}$ has $A - A$ containing an open interval. Combine with additivity and the $\mathbb{Q}$ -linearity result: if $f$ is bounded on $A$ , then $f$ is bounded on $A - A \supseteq (-\delta, \delta)$ for some $\delta > 0$ , hence bounded on a neighborhood of zero, hence linear by the boundedness case. Measurability + the Frechet 1913 trick reduces to this.

Why It Matters

The strength of these implications is striking. Any of the five conditions: even very weak ones like Lebesgue measurability: suffices to collapse the entire space of additive functions to the one-parameter family of linear functions $\{x \mapsto cx : c \in \mathbb{R}\}$ . The implication is robust to slight weakening: e.g., "bounded on a set of second category in the Baire sense" also works.

Failure Mode

Without any regularity, the conclusion fails badly. Pick a Hamel basis $H$ of $\mathbb{R}$ as a $\mathbb{Q}$ -vector space (this requires the axiom of choice). Define $f$ on $H$ arbitrarily and extend $\mathbb{Q}$ -linearly. Generically this $f$ is additive but not $\mathbb{R}$ -linear. Its graph is dense in $\mathbb{R}^2$ (Hamel 1905). It is not measurable, not bounded on any open set, and discontinuous everywhere.

report a correction →

The Standard Olympiad Move

For olympiad / Putnam problems featuring an unknown function $f$ , the workflow:

Plug in special values. $x = y = 0$ usually gives $f(0)$ . Symmetric or antisymmetric substitutions often collapse the equation.
Derive Cauchy's equation (or a multiplicative variant $g(xy) = g(x) g(y)$ , or Jensen's $f(\frac{x+y}{2}) = \frac{f(x) + f(y)}{2}$ ).
Use $\mathbb{Q}$ -linearity to determine $f$ on rationals.
Apply the problem's regularity (continuity, monotonicity, bound on an interval, etc.) to extend to $\mathbb{R}$ .
Verify the candidate solution back in the original equation.

Step 5 is essential. Many problems have $f \equiv 0$ or $f(x) = c x$ plus extra solutions you'd miss without checking. A famous trap: $f(x+y) = f(x) f(y)$ (the multiplicative Cauchy) admits both $f(x) = a^x$ and $f \equiv 0$ .

Worked Example: Olympiad-Style

Example

A continuous additive function

Problem. Find all continuous $f: \mathbb{R} \to \mathbb{R}$ with $f(x + y) = f(x) + f(y)$ for all $x, y$ .

Solution. By the $\mathbb{Q}$ -linearity theorem, $f(q) = q f(1)$ for all $q \in \mathbb{Q}$ . By continuity, for any $x \in \mathbb{R}$ , choose rationals $q_n \to x$ : $f(x) = \lim f(q_n) = \lim q_n f(1) = x f(1)$ . So $f(x) = c x$ where $c = f(1)$ .

Verification: $f(x + y) = c(x + y) = cx + cy = f(x) + f(y)$ . Done.

Without continuity. Pathological solutions exist (Hamel 1905); the problem is no longer well-posed in the olympiad sense. Olympiad problems always include a regularity condition.

Worked Example: Multiplicative Variant

Example

The multiplicative Cauchy

Problem. Find all continuous $g: \mathbb{R}^+ \to \mathbb{R}^+$ with $g(xy) = g(x) g(y)$ .

Solution. Take logarithms: let $f(t) = \log g(e^t)$ . Then $f(s + t) = \log g(e^{s+t}) = \log g(e^s e^t) = \log(g(e^s) g(e^t)) = f(s) + f(t)$ . So $f$ is additive, and continuous (composition of continuous functions). By the linearity theorem, $f(t) = ct$ for some $c$ . Inverting: $g(x) = e^{c \log x} = x^c$ .

So the continuous solutions are $g(x) = x^c$ for any real $c$ .

Note the trap. The trivial solution $g \equiv 1$ is the $c = 0$ case (i.e., $g(x) = x^0 = 1$ ). The trivial solution $g \equiv 0$ is not in our solution set because we restricted to $g: \mathbb{R}^+ \to \mathbb{R}^+$ .

Variants and Related Equations

Equation	Standard solution (with regularity)
$f(x + y) = f(x) + f(y)$	$f(x) = cx$
$f(xy) = f(x) + f(y)$	$f(x) = c \log x$ (on $\mathbb{R}^+$ )
$f(x + y) = f(x) f(y)$	$f(x) = a^x$
$f(xy) = f(x) f(y)$	$f(x) = x^c$
$f(\frac{x+y}{2}) = \frac{f(x)+f(y)}{2}$ (Jensen)	$f(x) = ax + b$
$f(x + y) + f(x - y) = 2f(x) + 2f(y)$ (parallelogram)	$f(x) = ax^2 + bx$
Pexider $f(x + y) = g(x) + h(y)$	$f(x) = ax + b + c$ , $g, h$ similar

Each can be reduced to Cauchy's via a logarithmic / exponential substitution or by combining with auxiliary hypotheses.

Common Mistakes

Watch Out

Forgetting regularity is necessary

The pathological-solution warning matters. An olympiad problem that asks "find all $f$ with $f(x + y) = f(x) + f(y)$ " without any regularity condition is under-specified: the solution set is uncountably infinite and includes wild non-linear functions. Always check what regularity the problem assumes.

Watch Out

Confusing Cauchy with Jensen

Cauchy: $f(x + y) = f(x) + f(y)$ . Jensen: $f(\frac{x+y}{2}) = \frac{f(x) + f(y)}{2}$ .

Both have linear solutions under regularity, but Jensen's allows a constant: $f(x) = ax + b$ . The two equations are related but not equivalent.

Watch Out

Forgetting to verify the candidate solution

Even after deriving $f(x) = cx$ , you must plug it back in to the original problem to verify it works (and find the constraints on $c$ ). Sometimes additional restrictions kick in: e.g., the original problem says $f$ is increasing, which forces $c > 0$ .

Watch Out

Asserting linearity from continuity at no specific point

"Continuity" must be at some specific point $x_0$ : the proof bootstraps continuity at one point to continuity everywhere. Saying "f is continuous" without specifying where is fine in practice, but understanding the bootstrapping matters for non-standard variants.

Exercises

ExerciseCore

Problem

Find all continuous $f: \mathbb{R} \to \mathbb{R}$ with $f(x + y) = f(x) + f(y) + xy$ for all $x, y$ .

ExerciseAdvanced

Problem

(IMO Shortlist 2002 A5) Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x) + y) = f(x^2 - y) + 4 f(x) y$ for all $x, y$ .

ML and Production Connections

Linear maps in deep learning. The Cauchy functional equation under continuity + boundedness gives $f(x) = cx$ , which is exactly the affine-without-bias case. Neural network "linear layers" implement (vector-valued) $\mathbb{R}$ -linear maps; the underlying claim is that any continuous additive function is necessarily linear.
Equivariance and homomorphisms. The condition $f(g \cdot x) = g \cdot f(x)$ is a group-equivariance condition (cf. equivariant-neural-networks-foundations); with abelian groups and continuity, the analog of Cauchy's theorem applies, and equivariant linear layers are forced by the group structure.
Information theory. Shannon's entropy axioms: additivity on independent sources, continuity, normalization: uniquely determine $H(p) = -\sum p_i \log p_i$ via a functional-equation argument: the same Cauchy applied to the entropy function.
Random feature theory. The Bochner-theorem side of Rahimi-Recht 2007 random features uses functional-equation arguments to characterize positive-definite kernels via Fourier representations; the Cauchy structure is implicit in the multiplicative-functional setup.

Cross-Network Links

ProofsPath: pigeonhole-principle-elementary and extremal-principle are sibling proof techniques; invariants-and-monovariants gives the parallel structural-argument toolkit; vieta-jumping is the Diophantine companion.
TheoremPath: information-theory-foundations uses functional-equation arguments to derive entropy uniquely; equivariant-neural-networks-foundations is the modern equivariance application.
CalculusPath: convex-functions-foundations is the convex-analytic companion (Jensen's inequality).

References

See structured references block. Primary entry points: Cauchy's Cours d'analyse (1821) for the original; Aczel (1966) for the standard functional-equations textbook; Engel "Problem-Solving Strategies" Ch 11 for the olympiad treatment; Hamel (1905) for the construction of pathological solutions.