Can a non-constant function satisfy $f\left(\frac{x+y+1}{2}\right) = \frac{f(x)+f(y)}{2}$?

Question

I'm trying to find functions $f$ over $\mathbb R$ which satisfy $$ f\left(\frac{x+y+1}{2}\right) = \frac{f(x)+f(y)}{2} $$ for all real $x$ and $y$. One thing I immediately note is that shifting any potential function by a constant preserves the above identity, so I can assume $f(0) = 0$ and so $f(\frac{x+1}{2}) = \frac{f(x)}{2}$ for all real $x$. I can then rewrite the left side of the above equality to deduce $$f(x + y) = f(x) + f(y)$$ for all real $x$ and $y$, which is the well-known Cauchy functional equation. Clearly, non-zero affine solutions do not work, so $f$ is either the zero function or is some highly pathological function.

Can such a pathological solution be compatible with the requirement that $f(\frac{x+1}{2}) = \frac{f(x)}{2}$?

Answer 1

Consider $\mathbb R$ as a vector space over $\mathbb Q$. Let $1$ be a basis element and $\vec v$ another one, and consider projection onto the line through $\vec v$. That satisfies $$f\left( \frac {x+1}2\right)=\frac {f(x+1)}2=\frac {f(x)}2$$

And it's "weakly" linear in the manner of Cauchy's equation. That is to say, it is linear as a map over $\mathbb Q$, though not over $\mathbb R$. Of course, it's not identically $0$ since $f(\vec v)=1$.

Answer 2

Un example of $f$ non-constant (it is not so pathological!)

At first sight we have $f\left(x+\dfrac12\right)=f(x)$ for all $x$ so $f$ is periodical with period $\dfrac12$. It follows we must have $f\left(\dfrac{x+y}{2}\right)=\dfrac{f(x)+f(y)}{2}$.

The function is entirely defined if we take $f(x)$ for $0\le x\lt\dfrac12$ and we consider the displacements of a periodic function. Giving to $f(0)$ an arbitrary real value, say $f(0)=a$, we cannot choose $f(x)$ arbitrarily because we need $f\left(\dfrac y2\right)=\dfrac{a+f(y)}{2}$ with $0\le y\lt\dfrac12$. A little calculation gives $\boxed{f(x)=a+\dfrac x2\text{ for } 0\le x\lt\dfrac12}$ which satisfies this condition, so complete definition of $f$ becomes $$\boxed{f(x)=\frac{4f(0)-n}{4}+\frac x2;\space \left\{\frac n2\le x\lt \frac{n+1}{2}\right\}\text{ with} \space x,f(0)\in\mathbb R; n\in\mathbb Z}$$ We add an example for $f(0)=3$, limiting us to two "segments" with points of the graph and four "segments" with complete graph.