Problem Determine all functions $f$ such that $$f(2x)-2f(x+y)+f(2y)=(x-y)^2$$
where $f$ is continuous and defined for all $x,y\in{\mathbb{R}}$.
I've found a particular solution:
$$f(x)= \frac{x^2}{2} + bx + c $$
where $b,c\in{\mathbb{R}}$.
Are there more solutions, or is that the only one and if so how do I prove it? I have tried doing it like this:
Suppose there exists another function $g(x)$ that satisfies $g(2x)-2g(x+y)+g(2y)=(x-y)^2$ and that $f(x)=\frac{x^2}{2} + bx + c $. Then
$$f(2x)-2f(x+y)+f(2y)-(g(2x)-2g(x+y)+g(2y)) = 2x^2+2bx+c-(x^2+2xy+y^2+2bx+2by+2c)-(g(2x)-2g(x+y)+g(2y))=x^2-2xy+y^2-(x-y)^2=0.$$
But that feels weird and kind of circular.
Best Answer
This equation has a unique smooth solution, just by the continuity assumption of $f$.
Starting from your last step, defining the function $g(x)=f(x)-x^2/2$ and substituting in the functional equation yields the homogeneous linear equation
$$g(2x)-2g(x+y)+g(2y)=0$$
Setting $x\to x/2~,~ y\to x/2+h/2$ and rearranging leads to the equivalent
$$g(x+h)-g(x+h/2)=g(x+h/2)-g(x)$$
We notice that the function $G_h(x)=g(x+h/2)-g(x)$ is by our hypothesis continuous and satisfies $G_h(x+h/2)=G_h(x)$. Fixing any $h\in\mathbb{R}$, we find that $G_h(x)$ is periodic for any given value of $h$, with period $T=h/2$. Furthermore, it is easy to show that with $a\in \mathbb{R}$ $$G_h(x+a/2)=G_{a+h}(x)-G_a(x)$$ and then by setting $a=h$ one can also show importantly that $$G_{2h}(x)=2G_h(x)$$
By iterating this relation we can show that $G_h(x)=2^nG_{h/2^{n}}(x)$ and we observe $G$ is continuous with arbitrarily small period and is therefore constant: $G_h(x)=G_h(0)$ (to show this, show $f$ is bounded everywhere by its values on an appropriate interval around the origin; shrink the interval and in the limit upper and lower bounds tend to the value of the function at $x=0$ proving constancy).
We can write henceforth that $G_h(x)=C(h)$, where $C(h)$ is x-independent. Note that $C(0)=0$. Substituting into the 3rd equation we find that the continuous (due to the continuity of $g$) function $C$ obeys Cauchy's equation
$$C(a+h)=C(a)+C(h)$$
and is therefore linear: $C(h)=-Ah/2$. With this we can write that
$$g(x+h/2)-g(x)=-Ah/2$$
Setting $\Omega(x)=g(x)-Ax$, we find that
$$\Omega(x+h/2)=\Omega(x)$$
from which we finally arrive to the conclusion that $\Omega(x)=B$ is constant and hence the unique solution is
$$g(x)=Ax+B\Rightarrow f(x)=x^2/2+Ax+B$$