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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.

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  • $\begingroup$ What number of problem is that? That book has around 3,200 problems...0 $\endgroup$ – DonAntonio Apr 5 at 20:00
  • $\begingroup$ @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 $\endgroup$ – Yan Kardziyaka Apr 5 at 20:13
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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 : $$f(x) = ax^2 + bx + c$$

A (probable) alternative approach :

For $y=0$ the equation holds. For $y \neq 0$ rewrite as :

$$\frac{f(y+x) - f(x)}{y} = f'\left(x + \frac{y}{2}\right)$$

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