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A real function $f(x,y)$ on $R^2$ satisfies $f(x+y,y) = f(x,0)+qy$ for some real number $q$. What form should $f$ assume without $f$ being continuous? Is the linear solution $f(x,y)=ax+(q-a)y$ for some constant real number $a$ the unique continuous or differentiable solution?

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  • $\begingroup$ Let $g$ be an arbitrary function on $\mathbb{R}$. Then $f(x,y)=g(x-y)+qy$ is, if I am not wrong, a solution. $\endgroup$ – Kelenner Sep 8 '15 at 8:02
  • $\begingroup$ @Kelenner: You are right. Could you explain a bit how you came up with this particular solution? What is the most general solution then? $\endgroup$ – Hans Sep 8 '15 at 8:19
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    $\begingroup$ If $f$ is a solution, put $g(x)=f(x,0)$; Then replace in the equation $x$ by $x-y$. You get $f(x,y)=g(x-y)+qy$. So this is the general solution. $\endgroup$ – Kelenner Sep 8 '15 at 8:23
  • $\begingroup$ @Kelenner: I actually got $f(x,y)=f(x-y,0)+qy$ but did not go further and realize $f(x,0)$ could just be taken to be an arbitrary function. Thank you. $\endgroup$ – Hans Sep 8 '15 at 16:40
  • $\begingroup$ @Kelenner please write down your correct answer, and close this question with approve the answer by Hans $\endgroup$ – user48941 Sep 8 '15 at 19:28
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a) If $f$ is a solution, put $g(x)=f(x,0)$; Then replace in the equation $x$ by $x−y$. You get $f(x,y)=g(x−y)+qy$.

b) Now let $g$ be an arbitrary function on $\mathbb{R}$. Put $f(x,y)=g(x-y)+qy$. It is easy to see that $f$ is a solution.

c) Hence the general solution of the functional equation is $f(x,y)=g(x-y)+qy$, with $g$ an arbitrary function.

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