A continuous linear function must have the form f(x) = ax. Discontinuous linear functions look dreadful.

Assuming that the function f is also continuous I plan to show that f(x) = ax for some real a. Please note that if indeed f(x) = ax then a = f(1) which provides a starting point for the proof. But first let me note that (*) contains an unknown which, as we are going to establish, is equal to f(x) = ax. In other words, (*) serves as an example of a functional equation - an equation whose unknown is a function.

Proof

The proof proceeds in several steps.

1. x is 0.

   f(0) = f(0 + 0) = f(0) + f(0) = 2f(0).

   Therefore f(0) = 2f(0) and finally f(0) = 0.

2. x is negative.

   Let x be negative, e.g., let x + y = 0, where y is positive; so that -x = y. Then

   0 = f(0) = f(x + y) = f(x) + f(y).

   Therefore f(-x) = f(y) = -f(x).

3. x is an integer.

   We have f(2) = f(1 + 1) = f(1) + f(1) = 2f(1). By induction, assume f(k - 1) = (k - 1)f(1). Then

   f(k) = f(1 + (k-1)) = f(1) + (k-1)f(1) = kf(1).

   Let's denote a = f(1). We have shown that for all integers n, f(n) = an.

4. x is rational

   First of all, for any integer n0, we have 1 = n/n. Then, as before, a = f(1) = f(n/n) = nf(1/n). Hence, f(1/n) = a/n = a(1/n). For p = m/n we similarly have

   f(p) = f(m/n) = mf(1/n) = m·a/n = a(m/n) = ap.

   
   
5. x is irrational

   Any irrational number r can be approximated by a sequence of rational numbers p_i. The closer p_i is to r, the closer ap_i is to ar. However, since ap_i = f(p_i) and assuming f continuous we must necessarily get  f(r) = ar.

Continuity of the function is quite essential as it's possible to show [Ref. 1, 2] that the graph of any discontinuous solution to (*) is dense in the plane R². For the sake of reference, the graph of a function f: RR is defined as a set of pairs (x, y), i.e. elements of R² such that y = f(x). Formally, graph(f) = {(x, y)R²: y = f(x)}.

Remark

Generally speaking, a function that satisfies (*) is called additive. The function that satisfies f(x) = ax for some a is said to be homogeneous. A function is said to be linear if it's both additive and homogeneous. We have just shown that a continuous additive function is necessarily linear.

References

1. B. R. Gelbaum and J. M. H. Olmsted, Counterexamples in Analysis, Holden-Day, 1964
2. B. R. Gelbaum and J. M. H. Olmsted, Theorems and Counterexamples in Mathematics, Springer-Verlag, 1990