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(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.)

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I would suggest not introducing a name for that class of functions... –  Mariano Suárez-Alvarez Feb 8 '10 at 18:55
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I agree with Mariano - for example, it seems likely that the sum of two such maps will not necessarily be another such map, and so the collection of such functions will not form a vector space as we would usually want. –  Zev Chonoles Feb 8 '10 at 19:09
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Please do not edit the question so much that existing answers to it stop being answers. If you want to ask another question, ask another question! –  Mariano Suárez-Alvarez Feb 8 '10 at 19:25
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2 Answers

up vote 9 down vote accepted

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:

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

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Sorry, I forgot the requirement f(0)=0. –  Samuel Feb 8 '10 at 19:00
    
+10 if I could! +1 will have to do –  HJRW Feb 8 '10 at 19:01
    
Samuel, idded 1 does not work anymore, but the other function works... –  Dmitri Feb 8 '10 at 19:11
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$F(v) = |v|$ 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 $$F(u+v) < F(u) + F(v).$$

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One can say that the triangle inequality happens only if the points are colinear. –  Hailong Dao Feb 8 '10 at 19:15
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Any strictly convex norm on $\mathbb{R}^2$ will do, because then the triangle inequality is strict $\|x+y\| \lt \|x\|+\|y\|$ whenever $x$ and $y$ are linearly independent. –  Konrad Swanepoel Feb 8 '10 at 19:16
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