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How can we find $F$ satisfying: exists a $c$ such that $$F(x,x)+F(x,c-x)-F(c-x,x)-F(c-x,c-x)=0 \text{ for all } x,y $$

Several quadratic polynomials in $x,y$ satisfy the above property. I'm trying to generate more examples of functions that satisfy it. I will be very grateful for any other examples.

The property says there is one $c$ that works for all $x$ but given $F$ we have freedom to find a $c$ that works. For example:

  1. If $F(x,y)=Ax^2+By^2+Cxy+Dx+Ey+F$ and $A+C\neq 0$ we set $c=\frac{-2D}{A+C}$ and one can easily verify $F$ satisfy the functional equation.
  2. Also if $F(x,y)=(A+By)x^2+Cy^2+Dxy+Ex+Fy+H$ there are two values of $c$.
  3. As Jeb pointed in the comments if $F(x,y)$ is even in $x$ and also even in $y$ then $c=0$ works.

My first idea was to try to transform the equation into a PDE by differentiating wrt $x$ but we do not get a PDE because we are evaluating $F$ at different points: $x$ and $c-x$. Honestly, I'm not even sure of what tags to use.

Martin Sleziak
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Sergio Parreiras
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3 Answers3

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Not sure if this is what you’re looking for, but here is a very general construction (in fact it is a full characterization of all solutions). Partition ${\mathbb R}^2$ as follows : ${\mathbb R}^2=A \cup B_- \cup B_+ \cup C$ where

$$ \begin{array}{lcl} A &=& \lbrace (x,c-x) | x\in{\mathbb R} \rbrace \\ B_{-} &=& \lbrace (x,x) | x < \frac{c}{2} \rbrace \\ B_{+} &=& \lbrace (x,x) | x > \frac{c}{2} \rbrace \\ C &=& {\mathbb R}^2 \setminus (A \cup B_- \cup B_+) \end{array} $$

So $A$ is a full straight line, and $B_-$ and $B_+$ are half-lines.

Let $f:{\mathbb R}\to \mathbb R$ and $g:(-\infty,\frac{c}{2}) \to \mathbb R$ be maps. Then, put

$$ F(x,y)=\left\lbrace\begin{array}{lcl} f(x) &\text{if} & (x,y)\in A \\ g(x) &\text{if} & (x,y)\in B_{-} \\ g(c-x)+f(c-x)-f(x) &\text{if} & (x,y)\in B_{+} \\ \text{anything you like } &\text{if} & (x,y)\in C \\ \end{array}\right. $$

This $F$ will satisfy your functional equation (and in fact any solution is of this form).

Ewan Delanoy
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For any function, $F(x,y)$, define the function $g(x,c)$ satisfying

$$ g(x,c) = F(x,x)+F(x,c-x)-F(c-x,x)-F(c-x,c-x) $$

We therefore require that $\exists c$ s.t. $\forall x,\ g(x,c)=0$.

If we let $x=z+\frac{c}2$ and $\hat F(x,y)=F(x+\frac{c}2,y+\frac{c}2)$, then we have $$ \hat g(z,c)=g(z+\frac{c}2,c)=\hat F(z,z)+\hat F(z,-z)-\hat F(-z,z)-\hat F(-z,-z) $$ Therefore, without loss of generality, we can consider the case of $c=0$, for which we require that $\hat g(z,0)=0$. In this case, we may split the even and odd components of our function in each of the two dimensions, thus giving us $$ \hat F(x,y) = A(x,y)+B(x,y)+C(x,y)+D(x,y) $$ satisfying $$A(x,y)=A(-x,y)=A(x,-y)\\B(x,y)=-B(-x,y)=B(x,-y)\\C(x,y)=C(-x,y)=-C(x,-y)\\D(x,y)=-D(-x,y)=-D(x,-y)$$ From which (letting $\hat g(x,c)=\hat g(x,0)=G(x)$), we may determine that $$\begin{align} G(x)=&A(x,x)+A(x,-x)-A(-x,x)-A(-x,-x)\\&+B(x,x)+B(x,-x)-B(-x,x)-B(-x,-x)\\&+C(x,x)+C(x,-x)-C(-x,x)-C(-x,-x)\\&+D(x,x)+D(x,-x)-D(-x,x)-D(-x,-x)\\ =&4B(x,x) \end{align}$$ Therefore, the requirement is that $B(x,x)=0$. Recall that $B(x,y)$ represents the component of $\hat F(x,y)$ that is even in $y$ and odd in $x$. This provides us with a "natural" method of creating a solution. Given a function $\hat F(x,y)$, we can find $B(x,y)$ as $$ B(x,y)=\frac{\hat F(x,y)-\hat F(-x,y)+\hat F(x,-y)-\hat F(-x,-y)}4 $$ and therefore, we may define a modified function $$ \tilde F(x,y)=\begin{cases}\frac{3\hat F(x,x)+\hat F(-x,x)-\hat F(x,-x)+\hat F(-x,-x)}4 & x=y\\ \hat F(x,y) & \text{otherwise} \end{cases} $$ Note that this leads to a result identical to that given by Ewan Delanoy, it is merely expressed in a different manner. For a more "natural" solution, we can require that $B(x,y)=0$, and thus define $$ \tilde F(x,y) = \frac{3\hat F(x,y)+\hat F(-x,y)-\hat F(x,-y)+\hat F(-x,-y)}4 $$ From here, we may choose any $c$, and shift the function so that $$ F(x,y)=\tilde F\left(x-\frac{c}2,y-\frac{c}2\right) $$ and we have a solution for a given $c$. Notice that, for $$ F(x,y)=Ax^2+By^2+Cxy+Dx+Ey+F $$ with $c=\frac{-2D}{2A+C}$ we have $$ \hat F(x,y) = Ax^2+By^2+Cxy-\frac{2BD-CD+E(2A+C)}{2A+C}y+\frac{C^2F+4ACF+4A^2F−CDE−2ADE+BD^2−AD^2}{(2A+C)^2} $$ which, you can see, does not have the $x$ term, which is the only term in a bivariate quadratic polynomial that is even in $y$ and odd in $x$. Note the $2A+C$ on the denominator of $c$ - the value given in the question is incorrect.

Glen O
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  • If you want an acknowledgement in an academic paper, please send your complete email to my UNC email address (publicly available). Thanks again for your help. – Sergio Parreiras Jun 05 '14 at 22:26
  • Thank you for the offer, but it is unnecessary. I'm glad I could help. – Glen O Jun 06 '14 at 03:10
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Do you mean to replace the second $x$ with $y$? On second glance this only seems true for linear functions. Take $F(x,y) = xy$ , you'll end up with $c^2$ term that will not vanish.

Jeb
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    Thanks! I corrected not all quadratic polynomials but second glance not correct: let $F(x,y)=Ax^2+By^2+Cxy+Dx+Ey+F$ and set $c=\frac{-2D}{A+C}$ and it will satisfy the property. In the case $F(x,y)=xy$ it works with $c=0$. Notice the property requires that exist some $c$ for all $x$ and $y$. – Sergio Parreiras Apr 15 '14 at 23:13
  • You should impose some restriction on what $c$ should be. $c=0$ will cause the above statement to be trivial for any function. – Jeb Apr 16 '14 at 00:11
  • c=0 implies only F(x,x)+F(x,-x)=F(-x,x)+F(-x,-x) this is not trivial. Not all functions satisfy it. – Sergio Parreiras Apr 16 '14 at 00:45
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    Sorry, you're right! I forgot about the sign change. In any case, we now know that any even function(in $x$ or $y$)works. – Jeb Apr 16 '14 at 00:53
  • you mean any even function in $x$ and in $y$ that is $F(x,y)=F(-x,y)$ and $F(x,y)=F(x,-y)$, right? If it is even only in $x$ (or only in $y$) I don't see how it would work. – Sergio Parreiras Apr 16 '14 at 14:15