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Find a continuous function , if exists , (explicitly in terms of known functions) $f: \mathbb{R}\times\mathbb{R} \to \mathbb{R}$, such that $f$ is not always positive , not even always negative and $f(u, v) > 0$ and $f(w, z) > 0$ implies $f(u+w, v+z) > 0$ and $f(uw - vz, uz+vw) > 0$.

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Let $f$ be the constant $-1$. –  Berci Oct 11 '12 at 10:07

2 Answers 2

I guess you want it to be continuous, but in case, not, consider this: $$f(x,y):=\left\{ \begin{align} 1 &\text{ if }x,y\in\Bbb Z \\ -1 &\text{ else} \end{align} \right. $$ If you insist, you can reformulate it, for example using signum and integer part functions..

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Yeah I need a continuous one –  Souvik Dey Oct 11 '12 at 10:17

Consider a function $f:\mathbb{C}\rightarrow{R}$ such that $$ \left\{\begin{align} f(x)&>0\\ f(y)&>0 \end{align}\right. \quad\Rightarrow\quad f(x+y)>0, \; f(xy)>0 $$

It has three remarkable properties:

  1. $f(x)>0$ implies $f(nx)>0$ and $f(x^n)>0$ for every natural $n>0$
  2. $f(x)\leq 0$ implies $f(x/2^n)\leq0$ and $f(\sqrt[2^n]x)\leq0$ for every natural $n\geq0$
  3. $f(x)>0$ with $f(y)\leq 0$ implies $f(y-x)\leq 0$

Let $B(x)$ be the open unit ball centered on $x$.

If $f$ is continuous

  1. $f(x)> 0$ with $x\in B(0)$ implies $f(0)=\lim_{n\rightarrow\infty}f(x^n)>0$
  2. $f(x)\leq 0$ implies $f(0)=\lim_{n\rightarrow\infty}f(x/2^n)\leq0$

So, since we require the existence of a $z$ such that $f(z)\leq0$, continuity implies $f|_{B(0)}\leq0$.

Now let $x\notin B(0)$ and also not on the unit circle or the axes. Suppose $f(x)>0$. We would have $f(x^{2^n})>0$ and $f|_{B(-x^{2^n})}\leq0$ for every natural $n>0$. But then continuity would lead to $f(x)=\lim_{n\rightarrow\infty}f(\sqrt[2^n]{-1}x)\leq0$, contradicting $f(x)>0$.

$f$ cannot be positive.

The functions satisfying the required properties are all the continuous, nowhere positive and not always negative ones (note this is simply the family of functions that make the properties vacuous).

Going back to your notation now. An example is $f(u,v)=0$. Another one is $f(u,v)=-|u+v|$. Yet another one is $f(u,v)=1-\sin(u)\cos(v)$.

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