Goofy FE problem

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Goofy FE problem

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#1 h Apr 7, 2025, 8:42 PM

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Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+f(y)) = y+f(x)$ over $\mathbb{R}$ .

This post has been edited 1 time. Last edited by Bread10, Apr 7, 2025, 8:44 PM
Reason: improvement

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#2 h Apr 7, 2025, 9:12 PM

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move to hso pls
clearly $f$ is surjective, swapping $x,y$ gives $f(y+f(x)) = x+f(y)$ so $f(f(x+f(y)) = x+f(y)$ for all $y.$ by varying $x$ we get $f(f(x)) = x.$ now let $y = f(t)$ to get $f(x+t) = f(x) + f(t)$ so $f$ is an additive involution. its easy to check that all such functions work so we are done

@2below the answer is all functions $f$ such that $f(x+y) = f(x) + f(y)$ for $x,y \in \mathbb{R}$ and $f(f(x))$ for all $x \in \mathbb{R}.$ there exist pathological functions $f$ which satisfy these two conditions

This post has been edited 1 time. Last edited by zoinkers, Apr 7, 2025, 9:17 PM

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#3 h Apr 7, 2025, 9:15 PM

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Here is what I have so far:

Claim: f is injective(f(a) = f(b) implies a = b and likewise)

Plug in x = 0 to get f(f(y)) = y + f(0).

Then we get f(f(y)) - f(0) = y.

Let f(a) = f(b) for real numbers a,b.
Now plug in y = a and y = b to obtain f(f(a)) - f(0) = a, and f(f(b)) - f(0) = b.

The LHS of both equations are equal, so a=b.

Now plug in x = y = 0. We obtain f(f(0)) = f(0). Since f is injective, we can remove the outside f from both sides to get f(0) = 0.

Now plug in x = 0 to get f(f(y)) = y. This is an involution, implying that f is bijective.

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#4 h Apr 7, 2025, 9:16 PM

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zoinkers wrote:

move to hso pls
clearly $f$ is surjective, swapping $x,y$ gives $f(y+f(x)) = x+f(y)$ so $f(f(x+f(y)) = x+f(y)$ for all $y.$ by varying $x$ we get $f(f(x)) = x.$ now let $y = f(t)$ to get $f(x+t) = f(x) + f(t)$ so $f$ is an additive involution. its easy to check that all such functions work so we are done

What exactly are the functions here? Like what's your final answer.

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#5 h Apr 7, 2025, 10:54 PM

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I didn't find this exact problem, but it's probably already posted on aops so look for it ig
bad solution

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