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The following has come up in the course of my research. I'm looking for a function $ f : \mathbb Z ^ \star \to \mathbb R $ such that $$ 2 f ( i ) - f ( i + j ) - f ( i - j ) = \lambda j $$ for all $ i \ge 0 $ and all $ j $ such that $ 0 \le j \le i $, where $ \lambda $ is a positive real parameter.

If I let $ f ( k ) = - \frac 1 2 \lambda k ^ 2 $ then I get $ 2 f ( i ) - f ( i + j ) - f ( i - j ) = \lambda j ^ 2 $, but I haven't been able to guess a function where it evaluates to $ \lambda j $. I have a suspicion that no such function exists, but how can I show this? Alternatively, if there is such a function, what is it?

Mohsen Shahriari
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N. Virgo
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1 Answers1

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We have $$\begin{align}2\lambda &= 2f(2)-f(0)-f(4)\\&=(2f(1)-f(0)-f(2))+2(2f(2)-f(1)-f(3))+(2f(3)-f(2)-f(4))\\ &=\lambda+2\lambda+\lambda \end{align}$$ hence necessarily $\lambda=0$. With $\lambda=0$, $f(n)=an+b$ is a valid solution.

Hagen von Eitzen
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