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I encountered in my research a functional equation that I'm not sure how to solve in general. It is similar to Cauchy's functional equation but includes an extra constant term of $-f(0)$. I'm not an expert in functional equations, so any help would be appreciated.

The functional equation is $f(x+y)=f(x)+f(y)-f(0)$

This is actually a part of a pair of functional equations that $f$ has to satisfy, but I'm at first interested in just solving this first functional equation. The other equation is $f(-x)=-f(x)+2f(0)$, and $f$ has to satisfy both the above one and this other one.

Clearly $f(x) = x$ solves the functional equations, as does any function of the form $f(x)=ax+b$, but are there any other solutions?

Any help is much appreciated.

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Define $g(x) = f(x) - f(0)$. Note that the condition $f(x + y) = f(x) + f(y) - f(0)$ is equivalent to the condition $g(x + y) = g(x) + g(y)$. Thus, the solutions for $f$ are precisely the solutions for Cauchy's functional equation, plus an arbitrary constant.

Your second condition is simply a special case of your first condition, incidentally, substituting $-x$ for $y$.

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  • $\begingroup$ That's right. I thought it might have been very simple. Thanks very much! $\endgroup$ – Janne Gustafsson Nov 29 '17 at 12:09
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Note $g(x)=f(x)-f(0)$ satisfies Cauchy's functional equation.

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