Trying to make sense of Igor's argument. Assume further that $f$ is defined on 0 and relaxt the continuity assumption by only assuming that the set $A=\{x>0:f (x)\neq f (0)\}$ is open.

It is shown that $A=\emptyset $ by showing that

1. $A $ is dense in $A+\mathbb R^+$, and
2. $A\cap 2A=\emptyset $.

Proof. (1) Let $(a,b) $ be a connected component of $A $. For every $0 <r <b-a $, $a+r\in A $ since the arithmetic average of $a\pm r $ is not in $A $ whereas the geometric is. It follows that the next connected component contains $(a,2a-b) $ and, by applying the same argument on the next component, the elements of $[a,\infty )\setminus A $ are at least $b-a $ apart from one another.

(2) Let $a\in A $. $a $ is the arithmetic average of 0 and $2a $ and 0 is their geometric..

Trying to make sense of Igor's argument. Assume further that $f$ is defined on 0 and relaxt the continuity assumption by only assuming that the set $A=\{x>0:f (x)\neq f (0)\}$ is open.

It is shown that $A=\emptyset $ by showing that

1. $A $ is dense in $A+\mathbb R^+$, and
2. $A\cap 2A=\emptyset $.

Proof. (1) Let $(a,b) $ be a connected component of $A $. For every $0 <r <b-a $, $b+r\in A $ since the arithmetic average of $b\pm r $ is not in $A $ whereas the geometric is. It follows that the next connected component contains $(b,2b-a) $ and, by applying the same argument on the next component, the elements of $[a,\infty )\setminus A $ are at least $b-a $ apart from one another.

(2) Let $a\in A $. $a $ is the arithmetic average of 0 and $2a $ and 0 is their geometric.