A nested functional equation: getting from involution to identity

frostbyte_93·Aug 3, 2025(edited)

Putting x=0 suggests

f(f(y))=y

, but only if I can cancel f(y). I'm worried that this discards a whole branch of solutions. Problem: Let

\mathbb{R}

denote the set of real numbers. Find all functions

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

such that

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

for all

x, y \in \mathbb{R}

. Source: Baltic Way 2011 Problem Shortlist · MNET-0185

Open MNET-0185 in the question bank →

7 replies

respawn_pending·Aug 5, 2025(edited)

First handle f(0)=c. At x=y=0,

cf(c)=0

. If c≠0, take y=c and x=c−t; since f(c)=0, the equation gives f(t)=t−c. At t=0 that contradicts c≠0. Thus f(0)=0.

frostbyte_93·Aug 5, 2025

And if some nonzero y0 has f(y0)=0, the equation becomes

y_0f(y_0-x)=0

for every x, so f is identically zero. That branch really does have to be separated.

respawn_pending·Aug 5, 2025

For the remaining branch, only zero maps to zero. Now x=0 gives the involution. Rewrite the equation as

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

shulk_said_no·Aug 5, 2025

Set x=t, y=f(t)−t. If f(t)≠t, the right side vanishes and involution gives

f(t)-f(-t)=2t

. If f(t)=t but f(-t)≠−t, use the same argument at −t to get the identity again. If both are fixed, it is immediate.

frostbyte_93·Aug 6, 2025

f(t)-f(-t)=2t

holds for every t, without assuming oddness.

respawn_pending·Aug 6, 2025

Next x=−y gives

yf(2y)=y^2+f(y)^2

. Apply this at −y too and add. Using the difference identity twice gives

4y^2=2y^2+f(y)^2+(f(y)-2y)^2

. Hence

(f(y)-y)^2=0

frostbyte_93·Aug 6, 2025

That leaves f=0 and f(x)=x. Substituting them into the original equation works. The zero branch was exactly the issue I would have lost by canceling at the start.

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A nested functional equation: getting from involution to identity

frostbyte_93·Aug 3, 2025(edited)

Putting x=0 suggests

f(f(y))=y

, but only if I can cancel f(y). I'm worried that this discards a whole branch of solutions. Problem: Let

\mathbb{R}

denote the set of real numbers. Find all functions

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

such that

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

for all

x, y \in \mathbb{R}

. Source: Baltic Way 2011 Problem Shortlist · MNET-0185

Open MNET-0185 in the question bank →

7 replies

respawn_pending·Aug 5, 2025(edited)

First handle f(0)=c. At x=y=0,

cf(c)=0

. If c≠0, take y=c and x=c−t; since f(c)=0, the equation gives f(t)=t−c. At t=0 that contradicts c≠0. Thus f(0)=0.

frostbyte_93·Aug 5, 2025

And if some nonzero y0 has f(y0)=0, the equation becomes

y_0f(y_0-x)=0

for every x, so f is identically zero. That branch really does have to be separated.

respawn_pending·Aug 5, 2025

For the remaining branch, only zero maps to zero. Now x=0 gives the involution. Rewrite the equation as

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

shulk_said_no·Aug 5, 2025

Set x=t, y=f(t)−t. If f(t)≠t, the right side vanishes and involution gives

f(t)-f(-t)=2t

. If f(t)=t but f(-t)≠−t, use the same argument at −t to get the identity again. If both are fixed, it is immediate.

frostbyte_93·Aug 6, 2025

f(t)-f(-t)=2t

holds for every t, without assuming oddness.

respawn_pending·Aug 6, 2025

Next x=−y gives

yf(2y)=y^2+f(y)^2

. Apply this at −y too and add. Using the difference identity twice gives

4y^2=2y^2+f(y)^2+(f(y)-2y)^2

. Hence

(f(y)-y)^2=0

frostbyte_93·Aug 6, 2025

That leaves f=0 and f(x)=x. Substituting them into the original equation works. The zero branch was exactly the issue I would have lost by canceling at the start.