# If $\varphi$ is bounded and $\,|\varphi(x)-\varphi'(x)| \le 1,$ for all $\,x \in \mathbb{R}$, then $|\varphi(x)| \le 1,$ for all $\,x \in \mathbb{R}$

If $\varphi \in C^{1}(\mathbb{R})$ is bounded and $\,|\varphi(x)-\varphi'(x)| \le 1,$ for all $\,x \in \mathbb{R}$, then $|\varphi(x)| \le 1,$ for all $\,x \in \mathbb{R}$.

How to prove it, or find a counterexample please?

Assume that $|\varphi(x)|\le M$. We have $$\big|\mathrm{e}^{-x}\big(\varphi'(x)-\varphi(x)\big)\big|\le \mathrm{e}^{-x}$$ or $$\big|\big(\mathrm{e}^{-x}\varphi(x)\big)'\big|\le \mathrm{e}^{-x}.$$ If $\varphi(a)<-1$, for some $a\in \mathbb R$, then for $a<b$, $$\mathrm{e}^{-b}\varphi(b)-\mathrm{e}^{-a}\varphi(a)\le \mathrm{e}^{-a} -\mathrm{e}^{-b}$$ and hence $$\varphi(b)-\mathrm{e}^{b-a}\varphi(a)\le \mathrm{e}^{b-a} -1,$$ hence, $$\varphi(b)+1\le \mathrm{e}^{b-a}\big(\varphi(a)+1\big),$$ So if $\varphi(a)<-1$, then the RHS tends to $-\infty$, when $b\to\infty$, while the LHS is bounded below by $-M+1$. Contradiction. The possibility $\varphi(a)>1$ is dealt with similarly.
• Thank you. by $\big|\big(\mathrm{e}^{-x}\varphi(x)\big)'\big|\le \mathrm{e}^{-x}$, I get $e^{-x}\le e^{-x}\varphi(x)\le-e^{-x}$, Then maybe it is $\mathrm{e}^{-b}\varphi(b)-\mathrm{e}^{-a}\varphi(a)\le-\mathrm{e}^{-b}- \mathrm{e}^{-a}$. but the same result. Thank you. – xunitc Jul 9 '17 at 9:09
Let $C_\varphi=[\inf(\varphi'^{-1}(0)),\sup(\varphi'^{-1}(0))]$
For each intervall $I\subset C_\varphi$, we get that for every point $x\in I$ there are two points $a,b$ s.th a,b are $\varphi'(a)=\varphi'(b) =$ of $|\varphi|$ and $\varphi|_{[a,b]}$ is monotone. W.l.o.g let's assume it is increasing. Then $\varphi(a)\leq\varphi(x)\leq\varphi(b)$. Since $\varphi|_{[a,b]}$ is monotone and $1\geq|\varphi(a)-\varphi'(a)| = |\varphi(a)|$ (and the same for b) and we get that $\varphi\in[\varphi(a),\varphi(b)]\subset[-1,1]$
If $C_\varphi = \mathbb R$ we are already done, else we just have the points left and right of $C_\varphi$ . Lets wlog look at the right part. Let $\hat C_\varphi = [\sup \varphi'^{-1}(0),\infty)$. We already now, that $\varphi|_{\hat C_\varphi}$ is monotone since the derivativ is nowhere zero. Wlog let's again assume it is increasing. Now we prove the rest by contradiction. If there exists a point $x \in \hat C_\varphi$ s.th $\varphi(x)>1$, then due to the monotonie $\lim_{x\to \infty} \varphi(x) =: L > 0$ (the limit exists due to the monotonic behavior and the boundedness). Furthermore, we know that $\lim_{x\to \infty}\varphi'(x) = 0$. Therefor for every $\epsilon< L$ we can find a point $\xi\in\hat C_\varphi$ and $c_1,c_2>0$ s.th $\varphi(\xi)1+\epsilon+c_1$ and $\varphi'(\xi)=\epsilon-c_2$. Therefor $$1\geq|\varphi(\xi)-\varphi'(\xi)| = |1+\epsilon+c_1-\epsilon+c_2| = |1+c_1+c_2| > 1$$ which is a contradiction.