# 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 \( f( f(x) + f(y)) = xyf(x+y) \).