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In Contests in Higher Mathematics: Miklos Schweitzer Competitions, 1962-1991 by Gabor J Szekely, problem F.57 there is the study of $f~:~[0,\infty)\to (0,\infty)$ such that: $\exists c>0, \forall x>0$, $f'(x)=cf(x+1)$.

It states, without proof, that for this equation has a solution if and only if $c\leq 1/e$.

There is a reference to an unpublished paper: T. Krisztin, Exponential bound for positive solutions of functional differential equations, unpublished manuscript

Does anyone have this paper (or a proof of $c\leq 1/e$)?

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  • $\begingroup$ You could try writing to him. See his web page. $\endgroup$ – Dave L. Renfro May 8 '15 at 20:43
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    $\begingroup$ I did, several months ago, no answer. $\endgroup$ – user239036 May 8 '15 at 21:20
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I'm going to prove the "only if" part of the problem. i.e. If $f : [0,\infty) \to (0,\infty)$ is a solution of the equation

$$f'(x) = c f(x+1),\quad c > 0\tag{*1}$$

then $c \le \frac{1}{e}$, the other direction is leave as an exercise.

For any solution $f(x)$ of the problem, let $\displaystyle\;\delta = \inf\left\{ \frac{f(x+1)}{f(x)} : x > 0 \right\}\;$.

Notice $f'(x) = cf(x+1) > 0$ implies $f(x)$ is strictly increasing. We have $f(x+1) > f(x)$ for all $x > 0$. This means $\delta$ exists and $\ge 1$. As a result, over the interval $(0,\infty)$, we have $$\begin{align} & f'(x) = cf(x+1) \ge c\delta f(x)\\ \implies & (\log f(x))' \ge c\delta\\ \implies & \log f(x+1) - \log f(x) \ge \int_x^{x+1} c\delta dx = c\delta \\ \implies & \frac{f(x+1)}{f(x)} \ge e^{c\delta} \end{align} $$ This leads to $$\delta = \inf\left\{ \frac{f(x+1)}{f(x)} : x > 0 \right\} \ge e^{c\delta} \implies c \le y e^{-y}\quad\text{ for } y = c\delta > 0$$ From this, we can conclude $$c \le \sup\big\{ y e^{-y} : y > 0 \big\} = \frac{1}{e}$$

BTW, if one remove the restriction that $f(x)$ is positive, there are solutions for the delayed ODE $(*1)$ even when $c > \frac{1}{e}$.

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This is a partial solution:

You can see that if $c \leq 1/e$ the equation $\ln(a) = c*a$ has a real solution for $a$. This means that the function defined as $f(x) = a^x$ satisfies the functional equation because $f'(x) = \ln(a) * f(x) = c*a*f(x) = c*f(x+1)$.

Now we still have to prove that there are no solutions if $c>1/e$.

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  • $\begingroup$ I have a proof that in order to have a solution $c$ has to be smaller than 1 $\endgroup$ – Ward Beullens May 8 '15 at 21:23

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