7

Find all $f \in \mathscr{C^\infty}(\mathbb R)$ that satisfy the equation $$f'(x) = f(x+1) - f(x).$$

The 'obvious' answer is the set of all affine maps, but I'm not entirely sure.

Some progress: For any $x \in \mathbb{R}$, we have $$f(x+h) = f(x) + f'(x)h + r(x,h),$$ where $r(x,h) = o(h)$. Letting $h = 1$, $$f(x+1) = f(x) + (f(x+1) - f(x)) + r(x,1).$$ This implies $r(x,1) = 0$ for all $x$, so the linear approximation is exact for an increment of 1.
I thought of Taylor's theorem as well. For any $x \in \mathbb R$, $$f(x+h) = f(x) + f'(x)h + \frac{f''(c)}{2}h^2$$ for some $c \in (x,x+h)$. Again letting $h = 1$ we get that $f''(c) = 0$, but that only shows that the second derivative vanishes at some point, not all points.

I have a feeling that the first proof just needs one extra step to finish the job. This problem is not for a class, so I would be interested in seeing the solutions if the $\mathscr{C}^\infty$ condition were weakened (my first proof only requires differentiability, so I would assume a proof requiring $\mathscr{C}^1$ would be on a different route).

user217285
  • 5,825
  • 1
  • 19
  • 32
  • This seems somewhat related to http://math.stackexchange.com/questions/950961/how-to-proof-the-following-function-is-always-constant-which-satisfies-f-left – PhoemueX Jul 26 '15 at 07:08
  • Might it be useful to note that $\displaystyle f(x+1)=\sum_{k=0}^\infty\frac{f^{(k)}(x)}{k!}$? (This is the Taylor series expansion of $f(x+1)$ centered on the point $x$.) – Akiva Weinberger Jul 26 '15 at 07:09
  • The original equation forces $f$ to be infinitely differentiable. RHS is differentiable, so is LHS. Now use induction. – Mohsen Shahriari Jul 26 '15 at 07:11
  • @MohsenShahriari I did not notice that the infinite differentiability is immediate, though I have tried differentiating both sides. Where does the induction come in play? I want to show that every solution is of the form $ax + b$. – user217285 Jul 26 '15 at 07:16
  • I just wanted to note that by a simple induction you can show that $f$ is infinitely differentiable. I'm still thinking about the solution. I think the @PhoemueX 's comment may be useful. – Mohsen Shahriari Jul 26 '15 at 07:24
  • 3
    It seems to me that there are a lot of solutions, let's take an aribtrary function $h$ defined on $[0,1]$ (with some restrictions) then we can extend this function into a solution for $\mathbb{R}^+$ using the equation $f(x+1)=f(x)+f'(x)$ and we can do the same for $R^{-}$ – Elaqqad Jul 26 '15 at 12:11
  • @Elaqqad I think so too. And the restriction is $h'(0)=h(1)-h(0)$, $h\in C^\infty$ I guess, in order to be able to stitch smoothly. It's some condition missing in the problem formulation. Maybe $f$ is bounded, for example? – A.Γ. Jul 26 '15 at 12:17
  • @A.G. not only that, $h''(0)=h'(0)-h'(1)$ must hold and so on – Elaqqad Jul 26 '15 at 14:02
  • But if we continue to make those restrictions, will we get a first degree polynomial? – user217285 Jul 26 '15 at 19:31
  • 10
    This question has a (long) answer here: http://math.stackexchange.com/questions/389847/when-fx1-fx-fx-what-are-the-solutions-for-fx – user217285 Jul 28 '15 at 06:18

0 Answers0