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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f'$ is bounded.

Show that there exists a constant $c$ such that $\varphi: \mathbb{R} \rightarrow \mathbb{R}$, $\varphi(x)= x+cf(x)$ is bijective.

I'm not sure about my proof of the surjection. Here is my whole attempt:

Since $f'$ is bounded, there exists an $M$ such that $|f'(x)|\leq M$. Taking $c=1/2M$, define $\varphi:\mathbb{R} \rightarrow \mathbb{R}$ by $\varphi=x+c.f(x)$. Note that $\varphi$ is differentiable, since $f$ is. Hence, $\varphi'(x)=1+\frac{f'(x)}{2M}$. Since $|f'(x)|\leq M$, then: $$0<1/2=1-\frac{1}{2M}M \leq1+\frac{f'(x)}{2M}=\varphi'(x)\leq1+\frac{1}{2M}M=3/2$$ Since $\varphi'(x)>0$, it follows that $\varphi$ is increasing and hence it is injective. Now consider $\frac{\varphi(x)}{x} =\frac{ x+c.f(x)}{x}$. Evaluating the limit as $x\rightarrow \infty$, by the L'hospital rule then: $$1/2\leq \lim_{x\to\infty} \frac{\varphi(x)}{x} = \lim_{x\to\infty} \varphi'(x) \leq 3/2$$

So we conclude that $\lim_{x\to\infty} \varphi(x)=\infty$ (i concluded this since $\lim_{x\to\infty} \frac{\varphi(x)}{x}>0$ (which means that $\varphi$ doesn't approach any finite number as $x$ goes to infinity). Similarly $\lim_{x\to-\infty} \varphi(x) = -\infty$, and therefore $\varphi(\mathbb{R})=\mathbb{R}$, which shows that $\varphi$ is a bijection.

How do i show that $\lim_{x\to+-\infty} \varphi(x) = +-\infty?$

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    $f'$ doesn't need to have a limit at $\pm\infty$, hence the same problem holds for $\varphi'$. – 2017-02-09
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    $\varphi $ is bounded from below and above by a nonconstant affine functions, so it has to be onto. (This way you can get rid of the unnecessary L'Hospital) – 2017-02-09
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    @Thomas, can you give some hints on why this is true? – 2017-02-09
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    Consider for example $f(x)=x+\sin(x)$ : $f$ is differentiable, $f'(x)=1+\cos(x)$, si $f'$ is bounded. We can take $M=2$ and so we have $\varphi(x)=x+\frac14(x+\sin(x))=\frac54x+\frac14\sin(x)$. Clearly, $\varphi'$ doesn't have a limit at $\pm\infty$ – 2017-02-09
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    If you have a lower bound given by an increasing function of the form $\varphi(x)> ax + c$ with $a\neq 0 $ then this shows that $\varphi \rightarrow \infty $ if $x\rightarrow \infty$ (since the lower bound has this behavior). Similar with an upper bound for $-\infty$ Then apply the mean value theorem. – 2017-02-09
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    you mean intermediate value theorem? – 2017-02-09
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    As your question is stated, I would go with $c=0$. :) – 2017-06-18

1 Answers 1

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Suppose for all constants $c$ ,$ \phi(x)=x+cf(x)$ is not bijective,so $\phi(x)$ is not injective .Then $\exists x_1\neq x_2$ such that $\phi(x_1)=\phi(x_1)\implies (x_1-x_2)=c(f(x_1)-f(x_2))\implies \dfrac{f(x_1)-f(x_2)}{x_1-x_2}=\dfrac{1}{c}$.Applying Mean Value Theorem to $f$ on $[x_1,x_2]$ ,$\exists p$ such that $f^{'}(p)=\dfrac{f(x_1)-f(x_2)}{x_1-x_2}=\dfrac{1}{c}$.Since the equation holds for all $c\in \Bbb R$ so $|f^{'}(p)|\to \infty $ as $c\to 0$ which contradicts that $f^{'}$ is bounded.

Suppose that $\phi$ is not bijective. If $\phi$ is not injective then go to the shaded part for the contradiction.

If $\phi$ is injective but not onto, then $\text{Range }(\phi)\subsetneq \Bbb R$. Let $A=\text{Range }(\phi)$.

Then since $\phi :\Bbb R\to A$ is continuous and bijective so it has a continuous inverse and hence $\Bbb R$ is homeomorphic to $A$ and hence isometric too.

Since $\Bbb R$ is complete so is $A$.The only subspace of $\Bbb R$ homeomorphic to $\Bbb R$ and complete is

$\Bbb R$

and hence $A=\Bbb R\implies \phi $ is onto.

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    Ok, but this only proves that $\phi$ must be injective. If $\phi$ is not bijective, or it is not injetive or it is not surjective. – 2017-02-09
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    Here I come to your rescue!! – 2017-02-10