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Is it possible to choose $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that

$$ a+yf(x,y) = 0, \;\;\;a>0 $$

is a bounded curve?

EDIT: I am very sorry for not being precise enough. I've been searching for an answer for some time now and I've come to think of some things for granted... Anyway:

ADDENDUM: $f$ should be continously differentiable with respect to both its variables. The trivial $f(x,y)=0$ is not accepted.

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    What do you mean by bounded curve, bounded as a subset of $\mathbb{R}^2$ or $|y(x)|\leq K$ for the curve $y(x)$ ?2012-09-25
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    Also, $f(x,y) : \mathbb{R}^{2} \rightarrow \mathbb{R}$.2012-09-25
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    @pritam As a subset of $\mathbb{R}^2$, eg like the circle $x^2+y^2=1$ but NOT like $y=x^2$2012-09-25
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    I'm not sure if this is what you want, but if you choose $f = 0$, you get the null set as the "curve", which is certainly bounded. Are there further restrictions? :)2012-09-25
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    @Yoni: I assume that he means a curve as a 1 dimensional subset of $\mathbb{R^2}$ so I'm not sure the empty set would work.2012-09-25

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There is no single continuous $f:\ {\mathbb R}^2\to{\mathbb R}$ that produces a nonempty bounded curve for all $a>0$.

Proof. Let $g(x,y):=y\ f(x,y)$. Then $\inf g(x,y)=-\infty$, or else the solution set $$S_a:=\{(x,y)\ |\ a + y\ f(x,y)=0\}$$ would be empty for large enough $a$. Therefore for arbitrary $R>a$ one can find a point $(\xi,\eta)\in{\mathbb R}^2$ with $\xi^2+\eta^2>R^2$ and $g(\xi,\eta)<-R$. Connect the point $(R,0)$ with $(\xi,\eta)$ by a curve $\gamma:\ t\to z(t)$ $\ (0\leq t\leq 1)$ such that $|z(t)|\geq R$ for all $t\in[0,1]$. As $g(R,0)=0$ and $g(\xi,\eta)<-R$ there has to be a $\tau\in[0,1]$ with $g\bigl(z(\tau)\bigr)=-a$. It follows that $z(\tau)\in S_a$ and at the same time $|z(\tau)|\geq R$. As $R>a$ was arbitrary, $S_a$ is unbounded.$\quad {}_\square$

When $f$ is allowed to depend on $a$ (or $a$ is bounded away from $\infty$) then one can construct examples: For a fixed $b>0$ consider the function $$f(x,y):={2b\over 1+x^2+y^2}\ .$$ I claim that the solution set $S_a$ is a circle for any $a$ with $0

Proof. Since $1+x^2+y^2>0$ for all $(x,y)$ the equation $a+y f(x,y)=0$ is equivalent with $$(1+x^2+y^2) + 2y{b\over a}=0$$ or $$x^2 +\Bigl(y+{b\over a}\Bigr)^2={b^2-a^2\over a^2}\ .$$ When $0 this is the equation of a circle in the $(x,y)$-plane. $\quad{}_\square$

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    Your answer is great, thanks alot.2012-09-26
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Seems like $f(x,y)=\frac{1}{y}$ (except $f(x,0)=0)$ should do the trick $\ldots$

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    but this isn't $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ (and it also yields a curve with no points!)2012-09-25
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    Sure it is! Tell me two real numbers $x$ and $y$. I'll evaluate $\frac{1}{y}$, ignoring $x$, and send that real number back to you.2012-09-25
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    a curve has dimension 1, while no points $(x,y)$ satisfy the equation $a+1 = 0$.2012-09-25
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    I think I misunderstood the ultimate goal of your question. You're looking for an expression in $x$ and $y$ (your $f(x,y)$) that yields another expression when substituted into that $a+y\cdot f(x,y)=0$ thingy, whose solution set is a bounded plane curve (for an arbitrary $a>0$). You should pose it that way. And the fact remains that $f(x,y)=\frac{1}{y}$ *is* a function from $\mathbb{R}^2\to\mathbb{R}$, it just doesn't satisfy the rest of your conditions.2012-09-25
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    Sorry, I missed $f(x,0) = 0$!! But it still won't do since it is an empty set, I' ll also edit the question to be more precise, please have a look2012-09-25
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What are the most basic types of bounded curves we know? Ellipses.

If you choose $f(x,y) = \frac{y}{x^2-1}$ it should work.

Define the value of $f(1,y)$ to be whatever you want. This way you do get a curve, not the empty set.

This looks a lot like homework.

Edit: sign mistake

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    I guess that if this can be made to be continouously differentiable with respect to x and y then it could be acceptable! (please take a look at the question again, since I edited!)2012-09-25
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For $1 \leq x \leq 2$ define $f(x, y) = -a$. For other values of $x$ define $f(x, y) = 0$.

The resulting set is an interval in $\mathbb{R}^2$, between the points $(1, 1)$ and $(2, 1)$.