Is $f ≡ 0$ if $f(0)=0$ and $f'(x) \le f(x)$?

Question

Is the following statement true or false?

Let $f$ be a non-negative function contnious on $[0, \infty)$ and differentiable function on $(0, \infty)$ such that $f(0)=0$ and such that $f'(x) \le f(x)$ for all $x \in (0, \infty)$. Then $f ≡ 0$.

I think this statement is false. Take $f(x)= x+e^x-1$ so that $f(0)=0$ and $f'(x)=e^x +1 \le x + e^x-1$, but $\lim_{x \rightarrow \infty} f(x)=\infty \neq 0$.

Are you sure that $1 \le x$ for all $x \in (0,\infty)$? – Umberto P. Mar 21 '20 at 18:01 — Umberto P., Mar 21 '20 at 18:01
Your new example is even worse :( – Umberto P. Mar 21 '20 at 18:03 — Umberto P., Mar 21 '20 at 18:03

score 0 · Accepted Answer · answered Mar 21 '20 at 18:08

The statement is true.

Let $g(x)=e^{-x}\bigl(f'(x)-f(x)\bigr)$. The $g(x)\leqslant0$ for each $x\geqslant0$. Now, let $h(x)=e^{-x}f(x)$. Then $h'(x)=g(x)\leqslant0$. So, $h$ is decreasing. It is also non-negative. But $h(0)=0$. So, $h$ is the null function, and therefore $f$ is the null function.

score 0 · Answer 2 · answered Mar 21 '20 at 18:08

The statement is true, and here is a significant hint. Your stated hypothesis implies $ [e^{-x} f(x)]' = e^{-x}f'(x) - e^{-x}f(x) \le 0$ for all $x > 0$, so that $$e^{-x} f(x) = e^{-x}f(x) - e^{-0}f(0) = \int_0^x [e^{-t}f(t)]'\, dt \le 0$$ for all $x > 0$.

Is $f ≡ 0$ if $f(0)=0$ and $f'(x) \le f(x)$?

2 Answers2