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Samuel
Samuel
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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. Should I delete this question and make a new one?

(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.)

(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. Should I delete this question and make a new one?

(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.)

(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.)

deleted 353 characters in body; added 30 characters in body
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Samuel
Samuel
  • 365
  • 4
  • 9

Does there exist a continuous function $f:\mathbb R^2\to\mathbb R$ such that $f(0)=0$ and $f(x+y)\neq f(x)+f(y)$ for all linearly independent $x$ and $y$? (On a sidenote: what's a good name forI've edited this? My suggestion would be contra-linear or anti-additive; I thought also of anti-linear, nowhere linear and globally non-linear, but these were misleading question).

More generally 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 $v$. Upon failure to exhibit such an example

My original question was about the special case $x=0, f(x)=0$ for merely continuous functions, I soughtwhich turned out to prove the nonexistance of the above family of functionsbe trivial. Should I delete this question and make a new one?

(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 anti-additive functions such as the above would imply the existence ofthat one cannot always find a parallelogram in the graph of any continuousa continuously differentiable function.)

(I've edited the original formulation; I forgot the requirement $f(0)=0$.)

Does there exist a continuous function $f:\mathbb R^2\to\mathbb R$ such that $f(0)=0$ and $f(x+y)\neq f(x)+f(y)$ for all linearly independent $x$ and $y$? (On a sidenote: what's a good name for this? My suggestion would be contra-linear or anti-additive; I thought also of anti-linear, nowhere linear and globally non-linear, but these were misleading).

More generally 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 $v$. Upon failure to exhibit such an example, I sought out to prove the nonexistance of the above family of functions.

(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 anti-additive functions would imply the existence of a parallelogram in the graph of any continuous function.)

(I've edited the original formulation; I forgot the requirement $f(0)=0$.)

(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. Should I delete this question and make a new one?

(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.)

added 78 characters in body; added 315 characters in body
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Samuel
Samuel
  • 365
  • 4
  • 9

Does there exist a continuous function $f:\mathbb R^2\to\mathbb R$ such that $f(0)=0$ and $f(x+y)\neq f(x)+f(y)$ for all linearly independent $x$ and $y$? (On a sidenote: what's a good name for this? My suggestion would be contra-linear or anti-additive; I thought also of anti-linear, nowhere linear and globally non-linear, but these were misleading).

More generally 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 $v$. Upon failure to exhibit such an example, I sought out to prove the nonexistance of the above family of functions.

(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 anti-additive functions would imply the existence of a parallelogram in the graph of any continuous function.)

(I've edited the original formulation; I forgot the requirement $f(0)=0$.)

Does there exist a continuous function $f:\mathbb R^2\to\mathbb R$ such that $f(0)=0$ and $f(x+y)\neq f(x)+f(y)$ for all linearly independent $x$ and $y$? (On a sidenote: what's a good name for this? My suggestion would be contra-linear or anti-additive; I thought also of anti-linear, nowhere linear and globally non-linear, but these were misleading).

(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 anti-additive functions would imply the existence of a parallelogram in the graph of any continuous function.)

Does there exist a continuous function $f:\mathbb R^2\to\mathbb R$ such that $f(0)=0$ and $f(x+y)\neq f(x)+f(y)$ for all linearly independent $x$ and $y$? (On a sidenote: what's a good name for this? My suggestion would be contra-linear or anti-additive; I thought also of anti-linear, nowhere linear and globally non-linear, but these were misleading).

More generally 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 $v$. Upon failure to exhibit such an example, I sought out to prove the nonexistance of the above family of functions.

(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 anti-additive functions would imply the existence of a parallelogram in the graph of any continuous function.)

(I've edited the original formulation; I forgot the requirement $f(0)=0$.)

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Samuel
Samuel
  • 365
  • 4
  • 9
Source Link
Samuel
Samuel
  • 365
  • 4
  • 9