Function-3

Upon observation, this functional equation resembles the famed Cauchy Equation (which admits linear functions as its solution). If $f(x)=px+q$ , then we have:

$2(px+q) + 2(py+q) = p(x+y)+q + p(x-y)+ q + 2y$ ;

or $2px + 2py + 4q = 2px + 2y + 2q$ ;

or $2py + 4q = 2y + 2q \Rightarrow p = 1, q = 0$

which gives us the unique linear function $\boxed{f(x) = x}.$

But were not done yet! If we have $x=y=0$ , then we obtain $2f(0) + 2f(0) = f(0)+f(0) + 2(0) \Rightarrow f(0)=0.$ Now taking a leap-of-faith, let's assume $f(x)$ is a differentiable function over all $x,y \in \mathbb{R}$ . Differentiating the above functional equation wrt $x$ and $y$ yields:

$2f'(x) = f'(x+y)+f'(x-y)$ (i)

$2f'(y) = f'(x+y)-f'(x-y) + 2$ (ii)

and adding (i) and (ii) next produces:

$2f'(x) + 2f'(y) = 2f'(x+y) + 2 \Rightarrow f'(x)+f'(y)=f'(x+y)+1$ (iii)

followed by differentiating (iii) wrt $x$ and $y$ :

$f''(x) = f''(x+y)$ (iv)

$f''(y) = f''(x+y)$ (v)

which leads to $f''(x) = f''(y) = A$ , and after integrating twice: $f(x) = \frac{A}{2}x^2 + Bx + C$ (where $A,B,C$ are arbitrary real constants). We know $f(0)=0 \Rightarrow C = 0,$ or $f(x) = \frac{A}{2}x^2 + Bx$ (vi). If we now substitute (vi) into the original functional equation, we obtain:

$2(\frac{A}{2}x^2 + Bx) + 2(\frac{A}{2}y^2 + By) = \frac{A}{2}(x+y)^2 + B(x+y) + \frac{A}{2}(x-y)^2 + B(x-y) + 2y$ ;

which expands & simplifies into:

$Ax^2 + Ay^2 + 2Bx + 2By = Ax^2 + Ay^2 + 2Bx + 2y \Rightarrow A = k, B = 1$

where $k \in \mathbb{R}$ . This gives the infinite family of functions $\boxed{f(x) = kx^2 + x}$ , which includes our above linear function (when $k = 0$ ), as the solution to the original functional equation.