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Find all differentiable functions $f:R \to R$ that have property that

$$ f'\left(\frac{y+z}2\right) = \frac{f(y)-f(z)}{y-z}$$ for all $y \neq z $

I have not been able to get any strong result. The only necessary condition I have managed to get is $f(x) = f(-x) + 2xf'(0)$. But, clearly this is not sufficient as $f(x)= x^4$ is a counter-example. In a more general form we can write it as $f(c+x)=f(c-x)+ 2xf'(c)$ by putting $y=c+x$ and $z= c-x$.

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Consider the last equation you wrote. By differentiating with respect to $x$, we have $$ f'(c+x) = -f'(c-x) + 2f'(c) $$ and differentiating with respect to $c$ gives $$ f'(c+x) = f'(c-x) + 2xf''(c). $$ If we put $c=x$ and subtract the two equations, we have $$ f'(x)-xf''(x) = f'(0) $$ for all $x\in \mathbb{R}$. Now consider the function $$ g(x) = \frac{f'(x)}{x} $$ for $x\neq 0$ and we can show that $g'(x) = -f'(0)/x^{2}$, so $f'(x) = f'(0) + Ax$ and $f(x) = B + f'(0)x + \frac{1}{2}Ax^{2}$ for some $A, B\in \mathbb{R}$. Hence $f$ is a polynomial with degree at most $2$ and you may found the constants $A$ and $B$.

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    $\begingroup$ @Martin He has eliminated f'(c+x) from first two equations. $\endgroup$ – Lord KK Jun 4 '18 at 10:55
  • $\begingroup$ Please describe what you did, e.g., by adding something like 'and subtract the two equations' just before the third equation. $\endgroup$ – CiaPan Jun 4 '18 at 10:59
  • $\begingroup$ @CiaPan I edited. $\endgroup$ – Seewoo Lee Jun 4 '18 at 11:10
  • $\begingroup$ My bad :) $\ \ $ $\endgroup$ – Martin Argerami Jun 4 '18 at 11:22
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    $\begingroup$ How do you know that $f'(c)$ is differentiable? $\endgroup$ – Somos Jun 4 '18 at 14:27
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Making $z = 0$

$$ f'\left(\frac{y}{2}\right) = \frac{f(y)-f(0)}{y} $$

or

$$ f'\left(y\right) = \frac{f(2y)-f(0)}{2y} $$

or

$$ 2y f'\left(y\right) = f(2y)-c_0 $$

now assuming that $f \in C^{\infty}$ and making $f(y) = \sum_{k=0}^{\infty} a_k y^k$ we have

$$ 2y\sum_{k=1}^{\infty}k a_k y^{k-1} = \sum_{k=0}^{\infty}a_k 2^k y^k-c_0 $$

hence $a_0 + c_0= a_3 = a_4 = \cdots = 0$

and finally

$$ f(y) = c_0+a_1 y + a_2 y^2 $$

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    $\begingroup$ In the question it was not given that function is infinitely differentiable. $\endgroup$ – Lord KK Jun 4 '18 at 13:54
  • $\begingroup$ Your first equation is wrong. Making $z=0$ you get $f'(y/2) = (f(y)-f(0))/y.$ In fact $f(y ) = a_0 + a_1 y + a_2 y^2$ is a solution. $\endgroup$ – Somos Jun 4 '18 at 14:20
  • $\begingroup$ You are right. I commit a little distraction. I will fix the answer. Thanks. $\endgroup$ – Cesareo Jun 4 '18 at 15:54
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In the paper

J. Aczél, A mean-value property of the derivative of quadratic polynomials - without meanvalues and derivatives, Math. Mag.58(1985), no. 1, 42–45

it is shown that $g(\frac{y+z}2) = \frac{f(y)-f(z)}{y-z}$ implies, without assumiñg any regularity, that $f$ is a quadratic function and $g=f'$.

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