(I've edited this question)


I'm searching for a continuously differentiable function $f:\mathbb R^2\to\mathbb R$ such that $f(x)+f(x+u+v)\neq f(x+u)+f(x+v)$ for all $x$ and all linearly independent $u$ and $v$.


My original question was about the special case $x=0, f(x)=0$ for merely continuous functions, which turned out to be trivial.


(I was lead to this question when investigating whether one can always find the vertices of a parallelogram (or more specifically, a square) in the graph of a continuously differentiable function $f:\mathbb R^2\to\mathbb R$. The nonexistence of functions such as the above would imply that one cannot always find a parallelogram in the graph of a continuously differentiable function.)



Existence of an "anti-additive" (or "never linear") map?
For your new question: functions that satisfy your inequality don't exist


Proof.


Suppose $f(x)+f(x+u+v)> f(x+u)+f(x+v)$ for all $x$ and all linearly independent $u$ and $v$.
Let us get a contradiction from it. Take any square insribed in a circle, and rotate it leaving insrcibed. Rotating continuously for the angle 90 degrees you can exchange to pairs of opposite vertices.


Here is the previous counterexample:


$(x^2+y^2)^{\frac{1}{4}}$


