# Functional equation in single-variable calculus

Suppose we know that $$f(x) \in C^2$$ and $$f(x)$$ defined for all real numbers. Furthermore, $$f(x)$$ has following property: $$\forall x,y \in \Bbb R \quad f(x+y) - f(x) = yf'(x + \frac{y}{2})$$ How to proof that $$f(x) = ax^2+bx +c \quad (a,b,c - const)$$?

Note: according to the textbook from which I get this task (B.P. Demidovich's Problems in mathematical analysis), it can be proved without using multivariable calculus and integration.

• What number of problem is that? That book has around 3,200 problems...0 – DonAntonio Apr 5 at 20:00
• @DonAntonio may be I have wrongly named the book, because at least two different calculus textbooks exist by B.P. Demidovich, but there are 4460 problems in book that I have :) If it is still important, it is problem #1246.2 – Yan Kardziyaka Apr 5 at 20:13

Differentiating with respect to $$y$$, we get :
$$f'(x+y) = f'(x+y/2) + y/2 f''(x+y/2)$$
We are able to carry out such a differentiation since $$f(x) \in C^2$$. I leave the simple $$'$$ notation since differentiation is symmetric, just work with $$y$$ for that first step and treat $$x$$ as constant. Continuing, set $$y=-2x$$ and yield the expression : $$f'(-x) = f'(0) - x f''(0) \xrightarrow{x := -x} f'(x) = f'(0)+x f''(0)\implies f(x) = c + f'(0) x + \frac{1}{2} f''(0) x^2$$ Now, let $$a= \frac{1}{2}f''(0)$$ and $$b=f'(0)$$. We have proven the desired expression : $$f(x) = ax^2 + bx + c$$
For $$y=0$$ the equation holds. For $$y \neq 0$$ rewrite as :
$$\frac{f(y+x) - f(x)}{y} = f'\left(x + \frac{y}{2}\right)$$