Welcome to mathimatikoi.org; a site of university mathematics! Enjoy your stay here!

## Zero function from an inequality

Real Analysis
Tolaso J Kos
Articles: 2
Posts: 853
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

### Zero function from an inequality

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $\left|f'(x) \right| \leq \left| f(x) \right|, \; \forall x \in \mathbb{R}$ and $f(0)=0$. Prove that $f$ is the zero function.
Imagination is much more important than knowledge.
Riemann
Articles: 0
Posts: 167
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

### Re: Zero function from an inequality

We note that

\begin{align*}
\left ( e^{-2x} f^2(x)\right ) ' &= 2e^{-2x}\left ( f(x)f'(x) - f^2(x) \right ) \\
&\leq 2e^{-2x}\left ( |f(x)f'(x)| - f^2(x) \right ) \\
&= 2e^{-2x}|f(x)|\left ( |f'(x)| - |f(x)| \right ) \\
&\leq 0
\end{align*}

hence the function $e^{-2x} f^2(x)$ is decreasing. Thus $e^{-2x} f^2(x) \leq f^2(0) =0$ and the result follows.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$