i need to find examples of functions $f(x,y):x,y $ belong to $R^2$ such that: $f$ is constant on the parabolas $y=c(x^2+1)$ c belongs to $R$. And the second one should be bounded on the unit circle, but not bounded on $x^2+y^2<4$. Any suggestions?

  1. If $f$ has to be constant on the parabolas $y = c(x^2+1)$ for a fixed $c$, then you need that $f(x,c(x^2+1)) = a$ for a constant $a$. A constant function would of course work, and a simple (non-constant) example would be $$f: \mathbb{R}^2 \to \mathbb{R}: (x,y) \mapsto y -c(x^2+1)$$ If you want a non-constant function independent of $c$, you could use $$f: \mathbb{R}^2 \to \mathbb{R}: (x,y) \mapsto \frac{y}{x^2+1},$$ which is well-defined since $x^2+1 > 0$ everywhere, and on the parabolas $y = c(x^2+1)$ it is constant, with its value being $c$.
  2. In polar coordinates, this would mean that $f(r,\theta)$ is bounded when $r = 1$ but unbouded for the area $r < 2$. A simple example would then be $$f: \mathbb{R}^+ \times [0,2\pi) \to \mathbb{R}: (r,\theta) \to \frac{1}{2-r}$$ which is bounded on the unit circle since it has the constant value $1$ there. But it is unbounded in the area $r < 2$ since $$\lim_{r \to 2} \frac{1}{2-r} = \infty$$ In Cartesian coordinates, this function would be $$f: \mathbb{R}^2 \to \mathbb{R}: (x,y) \to \frac{1}{2-\sqrt{x^2+y^2}}$$
  • $\begingroup$ For the first example i could've just said f=0 for all x,y and it would've been the same, isn't there an example without this c? $\endgroup$ – CodeHoarder Dec 25 '16 at 15:22
  • $\begingroup$ @CodeHoarder Indeed, constant functions would of course have done the trick. I've added an example of a non-constant function that does not depend on $c$ (but its constant value on the parabolas mentioned is $c$, though). $\endgroup$ – TastyRomeo Dec 25 '16 at 15:25
  • $\begingroup$ Thanks man, seems so obvious now haha. $\endgroup$ – CodeHoarder Dec 25 '16 at 15:29

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