Is there a rule for taking any function of both parts of an equation or an inequality?

Question

1) I have an inequality $f(x)\ge g(x)$.

Can I state (maybe with some stipulations) that for any function h(x) the following holds $$ h(f(x)) \ge h(g(x)) $$

2) I have an equation $f(x) = g(x)$.

Can I state (maybe with some stipulations) that for any function h(x) the following holds $$ h(f(x)) = h(g(x)) $$

What all those stipulations for cases 1 and 2 should be? Are there any limitations on h(x) function (its type, domain, range)?

I feel there shoukld be some general rule, but I am not sure about all correct stipulations about domain/range, etc...

Answer 1

For the first point (1): Let's assume that the functions $ f(x) $ , $ g(x) $ and $ h(x)$ have domains $D_f $ , $ D_g $ and $ D_h $ respectively and : $$ f(x) \ge g(x) \ \forall x \in D_f \cap D_g $$ (Assume $ D_f \cap D_g \neq \emptyset$)

Before moving it is necessary to write down the domains of the composite functions that appear in the new inequality. Those functions are:$$ h(f(x)) $$ and $$ h(g(x)) $$

By definition the domains of those are $ D_{h \circ f} = \{x \in D_f | \ f(x) \in D_h \}$ and $ D_{h \circ g} = \{x \in D_g | g(x) \in D_h \} $ respectively.

Now you CAN state that: $$ h(f(x)) \ge h(g(x))\ \forall x\in D_{h \circ f} \cap D_{h \circ g} $$

iff $ \ $i) h(x) is a monotone increasing (strictly or not) function and ii) $ D_{h \circ f} \cap D_{h \circ g} \neq \emptyset $

For the second point (2):

Suppose now h(x) is any function with random properties, then you can freely state that if

$$ f(x)=g(x) \ \ (Equation \ 1) $$

$$ then $$

$$ h(f(x))=h(g(x)) \ \ (Equation \ 2)$$

or in other words $ f(x)=g(x) \Rightarrow h(f(x))=h(g(x)) \ \ (3)$ as long as $ D_{h \circ f} \cap D_{h \circ g} \neq \emptyset $ .

The above (Equation 3) is always true because of the definition of a function.

However, be careful the equation (1) and (2) might not be equivalent (might not have the same solutions). Equation (3) can become:

$$ f(x)=g(x) \Leftarrow\Rightarrow h(f(x))=h(g(x)) \ (Equation \ equivalency) $$

if and only if h(x) is a 1-1 function. ($ \forall x,y \in D_h \Rightarrow h(x) \neq h(y)$)

Answer 2

For both cases you have to assume that $h$ is defined over at least a subset of the intersection of the codomain of $f$ and $g$, and subsequently limit the domain of $f$ and $g$.

For the first case, you have to assume that $h$ is an increasing function, at least over the common range of $f$ and $g$, otherwise you could have cases in which $$h\big(f(x)\big)<h\big(g(x)\big)$$ even if $$f(x)\geq g(x)$$ for example let $x,y\in (0,\infty)$ and $$x> y$$ now if we consider the function $f:(0,\infty)\to (0,\infty)$ defined as $$f(z)=\frac{1}{z}$$ then we have $$f(x)<f(y)$$

For the second case, if we call $z=f(x)=g(x)$ then is trivial to see that $$h\big(f(x)\big)=h(z)=h\big(g(x)\big)$$ with no particular restrictions