If $f(0,0) = f(\infty,0)=f(0,\infty) = 1$, but $f(\infty,\infty) = 0$, then what is $f(a,b)$?

Question

$f(a,b)=1$ if both $a$ and $b$ are 0, or if one of them is infinity and the other zero. However $f(\infty,\infty)=0$. What is $f(a,b)$?

I tried using exponentials, $f(a,b)=\exp(-a)+\exp(-b)$, but could not satisfy all constraints.

Answer 1

For a function $f(a,b)$ that's continuous on $\mathbb{R}^2$, with values for infinite $a$ and/or $b$ defined via appropriate limits, we could make a small adjustment to your sum of exponentials:

$$f(a,b) = \frac{e^{-a} + e^{-b}}{1+e^{-a}\,e^{-b}}$$

We have $$f(a,0) \;\equiv\; 1 \;\equiv\; f(0,b)$$ for all finite $a$ and $b$, hence $$f(\infty,0) := \lim_{a\to\infty}f(a,0) = 1 = \lim_{b\to\infty} f(0,b) =: f(0,\infty)$$ Also, $$f(\infty,\infty) := \lim_{a\to\infty,\;b\to\infty}f(a,b) = 0$$

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Note. You can simplify this to $$f(a,b) = \frac{e^a + e^b}{1 + e^{a+b}}$$ but the limiting behavior becomes slightly less evident.

Answer 2

You are almost there. How about: $$f(x)=\max(e^{-a},e^{-b})$$ The usage of the maximum makes $f(0,0)=1$, not 2 as would arise with your try of $e^{-a}+e^{-b}$. All other constraints remain the same.

Answer 3

$f(a,b) = 1 - \frac{a b}{(a+1)(b+1)}$

We could use some construct like $g(a, b) = 1 - a b$, in which we get $g(0,0) = g(1,0) = g(0,1) = 1$ and $g(1,1) = 0$.

So, we need to find a function that would make $0$s into $0$s, and $\infty$s into $1$s and form a composition with $g$.

One such function is $h(x) = \frac{x}{x + 1}$.

Forming a composition, we get: $f(a,b) = 1 - \frac{a b}{(a+1)(b+1)}$.

Answer 4

Your statement is a fine definition of a function. The domain is $\{0,\infty\} \times \{0,\infty\}$ and the range is $\{0,1\}$. What is the problem? If the domain is supposed to be $[0,\infty] \times [0,\infty]$ you can define $f(a,b)$ on the rest of the domain however you want. Maybe it is $\pi$.