Is $f(x)\cdot g(x)$ a sign-preserving transformation of $f(x)$?

Question

Here are what I learned: a transformation is just a real function $F$. If a real function $f(x)$ is transformed, then $f(x)$ becomes "$F[f(x)]$", which is another real function.

Assume $g(x)>0$. $x$ is a variable.

Question：

1. Is $f(x)\cdot g(x)$ a transformation of $f(x)$?

2. Is $f(x)\cdot g(x)$ a sign-preserving transformation of $f(x)$?

My answer:

1. $f(x)\cdot g(x)$ can not equal to $F[f(x)]$, so $f(x)\cdot g(x)$ is not a transformation.

2. Since $g(x)>0$, $f(x)\cdot g(x)$ is indeed a sign-preserving transformation.

So the answers seem to contradict each other: how can a sign-preserving transformation not a transformation?? I am very very confused.

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For Q1 we might be able to play a trick like $F(y)=g(f^{-1}(y))\cdot y$. But $f^{-1}$ can be ill-defined.

Answer 1

Your definition of "transformation" isn't informative. It's just notation. A more concrete definition would be something like:

Let $X$ and $Y$ be spaces of functions. (E.g. $X$ could be all functions from $\mathbb R$ to $\mathbb R$, and $Y$ could be all functions from $\mathbb R$ to $\mathbb R^2$.) A transformation is a function $F : X \to Y$, where evaluation is typically written $F[f]$ (with square brackets) instead of $F(f)$ (with parentheses).

If you use this (or other sane definitions), then you'll quickly see there is no contradiction. Yes, the map $f \mapsto f\cdot g$ is a transformation, and yes, it preserves sign if $g > 0$.