Additive Functions and the Hamel Basis

See how free a function that preserves addition can be. On the rationals it is completely pinned; at irrationals its values come loose.

An additive function obeys

f(x + y) = f(x) + f(y)

. On the rationals

f(q) = qf(1)

leaves no freedom. But viewing the reals as a vector space over the rationals,

1

and

\sqrt{2}

are independent, so

f(\sqrt{2})

may be chosen as anything. That freedom is the Hamel basis.

For example, a 📐 square with side 1 has diagonal

\sqrt{2}

. Knowing only the side pins nothing about the value assigned to the diagonal, because the diagonal is a new, independent direction. Impose continuity and the freedom vanishes, leaving only a straight line.

Examples & Exercises

For a function $f$ satisfying $f(x + y) = f(x) + f(y)$ for all reals, prove completely that $f(qx) = qf(x)$ for every rational $q$ and every real $x$ .

f(qx) = qf(x), \qquad q \in \mathbb{Q}

Rational scaling passes straight through the function.

Show proof

Integer multiples: $f(0) = f(0 + 0) = 2f(0)$ , so $f(0) = 0$ . Induction gives $f(nx) = nf(x)$ for natural $n$ . Also $0 = f(0) = f(x + (-x)) = f(x) + f(-x)$ , so $f(-x) = -f(x)$ , covering negative integers.
Reciprocals: for $n \ge 1$ , $f(x) = f\!\left(n \cdot \frac{x}{n}\right) = n f\!\left(\frac{x}{n}\right)$ , hence $f\!\left(\frac{x}{n}\right) = \frac{1}{n} f(x)$ .
Combining with $q = \frac{p}{n}$ : $f(qx) = p f\!\left(\frac{x}{n}\right) = \frac{p}{n} f(x) = qf(x)$ . With $x = 1$ , $f(q) = qf(1)$ : on the rationals the function is completely pinned.

Prove completely that if an additive function $f$ is continuous at a single point $x_0$ , then $f(x) = cx$ for every real $x$ , where $c = f(1)$ .

f \text{ additive } + \text{continuous at one point} \ \Longrightarrow \ f(x) = cx

Continuity removes the freedom and leaves only the line.

Show proof

Continuity moves to $0$ : $f(x_0 + h) - f(x_0) = f(h)$ , so the left side tending to $0$ as $h \to 0$ — continuity at $x_0$ — says exactly $f(h) \to 0 = f(0)$ , continuity at $0$ . The same identity gives continuity everywhere.
Take a rational sequence $q_n \to x$ for an arbitrary real $x$ . The previous problem gives $f(q_n) = cq_n$ , and continuity gives $f(q_n) \to f(x)$ . But $cq_n \to cx$ . Limits are unique, so $f(x) = cx$ .

Prove completely that if an additive function $f$ is bounded on some interval $(-\delta, \delta)$ , then $f$ is continuous, and hence of the form $f(x) = cx$ .

|f| \le M \text{ on } (-\delta, \delta) \ \Longrightarrow \ f(x) = cx

Boundedness on one small interval forces the whole graph to be a line.

Show proof

Say $|f| \le M$ on the interval. If $|h| < \frac{\delta}{n}$ then $nh \in (-\delta, \delta)$ , so $|f(nh)| \le M$ . But $f(nh) = nf(h)$ , hence $|f(h)| \le \frac{M}{n}$ .
Given $\varepsilon > 0$ , choose $n$ with $\frac{M}{n} < \varepsilon$ and set $\eta = \frac{\delta}{n}$ . Then $|h| < \eta$ gives $|f(h) - f(0)| = |f(h)| \le \frac{M}{n} < \varepsilon$ : $f$ is continuous at $0$ .
By the previous problem, $f(x) = cx$ . Even a whisper of good behaviour rules out the monstrous additive functions.