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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 $$ 2f(i) - f(i+j) - f(i-j) = \lambda j $$ for all $i\ge0$ 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 $2f(i) - f(i+j) - f(i-j) = \lambda j^2$, 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?

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  • $\begingroup$ @achillehui corrected. $\endgroup$ – Nathaniel Dec 27 '13 at 16:07
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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.

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