Strange simple inequality

Question

In an online PDF about continuity the author claim (without proof) that for $h$ small and $f$ continuous we have:

$|f(x)-f(x+h)|<\dfrac{|f(x+h)|}{2}$

I try to prove if that statement is true or false but i didn’t found any satisfactory way (Maclaurin series, inverse triangle inequality or mean value theorem). It’s very likely very simple (or very wrong) but i don’t found a way to check.

Thanks for your help

Answer 1

This assertion does not hold, but the following does.

Let $f:\mathbb R\rightarrow\mathbb R$ be continuous and let $x\in\mathbb R$. Then the following statements are equivalent.

• $x$ is not a root of $f$.
• There exists $\varepsilon>0$ such that $|f(x)-f(x+h)|<\frac12|f(x+h)|$ for all $h\in(-\varepsilon,\varepsilon)$.

Proof sketch: If $x$ is a root, then the second statement does not hold (as suggested by user23571113) since for all $h\in\mathbb R$ we have $$|f(x)-f(x+h)|=|f(x+h)|\ge\frac12|f(x+h)|.$$ If $x$ is not a root, then by the continuity of $f$ there exists $\varepsilon>0$ such that $|f(x)-f(x+h)|<\frac13|f(x)|$ for all $h\in(-\varepsilon,\varepsilon)$ because $|f(x)|>0$. This suggests that $|\frac12|f(x)|-\frac12|f(x+h)||\le\frac12|f(x)-f(x+h)|<\frac16|f(x)|$, which yields $\frac12|f(x+h)|>\frac12|f(x)|-\frac16|f(x)|=\frac13|f(x)|$. Hence, we have $|f(x)-f(x+h)|<\frac13|f(x)|<\frac12|f(x+h)|$ and thus the second statement holds.

So, this inequality does hold (for small enough $h$ where it depends on $x$ and $f$ what small enough is) if and only if $x$ is not a root of $f$, i.e. $f(x)\neq 0$.