# Functional Equation

For all real numbers $x,y,z$ we have: \begin{align} H(x,y,z)=&f(x,y+z) + g(y,z) \\=& h(y,x+z)+ j(x,z) \end{align} All functions f,g,h,j are continuous (or even continuously differentiable). I originally suspected $H(x,y,z)= \phi_1(x)+\phi_2(y)+\phi_3(z)+\phi_4(x+y+z)$ to be the solution. However I cannot make any progress using the usual methods to solve Cauchy-like and transitivity equations. However, I also could not find any obvious counterexamples.

• What is $\hat{f}$? – Paul Feb 1 '17 at 11:59
• an arbitrary function, i will edit to make it more clear – HRSE Feb 2 '17 at 4:13

We differentiate with respect to $x$ on both sides to obtain: \begin{align} f_1(x,y+z)=h_2(y,x+z) + j_1(x,z) \end{align} Setting $x=0$ gives us: \begin{align} f_1(0,y+z)=h_2(y,z) + j_1(0,z) \end{align} Forming antiderivatives with respect to $z$ on both sides gives us: \begin{align} h(y,z) = \phi(z) + \psi(y) + \chi(y+z) \end{align} where $\phi(z)=\int j_1(0,z) dz$, $\chi(y+z)=\int f_1(0,y+z)dz$, and $\psi(y)$ is an arbitrary continuously differentiable function. Solving in the same manner for $j(x,z)$ reduces our functional equation to an equation of the form: \begin{align} f(x,y+z) = h(y,x+z)+ u(x) + v(y) + w(x+z) + v(y+z) \end{align} which can be solved using standard methods.
Note: The above method (transform equation, set x=0, undo transformation) also seems to work for functions which are only integrable. Instead of differentiating with respect to x, one can use the Laplace transforms with respect to x and z, set the transformed variable corresponding to $x$ to 0, and undo the transformation with respect to $z$. However, there are quite a few details which have to be verified for this to work.