Solution of the functional equation $2f\left(\frac{x+y}{2}\right)+f\left(\frac{x-y}{2}\right)+f\left(\frac{y-x}{2}\right)=f(x)+f(y)$ Can someone gave me a solution to the functional equation $2f\left(\frac{x+y}{2}\right)+f\left(\frac{x-y}{2}\right)+f\left(\frac{y-x}{2}\right)=f(x)+f(y)$ ?
possibly a quadratic one, i.e., a solution $f$ with $f(ax)=a^2f(x)$.
Thanks in advance!
 A: The general solution $f:V\to W$, where $V$ and $W$ are $\mathbb Q$-vector spaces, can be expressed as
$$f(x)=B(x,x)+L(x)\text{ for all $x\in V$}$$
where $B:V\times V\to W$ is a bilinear map and $L :V\to W$ is a linear map. Any functional of this form can easily be checked to satisfy the functional equation.
Even if you are only interested in $V=W=\mathbb R$, the real numbers need to be considered as a $\mathbb Q$-vector space to get a general solution, because under the axiom of choice there are solutions of Cauchy's functional equation that are not $\mathbb R$-linear.
To show all solutions have this form, the main trick to use is to split into odd and even parts.
Odd part
Define $L(x)=(f(x)-f(-x))/2$. Taking the functional equation applied to $(x,y)$, and subtracting the functional equation applied to $(-x,-y)$, gives
$$2L(\tfrac{x+y}2)=L(x)+L(y)\text{ for all $x,y\in V$.}$$
Since $L$ is an odd function we know $L(0)=0$; taking $y=0$ gives $2L(x/2)=L(x)$ for all $x\in V$, so in fact $L$ is additive. An additive map between $\mathbb Q$-vector spaces is linear.
Even part
Define
$$Q(x)=(f(x)+f(-x))/2$$
$$B(x,y)=Q(\tfrac{x+y}2)-Q(\tfrac{x-y}2)$$
Note that $f(0)=0$ (plug in $x=y=0$ to the functional equation), so we do in fact have $f(x)=B(x,x)+L(x)$.
Taking the functional equation applied to $(x,y)$, and adding the functional equation applied to $(-x,-y)$, gives
$$2Q(\tfrac{x+y}2)+2Q(\tfrac{x-y}2)=Q(x)+Q(y)$$
Consider $x,y,z\in V$. We have
$$
\begin{align*}
&B(x,z)+B(y,z)\\
&=Q(\tfrac{x+z}2)-Q(\tfrac{x-z}2)+Q(\tfrac{y+z}2)-Q(\tfrac{y-z}2)\\
&=(Q(\tfrac{x+z}2)+Q(\tfrac{y+z}2))-(Q(\tfrac{x-z}2)+Q(\tfrac{y-z}2))\\
&=(2Q(\tfrac{x+y}2 + z)+2Q(\tfrac{x-y}2))-(2Q(\tfrac{x+y}2 - z)+2Q(\tfrac{x-y}2))\\
&=2Q(\tfrac{x+y}2 + z)-2Q(\tfrac{x+y}2 - z)\\
&=2B\left(\tfrac{x+y}2,z\right).
\end{align*}
$$
Taking $x=0$ gives $B(y,z)=2B(y/2,z)$ for all $y\in V$, so in fact we have
$$B(x,z)+B(y,z)=B(x+y,z)\text{ for all $x,y,z\in V$.}$$
As in the odd case, this shows that $B$ is linear in the first argument, and symmetrically linear in the second argument, and hence bilinear.
A: Since you only want "a" solution, then $f(x) =x$ works.
A: Assuming a power series expansion for $f(x)$ the most general solution is $f(x):=a_1 x+a_2 x^2$ where $a_1$ and $a_2$ are constants. The proof assumes that $f(x):=a_0 + a_1 x + a_2 x^2 +\dots$ and by substituting this definition in the functional equation we solve for $a_0=0$ and $a_n=0$ for $n>2\;$. All that is left is the degree 1 and 2 terms.
