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Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that $f(x,y)+f(y,z)+f(z,x) = 0$ for all real numbers $x, y$, and $z$. Prove that there exists a function $g : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x,y) = g(x)−g(y)$ for all real numbers $x$ and $y$.

Attempt: Can we just set $z = 0$ and solve for $f(x,y)$?

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    $\begingroup$ I appreciate vadim123's feedback , but I still don't understand my solution is wrong. Can someone provide more insight about this problem, so I can see it from a different angle? $\endgroup$ Dec 8, 2017 at 6:57
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    $\begingroup$ x,y, and z are not constants to be solved. This will be true for all possible values of x,y, and z. $\endgroup$
    – fleablood
    Dec 8, 2017 at 7:13
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    $\begingroup$ So it is true that $f(x,y) = -f(0,x) - f(y,0)$ for all $x,y$ but how can you set $g(x) = -f(x,0)$ and $g(y) = f(y,0)$? If $g(x) = -f(x,0)$ then $g(y) =-f(y,0) \ne f(0,y)$..... unless it does..... $\endgroup$
    – fleablood
    Dec 8, 2017 at 7:21
  • $\begingroup$ if $g(y) = -f(y,0)$ then why does this entail that $g(y) = f(0,y)$. How do they relate? $\endgroup$ Dec 9, 2017 at 5:04
  • $\begingroup$ This is the step I can't understand $g(y)=−f(y,0)≠f(0,y)$ .. $\endgroup$ Dec 9, 2017 at 5:09

2 Answers 2

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Hints:

  1. Prove that $f(x,x)=0$ for all $x$.

  2. Prove that $f(x,y)=-f(y,x)$ for all $x,y$.

  3. Set $g(x)=f(0,x)$, and then apply the initial attempt (since modified) to prove that $f(x,y)=g(x)-g(y)$.

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  • $\begingroup$ Here is my erroneous thinking, so you can fix it . If $f(x,y) = x^2 +y$ , so $f(x,0) = x^2$ which implies that $f(x,0) = g(x)$ because f(x,0) just depends on x. $\endgroup$ Dec 8, 2017 at 5:58
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    $\begingroup$ @Dan, this function you give doesn't satisfy the property. Just try following the hints. $\endgroup$
    – vadim123
    Dec 8, 2017 at 6:02
  • $\begingroup$ But the thing is I don't understand why I have to do what you are telling me to do, so I still can't go with your approach until I understand why mine is wrong. $\endgroup$ Dec 8, 2017 at 6:03
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    $\begingroup$ Your approach is wrong, because you do not have a proof structure leading from the hypotheses to the conclusion, rather some calculations. $\endgroup$
    – vadim123
    Dec 8, 2017 at 6:05
  • $\begingroup$ 3f(x,x) = 0 , so f(x,x) = 0 and for 2 we just set (x,y,z) = (x,y,x); however, I still don't understand why I am doing this. $\endgroup$ Dec 8, 2017 at 6:14
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This will not be true in general but it will be true for this unusual function that has the unusual property that:

$f(x,y) + f(y,z) + f(z,x) = 0$.

This means that $f(x,y) = -f(z,x)- f(y,z) $.

This is true for all $z$ so if we could define $g(w) = -f(c,w)$ and $g(w) = f(w,z)$ for some constant $c$ we would be done. (As $g(x) - g(y) = -f(c,x) -f(y,c)= f(x,y)$.)

But that would require that there be some constant $c$ so that $-f(c,w) =f(w,c)$ for all $w$ and that is not true in general.

But, $f$ is not a usual function. Maybe this IS true for $f$.

Can we prove that $f(w,z) = -f(z,w)$ for all $w,z$.?

Let $x = y = w$: Then $f(w,w) + f(w,z)+ f(z,w) = 0$ so $f(w,z) = -f(z,w) - f(w,w)$.

Shoot! That was so close but $-f(z,w) - f(w,w) \ne -f(z,w)$ unless $f(w,w) = 0$ and that is not true in general.

But, again, $f$ is not a general function. Maybe $f(w,w)$ does equal $0$ for all $w$.

Let $x = y = z = w$: Then $f(w,w) + f(w,w) + f(w,w) = 0$ so $f(w,w) = 0$.

That's it! We are done.

.....

If we set $g(w) = f(w,c)$ for some constant $c$ we get:

$f(x,y) + f(y,c) + f(c,x) = 0$ so

$f(x,y) = -f(c,x) - f(y,c) = -f(c,x) - g(y)$.

Now $f(c,x)+ f(x,x) + f(x,c) = 0$ so

$f(c,x) = -f(x,c) - f(x,x) = -g(x) - f(x,x)$ so

$f(x,y) = -f(c,x) - g(y) = g(x) + f(x,x) - g(y)$.

Now $f(x,x) + f(x,x) + f(x,x) = 0$ so $f(x,x) = 0$ so

$f(x,y) = g(x) +f(x,x) - g(y) = g(x) - 0 - g(y) = g(x) - g(y)$.

We are done.

Or for a third time:

$f(x,y) = -f(y,c) - f(c,x) $

$=-g(y) - (0 - f(x,x) - f(x,c))$

$= -g(y) + g(x) + f(x,x) $

$=g(x)-g(y) + \frac 13(f(x,x) + f(x,x) + f(x,x))$

$=g(x) - g(y) + \frac 13(0)$

$= g(x)-g(y)$.

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