Level 3, PSS, Day 1

Today, we discussed a single problem on Functional Equation( Agam’s Request ) from Bulgaria Olympiad, 2018. ( International Fest for Young Mathematicians, Sozopol).

Problem 1:

Find all functions f : \( [{0, \infty}) \rightarrow [{0, \infty}) \) such that for all x, y non-negative, f satisfies \( f( f(x) + f(y)) = xyf(x+y) \).

Discussion:

Well, there is always an algorithm to approach a functional equation problem. The function equation problem is a well defined mathematical problem. Just that we have

  • To be creative, observe the pattern by playing around with the numbers and
  • Get an intuition about the plausible solutions and
  • Therefore move towards proving that intuition, you have.

Always try to reduce the number of variables by putting in special numbers, so that the problem is more tractable.

Building the intuition, Observing the pattern :

Observe that \( f(x) = 0\) is a solution. This gives us the intuition, that we have to reach somewhere, where one of the solutions of f(0) is 0. Mbe ouay

We therefore put \( y = 0 \rightarrow ( f( f(x) + f(0)) = 0 \). Therefore the problem demands the value of f(0).

Step 1 ( Finding the f(0) ):

  • \( x=0, y=0 \rightarrow ( f(2f(0)) = 0 \).
  • \( x=0, y = 2f(0) \rightarrow ( f(f(0)) = 0 \).
  • \( x= f(0) , y = f(0) \rightarrow f(0) = 0 \).
  • \( f(0) =0 \rightarrow f(f(x)) = 0 \).

Step 2 ( Finding the f(x) ):

May be the conditiosn tell us that f(x) = 0 is a strong candidate for the solution. We will try to use that sixth sense. Assume that f(x) is not identically 0.

  • \( y = f(z) \rightarrow xf(z)f( x + f(z)) = 0 \). Added an extra dimension to make the f(x) avaiable.
  • Observe we have an equation in two variables. We will de-dimensionalize it by putting z = appropriate constant.
  • Take \( z_0 : f(z_0) \neq 0, \rightarrow xz_0f(x + f(z_0)) = 0\).
  • We know that \( f(f(z_0)) = 0\). Now \( x = y – f(z_0) \rightarrow (y – f(z_0))z_0f(y) = 0 \rightarrow f(y) = 0 \forall y \neq z_0\).
  • Hence,\((y) = 0 \forall y \in [{0, \infty}) \).

Food for Thought:

  • What if we remove 0 from the function’s domain and range?
  • Find all functions f : \( ({0, \infty}) \rightarrow ({0, \infty}) \) such that for all x, y non-negative, f satisfies \( f( f(x) + f(y)) = xyf(x+y) \).

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