Question

(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.)

Answer 1

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:

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

Answer 2

$F(v) = |v|$ works. To be more explicit, I mean $|(x,y)| = \sqrt{x^2 + y^2}$.

We have $F(u+v)^2 = F(u)^2 + F(v)^2 + 2 F(u) F(v) \cos \theta$, where $\theta$ is the angle between $u$ and $v$. When $u$ and $v$ are linearly independent, $\theta$ is nonzero, so $$F(u+v)^2 = F(u)^2 + F(v)^2 + 2 F(u) F(v) \cos \theta < F(u)^2 + F(v)^2 + 2 F(u) F(v)$$ and $$F(u+v) < F(u) + F(v).$$