Cauchy’s functional equations are very simple. The most familiar one has a simple formula:

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

But first, for the uninitiated, what is a functional equation after all? 

What is a functional equation?

Usually, functions appear as formulae. For example \( f(x) = x^2 \) is a function. It takes in a number and gives out one. If you wanted to do away with the formula, you describe the function as follows: square the input number. 

There is another way of describing a function. It is by describing abstractly what it does. For example, you could say the function f always gives a non-negative output.

But there is one problem with this method. There could be more than one function that fits that description. For example, 

“always gives a non-negative output”

is satisfied by 

\( f(x) = x^2, g(x) = e^x \)

This alternative description of a function is known as a functional equation (our example was more of a functional inequality). All the functions that satisfy a particular functional equation are known as solutions to that functional equation.

It is hard to find solutions to functional equations. That is because there can be multiple (weird) functions satisfying the given conditions. 

First Example

One of the most important examples of functional equation is Cauchy’s Functional Equation. We will examine it in a moment. But before that, let’s pick a geometric example. 

Suppose C is a unit circle with O as the center (you may take it to be the origin) on Euclidian Plane. Define \(f \) to be a function from \( \displaystyle{\mathbb{R}^2 – \{0\}\to \mathbb{R}^2 – \{0\}  }\) such that:

** f(P) = P if P is on C.

** f(P) is outside C if P is inside C and vice-versa.

** If P is on a circle orthogonal to C then f(P) is in that circle.

Show that such a function exists and is unique. Find a geometric description of f. Also find an explicit formula for it.

This function f is known as inversion.

Here is a hint. Pick a point P outside the circle C. Draw tangent PT and PT’ from P to the circle. Join OP and TT’. Suppose the intersection is P’.

Show that f(P) = P’ (and vice versa).

Any circle passing through PP’ is necessarily orthogonal to the circle C (why?!)

A Typical Problem

(INMO 2018) Let N denote the set of all natural numbers and let f: N→N be a function such that
(a) f(mn)=f(m)f(n) for all m,n in N ;
(b) m+n divides f(m)+f(n) for all m, n in N
Prove that there exists an odd natural number k such that \(f(n)=n^k\) for all n in N.

 This problem is immediately reminiscent of Cauchy’s functional equation. Therefore let’s work on it first.

Cauchy’s Functional Equation: Find f, such that f(x+y) = f(x) + f(y).

If Domain and Co-Domain are sets of Natural Numbers,  then clearly f(x) = cx where c is a constant (=f(1) ). Why?

This is readily showed using induction.

Suppose f(1) = c (afterall f(1) is SOME natural number; lets name it c )

Then f(2) = f(1+1) = f(1) + f(1) = c +c = 2c. Now use induction to show f(n) = n*c.

What happens if we extend the Domain and Co-Domain to sets of Integers. It is still easy to show f(x) = cx.

Hint: Show that f(x) is an odd function.

Next, extend the Domain and Co-Domain to sets of Rational Numbers. We will show that f(x) = cx where x is rational.

Suppose \( x = \frac{p}{q} \). Therefore \( qx = p \). Hence f(qx) = f(p). But f(p) =cp and f(qx) = qf(x).

Hence \( q f(x) = c \times p \rightarrow f(x) = c \times \frac{p}{q} = cx \)

So, up to rational numbers, Cauchy’s functional equation describes a family of linear functions (which look similar and only differ by constant multiplication).

It is not possible to extend the domain to sets of real numbers and conclude that the function is still linear. For that, we need at least one additional condition (like continuity or monotonicity or continuity at one point). 

Finally solve the INMO 2018 problem by using Cauchy’s Functional Equation.

Hint: Set f = g o log and apply Cauchy

This is the supplemental document for Cheenta Open Seminar on Functional Equation. Most of the discussions are sketches which are expanded in the live discussion.