# Putnam 2012: B1

#### jakncoke

##### Active member
Problem: Let S be the class of functions from $[0, \infty)$ to $[0, \infty)$ such that

1) $S_1 = e^{x} - 1, S_2 = ln(x + 1)$ are in S
2)if f(x), g(x) $\in S$, then f(x)+g(x), f(g(x)) are also in S
3)if f(x), g(x) $\in S$, and f(x) $\geq$ g(x) for $x \geq 0$, then f(x) - g(x) is in S

Prove that if f(x), g(x) $\in S$, then f(x)g(x) $\in S$.

#### jakncoke

##### Active member
This is also quite easy. It borders on realizing log(x) + log(y) = log(xy) and $e^{log(xy)} = xy$.

Assume f(x),g(x) $\in S$, then ln(f(x)+1) + ln(g(x) + 1) $\in S$, since this is just composing functions already in S.

ln(f(x)+1) + ln(g(x) + 1) = ln( (f(x) + 1)(g(x) + 1) ) = ln( f(x)g(x) + f(x) + g(x) + 1)

Composing again with $e^{x} - 1$, we get $f(x)g(x) + f(x) + g(x) + 1 - 1 = f(x)g(x) + f(x) + g(x)$.

Since f(x)g(x) $\geq 0$, f(x)g(x) + f(x) + g(x) $\geq$ f(x) + g(x) for all $x \geq 0$.
so

f(x)g(x) + f(x) + g(x) - f(x) - g(x) = f(x)g(x) $\in S$.

QED