Following this question and discussion recently
Is the derivative of a function bigger or equal to $e^x$ will always be bigger or equal to the function ?!
I decided to look at the different forms that a function which is always less than its derivative takes, so instead of looking at when $\frac{f'(x)}{f(x)} \ge 1$. I restated the question to this.
Let $g(x)$ be a function such that $g(x) \ge 1\space \forall x$ then solving the following the inequality will give us the functions we need.
$$\frac{f'(x)}{f(x)} = g(x)$$
Solving that differential equation you get that $f(x)$ takes on the form (where $k$ is a constant and $g(x) \ge 1 \space \forall x$)
$$f(x) = ke^{\int g(x)dx}$$
My question is do all functions that have derivatives greater than the function itself have this form or am i missing something?
Best Answer
We are looking for a simple criterion for a positive function $f$, which is equivalent to the differential inequality $f'(x)\geq f(x)$ for all $x$. One can say the following:
A differentiable function $f:{\mathbb R}\to{\mathbb R}_{>0}$ satisfies $f'(x)\geq f(x)$ for all $x$ iff the function $$g(x):={f(x)\over e^x}$$ is increasing.
Proof. One has $g'(x)=\bigl(f'(x)-f(x)\bigr)e^{-x}$, and this is $\geq0$ for all $x$ iff $f'(x)\geq f(x)$ for all $x$.