Find all Riemann integrable functions $f:[0,1] \to \mathbb{R}$ such that $\forall x \in [0,1]$ we have $\int_0^x f(t)dt =(f(x))^{2015}+f(x)$.

We consider $g:\mathbb{R} \to \mathbb{R}$, $g(x)=x^{2015}+x$. Then $g$ is differentiable and invertible. We denote $F(x)=\int_0^x f(t)dt$.

The hypothesis can be written as: $g^{-1}(F(x))=f(x)$ (1).

$f$ is Riemann integrable $\Rightarrow$ $F$ is continuous, so, by (1), we deduce that $f$ is continuous. Then $F$ is differentiable, so, again by (1), we have that $f$ is differentiable.

By differentiating $F(x)=g(f(x)), \forall x\in [0,1]$ or $g^{-1}(F(x))=f(x)$ we get that $f(x)=f'(x)g'(f(x))$. Can anyone help me to finish this problem? Thank you!

  • $\begingroup$ Where does this problem come from? The appearance of the number 2015 is typical of contest problems. Please credit your source. $\endgroup$ Apr 24 '15 at 14:58
  • $\begingroup$ source: Romanian NMO 2015 Shortlist! $\endgroup$
    – npatrat
    Apr 28 '15 at 19:30

In fact, there are no non-zero differentiable solutions to this equation. More generally: if $f:[0,1] \to \Bbb R$ with $|f'(x)| \leq C |f(x)|$ for some $C > 0$ for all $x \in [0,1]$ and $f(0) = 0$, then $f(x) = 0$.

Note also: plugging into original equation, we have $$ 0 = [f(0)]^{2015} + f(0) \implies f(0) = 0 $$

Back to where you were: we have a separable differential equation. In particular, $$ f(x) = f'(x)g'(f(x)) \implies\\ \frac{g'(f(x))}{f(x)} f'(x) = 1 $$ Integrating both sides (making the substitution $y = f(x)$), we have $$ \int_a^{f(x)} \frac{g'(y)}{y}\,dy = x+C \implies\\ \frac{2015}{2014}[f(x)]^{2014} + \ln[f(x)] = x+C $$ We run into a bit of a problem here, however, since $\ln[f(0)]$ is not defined. I'm not sure what to take away from this, honestly. Perhaps this implies that $f = 0$ is the only solution.

  • $\begingroup$ The differential equation has other solutions, but none of them satisfy $f(0)=0$ which is implied by the problem statement. (Putting $x=0$ into the original equation gives $0 = f(x)^{2015} + f(x) = f(x)(f(x)^{2014}+1)$ and the second factor is strictly positive.) You have, in effect, already argued that $f=0$ is the unique solution to the initial value problem with $f(0)=0$. Your "separation of variables" approach is invalid for $f$ with $f(0)=0$, since it involved a division by zero early on, so it is not surprising that this particular solution is not found by that method. $\endgroup$ Apr 24 '15 at 15:01
  • $\begingroup$ For those wanting a proof of the "more generally" statement in the first paragraph: see Gronwall's inequality. $\endgroup$ Apr 24 '15 at 15:04
  • $\begingroup$ @NateEldredge thanks for clarifying. The "more generally" statement is also a problem from baby Rudin. $\endgroup$ Apr 24 '15 at 17:13

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