For every real function $f$, is there a strictly increasing real function $g$ above $f$?

Question

I apologize if this question is too elementary, but I don't know how to prove the statement in my question. Suppose $f$ is a real function. Does there exist a real function $g$ such that $g \geq f$, and $g$ is strictly increasing? By $g \geq f$, I mean, $g$ is greater than or equal to $f$ pointwise.

Answer 1

We can show the following equivalence for functions $f: \Bbb R \to \Bbb R$. Similar arguments can be made if the domain of $f$ is a subset of $\Bbb R$.

A strictly increasing function $g: \Bbb R \to \Bbb R$ with $f \le g$ exists if and only if the sets $$ S_a = \{ f(x) \mid x \le a \} $$ are bounded above for all $a \in \Bbb R$.

Remark: A necessary condition is that $\limsup_{x \to -\infty} f(x) < +\infty$. For continous functions $f: \Bbb R \to \Bbb R$ this is also sufficient.

Examples: Such an “increasing majorant” $g$ exists for $f(x) = e^x\sin(x) $, but not for $f(x) = x^2\sin(x) $.

Proof of the equivalence: If an increasing function $g$ with $f \le g$ exists then $$ f(x) \le g(x) < g(a) $$ for $x \le a$, so that $S_a$ is bounded above.

Conversely, if $S_a$ is bounded above for all $a \in \Bbb R$ then the function $$ g: \Bbb R \to \Bbb R \, , \, g(a) = \sup \{ f(x) \mid x \le a \} + e^a $$ is well-defined and strictly increasing with $f \le g$. (Without the term $e^a$ the function $g$ would still be increasing, but not necessarily strictly increasing.)

Answer 2

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $\displaystyle\lim_{x \rightarrow -\infty} f(x) = +\infty$.

If $g : \mathbb{R} \rightarrow \mathbb{R}$ satisfies $g \geq f$, then one must have $\displaystyle\lim_{x \rightarrow -\infty} g(x) = +\infty$, which is impossible if $g$ is increasing.

Answer 3

No. If $f(x)=1/x$ when $x>0$ and if $g\ge f$ then $g(1)\ge f(1)=1 .$

So $g(1)>1/2g(1)>0 $

and $g(1/2g(1))\ge f(1/2g(1))=2g(1)>g(1) $

so $g $ is not increasing,