MHB Proving that $S_1 \cdot S_2 \in S$

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The discussion focuses on proving that the product of two functions from the class S, defined as functions from [0, ∞) to [0, ∞), also belongs to S. It establishes that both functions S1 = e^x - 1 and S2 = ln(x + 1) are members of S. The properties of S include closure under addition and composition, as well as the condition that if one function dominates another, their difference remains in S. The challenge is to demonstrate that the product of any two functions f(x) and g(x) in S also satisfies the criteria to be in S. Successfully proving this would reinforce the robustness of the class S regarding multiplication.
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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) $\in S$.
 
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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
 
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