Proving the uniqueness of solution of a functional equation.

Question

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.

Answer 1

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

$$g(x)=Ax+B\Rightarrow f(x)=x^2/2+Ax+B$$

Answer 2

I found a direct solution, (thanks to your previous analysis).

We put $$\forall x,y \in \mathbb{R}, \qquad f(2x) - 2 f(x+y) + 2f(y) = (x-y)^2 \qquad (E)$$ Of course this makes us think of the properties of the square function.

We can see that for $x \in \mathbb{R}$, $h \neq 0$ $$\begin{align*} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} &= \frac{f(2\frac{x+h}{2}) -2f(\frac{x+h}{2}+\frac{x-h}{2})+2f(\frac{x-h}{2})}{h^2}\\ & = \frac{\left(\frac{x+h}{2} - \frac{x-h}{2}\right)^2}{h^2} = \frac{h^2}{h^2} \\ &= 1 \end{align*}$$

So $f$ is two times derivable and $\forall x \in \mathbb{R}$, $f''(x) = 1$, which means that there exists $b,c \in \mathbb{R}$ such that $\forall x \in \mathbb{R}, \, f(x) = \frac{x^2}{2} + bx + c$.

Since you already checked that these solutions are valid, we can conclude that $$S(E) = \left\{x \mapsto \frac{x^2}{2} + bx + c, \, b,c \in \mathbb{R}\right\}.$$