Functional equations path · E02

Before this lesson: Functions, inverses and composition, How to write a proof

Your goal: Find all admissible functions and verify the complete family in the original equation.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Solving functional equations: the key idea

A functional equation asks for all functions on a stated domain satisfying an identity for every permitted input. Begin with legal substitutions such as 0,1, equal inputs or negatives. Record what each substitution proves. To establish injectivity, force equal outputs into an equation that reveals equal inputs; to establish surjectivity, construct an input for an arbitrary target. On integer domains, induction can propagate known values. Iteration can turn a functional equation into a recurrence. Polynomial guesses are candidates, not evidence that every solution is polynomial. For Cauchy’s additive equation on ℝ, additivity implies rational linearity but real linearity needs regularity, such as continuity at one point, monotonicity or boundedness on a nontrivial interval. Always finish by substituting every candidate into the original equation.

A worked example

Find all f:ℝ→ℝ satisfying f(x+y)=f(x)+f(y) and continuity at 0.

Set x=y=0 to get f(0)=0. Setting y=−x gives f(−x)=−f(x). Repeated addition gives f(nx)=nf(x) for integers n. For rational q=m/n, nf(q)=f(m)=mf(1), so f(q)=qf(1). For any real x choose rationals qₖ→x. Additivity and continuity at 0 imply f(x)−f(qₖ)=f(x−qₖ)→0. Hence f(x)=cx with c=f(1). Conversely every f(x)=cx is continuous and additive.

Your turn: change one thing

Find all f:ℤ→ℤ satisfying f(n+1)=f(n)+3 and f(0)=2.

Try this on paper before opening the explanation.

Compare your reasoning

Forward induction gives f(n)=3n+2 for n≥0. Rewriting f(n)=f(n+1)−3 gives the same formula for negative n by backward induction. Direct substitution verifies it for every integer.

Methods and connections

A substitution toolkit, with examples

Zero isolates constants: in f(x+y)=f(x)+y, x=0 gives the entire candidate family. Equal inputs reveal squares or doubles: additivity gives f(2x)=2f(x). Opposite inputs expose symmetry: additivity and f(0)=0 give oddness. Keep the unused variable free so the resulting statement remains an identity, not just a numerical example.

Injectivity, surjectivity and fixed points

If an equation gives f(f(x))=x, then f is injective: f(a)=f(b) implies a=f(f(a))=f(f(b))=b. It is also onto: any target y equals f(f(y)). This condition alone does not force f(x)=x; f(x)=−x also works, as do other involutions. A fixed point satisfies f(c)=c. When such a point is proved to exist, substituting c can simplify nested expressions; do not assume its existence from a graph.

Separate integer and real domains

On ℤ, an additive function has f(n)=nf(1), proved by induction and negatives. On ℚ, use division by positive integers. On ℝ, extending from dense rational inputs needs an explicitly stated regularity hypothesis. A polynomial ansatz is valid if the question requires a polynomial function, or after polynomiality has been proved. Trying f(x)=ax+b may discover solutions, but cannot by itself exclude nonlinear ones.

Iteration as a recurrence

If f(x+1)=2f(x) on the integers and f(0)=3, then f(n)=3⋅2nf(n)=3\cdot 2^{n} for every integer n. Forward induction handles n≥0, and f(n)=f(n+1)/2 handles negatives. On all real inputs, knowing f(0) and the same shift equation leaves the values between 0 and 1 undetermined: the domain changes the solution family.

Pause and check

A trap to avoid: Finding one function is not finding all; add D02, R03 or A01 before the corresponding advanced branch.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Find all f:ℝ→ℝ satisfying f(x+y)=f(x)+y for every real x,y.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Set x=0. Then f(y)=f(0)+y, so any solution must be f(t)=t+c for a real constant c. Conversely f(x+y)=x+y+c and f(x)+y=x+c+y agree for all inputs. Thus all real constants c give solutions, and the x=0 substitution proves completeness.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. What must be checked before substituting an input?

Foundation

  1. It is positive always
  2. It is an integer always
  3. It is an output
  4. It belongs to the stated domain
Hint

Domain restrictions apply to every variable.

Answer and reasoning

It belongs to the stated domain. A substitution outside the domain is not permitted by the equation.

2. If f is additive on ℝ, what is f(0)?

Foundation

  1. Any value
  2. 0
  3. 1
  4. f(1)
Hint

Set x=y=0.

Answer and reasoning

0. f(0)=2f(0), so f(0)=0.

3. If f is additive and f(1)=6, what is f(1/3)?

Core

  1. 6
  2. 18
  3. 2
  4. 3
Hint

Add 1/3 three times.

Answer and reasoning

2. 3f(1/3)=f(1)=6.

4. If f(x+y)=f(x)+y, the substitution x=0 gives what?

Core

  1. f(y)=yf(0)
  2. f(y)=y²
  3. f(y)=y+f(0)
  4. f(y)=0
Hint

Keep the remaining variable free.

Answer and reasoning

f(y)=y+f(0). The equation directly expresses every value in terms of f(0).

5. Does additivity alone on ℝ justify f(x)=cx for all real x?

Stretch

  1. It forces f(x)=x²
  2. No, a regularity condition is needed for that conclusion
  3. Yes always
  4. Only if c=0
Hint

Rational linearity does not by itself extend by limits.

Answer and reasoning

No, a regularity condition is needed for that conclusion. Continuity or another suitable regularity hypothesis is needed to justify the usual real-linear conclusion.

6. Why substitute a candidate into the original equation?

Stretch

  1. Because examples prove completeness
  2. To establish sufficiency and catch lost restrictions
  3. To prove uniqueness alone
  4. To avoid using the domain
Hint

Discovery and verification are different obligations.

Answer and reasoning

To establish sufficiency and catch lost restrictions. Necessary deductions can admit extraneous candidates, so each candidate must be checked in the full equation.

7. To find all functions, one must prove both what?

Foundation

  1. Only one example
  2. Only continuity
  3. Only a graph
  4. Necessity and sufficiency
Hint

Find candidates and verify completeness.

Answer and reasoning

Necessity and sufficiency. Every solution must have the proposed form, and every permitted candidate must work.

8. For additive f with f(1)=4, what is f(−2)?

Core

  1. 2
  2. −2
  3. −8
  4. 8
Hint

Use repeated addition and oddness.

Answer and reasoning

−8. f(2)=8 and f(−2)=−f(2)=−8.

9. Does f(f(x))=x imply injectivity?

Stretch

  1. Only for continuous f
  2. Only for integer x
  3. Yes
  4. No
Hint

Apply f to an equality f(a)=f(b).

Answer and reasoning

Yes. It gives f(f(a))=f(f(b)), hence a=b.

Choose your next step

Try a written problem in the challenge room. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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