Composing a scaling functional equation in two different orders

hollowbean·Jun 25, 2026(edited)

I tried linear and quadratic functions and got contradictory guesses for a. Is there a substitution that eliminates f before we start guessing examples? Problem: Find all real numbers

a

for which there exists a non-constant function

f: \mathbb{R} \to \mathbb{R}

satisfying the equations

1) \quad f(ax) = a^2 f(x)

2) \quad f(f(x)) = a f(x).

Source: Baltic Way 2016 · MNET-01CW

Open MNET-01CW in the question bank →

3 replies

respawn_pending·Jun 25, 2026(edited)

Compute the triple iterate two ways. One route gives

f(f(f(x)))=a^2f(x)

. Another gives

f(af(x))=a^2f(f(x))=a^3f(x)

hollowbean·Jun 25, 2026(edited)

a^2(1-a)f(x)=0

for all x. Nonconstant f isn't identically zero, giving a=0 or a=1. Does a=0 actually allow a nonconstant function?

respawn_pending·Jun 26, 2026(edited)

Take

f(x)=|x|-x

: it maps negatives to nonnegative values and nonnegatives to zero, so its second iterate is zero and f(0)=0. For a=1, take f(x)=x. Both candidates satisfy both equations.

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Composing a scaling functional equation in two different orders

hollowbean·Jun 25, 2026(edited)

I tried linear and quadratic functions and got contradictory guesses for a. Is there a substitution that eliminates f before we start guessing examples? Problem: Find all real numbers

a

for which there exists a non-constant function

f: \mathbb{R} \to \mathbb{R}

satisfying the equations

1) \quad f(ax) = a^2 f(x)

2) \quad f(f(x)) = a f(x).

Source: Baltic Way 2016 · MNET-01CW

Open MNET-01CW in the question bank →

3 replies

respawn_pending·Jun 25, 2026(edited)

Compute the triple iterate two ways. One route gives

f(f(f(x)))=a^2f(x)

. Another gives

f(af(x))=a^2f(f(x))=a^3f(x)

hollowbean·Jun 25, 2026(edited)

a^2(1-a)f(x)=0

for all x. Nonconstant f isn't identically zero, giving a=0 or a=1. Does a=0 actually allow a nonconstant function?

respawn_pending·Jun 26, 2026(edited)

Take

f(x)=|x|-x

: it maps negatives to nonnegative values and nonnegatives to zero, so its second iterate is zero and f(0)=0. For a=1, take f(x)=x. Both candidates satisfy both equations.