Functional Equations: The Four Substitutions That Crack the Problem
f(x + y) = f(x) + f(y) looks forbidding until you try x = y = 0. Most functional equations give up their secrets to a small set of standard substitutions.
Olympiad Desk · Core Learning
What you will take away
- Try 0, 1, −x, y = x and y = −x before anything else
- Injectivity or surjectivity is always a gift — grab it early
- Substituting a value that makes two terms cancel is the heart of the method
- Always verify your candidate function back in the original equation
A functional equation is a puzzle about a rule, not a number. You are given one identity and asked to identify the rule — so your tools are substitutions, not algebra.
The standard substitutions
- Set x = y = 0 to find f(0)
- Set y = 0 to relate f(x) to f(x + 0)
- Set y = −x to expose f(0) again
- Set y = x to convert a two-variable identity into one variable
Harvest injectivity and surjectivity
If you can show f(a) = f(b) forces a = b, you may cancel f on both sides of an equality — which is usually the decisive move. Surjectivity gives you a value you can choose freely, letting you rewrite the target in a friendlier variable.
Worked problems
- 1
Find all f: ℝ → ℝ with f(x + y) = f(x) + f(y) for all real x, y.
Show hint
Compute f(0) first, then look at f(x) + f(−x).
Show solution
With x = y = 0 we get f(0) = 2f(0), so f(0) = 0. With y = −x we get f(0) = f(x) + f(−x), so f(−x) = −f(x). By induction f(nx) = n f(x) for integers n, giving f(q) = cq for rationals with c = f(1). A full classification of all such functions requires linearity over ℝ and is exactly the Cauchy equation: without an extra regularity assumption (monotonicity, continuity or boundedness) wild non-linear solutions exist.
Answer: f(x) = cx for some constant c, assuming a regularity hypothesis such as monotonicity.
