About $M(\frac{x+y}{2}) = \frac{1}{2}(M(x) + M(y))$

Question

Suppose $M$ is map from vector space $X$ to vector space $Y$, $M(0) =0$, and $M(\frac{x+y}{2}) = \frac{1}{2}(M(x) + M(y))$.

Does this mean that $M$ is a linear map?

If not, could someone please give me an example?

It seems to me that some continuity condition is needed.

Thank you!

Answer 1

Here’s a counterexample.

The defining equation can be rewritten $M(x+y)=\frac12M(2x)+\frac12M(2y)$. Since $M(0)=0$, we have $M(x)=\frac12M(2x)$ for all $x\in X$, and hence $M(x+y)=M(x)+M(y)$. Take $X=Y=\Bbb R$, and let $M$ be any non-linear solution of the Cauchy functional equation. Then for all $x,y\in\Bbb R$ we have $M(x+y)=M(x)+M(y)=\frac12M(2x)+\frac12M(2y)$, since every solution of the Cauchy functional equation satisfies $M(qx)=qM(x)$ for $q\in\Bbb Q$.

Answer 2

$\mathbb R$ is a vaector space over $\mathbb R$, but not every map $M\colon\mathbb R\to\mathbb R$ with the given property is $\mathbb R$-linear. For example it might merely be $\mathbb Q$-linear.

Complex conjugation $\mathbb C\to\mathbb C, z\mapsto\bar z$ is not $\mathbb C$-linear.