### International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

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### Problem 4

Problem 4. Find all differentiable functions $\displaystyle f:(0,\infty)\to\RR$ such that

$\displaystyle f(b)-f(a)=(b-a)f'\left(\sqrt{ab}\right) \quad \text{for all} \quad a,b>0. \tag2$

(Proposed by Orif Ibrogimov, National University of Uzbekistan)

Solution. First we show that $\displaystyle f$ is infinitely many times differentiable. By substituting $\displaystyle a=\frac12t$ and $\displaystyle b=2t$ in (2),

$\displaystyle f'(t) = \frac{f(2t)-f(\tfrac12t)}{\tfrac32t}. \tag{3}$

Inductively, if $\displaystyle f$ is $\displaystyle k$ times differentiable then the right-hand side of (3) is $\displaystyle k$ times differentiable, so the $\displaystyle f'(t)$ on the left-hand-side is $\displaystyle k$ times differentiable as well; hence $\displaystyle f$ is $\displaystyle k+1$ times differentiable.

Now substitute $\displaystyle b=e^ht$ and $\displaystyle a=e^{-h}t$ in (2), differentiate three times with respect to $\displaystyle h$ then take limits with $\displaystyle h\to0$:

$\displaystyle f(e^ht)-f(e^{-h}t) - (e^ht-e^{-h}t)f(t) = 0$

$\displaystyle \left(\frac\partial{\partial h}\right)^3\Big( f(e^ht)-f(e^{-h}t) - (e^ht-e^{-h}t)f(t) \Big) = 0$

$\displaystyle e^{3h}t^3f'''(e^ht)+3e^{2h}t^2f''(e^ht)+e^{h}tf'(e^ht)+ e^{-3h}t^3f'''(e^{-h}t)+3e^{-2h}t^2f''(e^{-h}t)+e^{-h}tf'(e^{-h}t) - \qquad\qquad\qquad$

$\displaystyle \qquad\qquad\qquad\qquad - (e^ht+e^{-h}t)f'(t) = 0$

$\displaystyle 2t^3f'''(t) + 6t^2f''(t) = 0$

$\displaystyle tf'''(t)+3f''(t) = 0$

$\displaystyle (t\,f(t))''' = 0.$

Consequently, $\displaystyle tf(t)$ is an at most quadratic polynomial of $\displaystyle t$, and therefore

$\displaystyle f(t) = C_1t+\frac{C_2}{t}+C_3 \tag{4}$

with some constants $\displaystyle C_1$, $\displaystyle C_2$ and $\displaystyle C_3$.

It is easy to verify that all functions of the form (4) satisfy the equation (1).