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Does there exist a twice differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$f'(x) = f(x+1)-f(x)$$ for all $x$ and $f''(0) \ne 0$.

Some things that I have tried is that obviously any linear function satisfies the first condition but not the second. Any hints/motivation would be helpful. Thanks.

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I'm afraid that's unnacceptable behaviour ssilwa. You'll end up getting yourself arrested one of these days. –  Eulerian Adventurer Sep 15 at 18:25
@EulerianAdventurer Ok, cool story bro. –  Sandeep Silwal Sep 15 at 18:48

2 Answers 2

Since we are just looking for existence, this will do.

Assume $f(z)$ is a function on complex plane instead, and assume it have the form $f(z)=e^{cz}$ for some constant $c$. The condition $f''(z)\not=0$ simply means $c\not=0$, while the other condition give: $ce^{cz}=(e^{c}-1)e^{cz}$ which means we need to solve for $e^{c}-c-1=0$. Applying standard technique to reduce to: $-(c+1)e^{-(c+1)}=-\frac{1}{e}$ which allow us to use Lambert W-function to get $c=-1-W(-\frac{1}{e})$. Hence found the solution in complex number.

To reduce this back to real number just take the real part, so we get: $f(x)=e^{Re(c)x}\cos(Im(c)x)$ which satisfy all condition.

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Thanks but I am looking for a more "elementary" approach. –  Sandeep Silwal Jun 9 at 4:57

Since the question is about existence: you can think of the right side as a difference quotient. $$f^{'}(x) =\frac{f(x+1)-f(x)}{(x+1)-x}.$$ The right hand side is a slope of the secant line between the points with $x-$coordinates $x$ and $x+1$. So any affine function (with non-zero slope and non-zero constant term) will satisfy this relation. For sure they are twice differentiable.

By affine I mean: $f(x)=ax+b$

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@user88595 Thanks for being so prompt with pointing out mistakes. I had read the question but due to bad usage of wording I wrote linear instead of saying straight line functions. –  Anurag A Jun 2 at 22:18
Sorry, but I had one of the conditions wrong. Please look again. –  Sandeep Silwal Jun 2 at 22:34

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