0
$\begingroup$

Suppose that $f : \mathbb R \rightarrow \mathbb R$ is a differentiable function with continuous derivative. Suppose furthermore that there is $k \in \mathbb R$ with the property that $|f'(x)| \le k, \forall x \in \mathbb R$. Show that there is a constant $c > 0$ such that the function $x + cf(x)$ is a bijection.

Let's define $g(x)=x + cf(x)$. Is $g$ injective? Supppose it is not: $g(x)=g(y) \Rightarrow x \neq y$. So we have (for let's say $x \gt y$): $x + cf(x) = y + cf(y) \Rightarrow \frac{f(x)-f(y)}{x-y}=\frac{-1}{c}$. Now, using the mean value theorem we have that there is $x_0 \in (y,x): f'(x_0)=\frac{f(x)-f(y)}{x-y}$ $\Rightarrow f'(x_0)=\frac{-1}{c}$ $\Rightarrow |f'(x_0)|=|\frac{1}{c}|$. So, if I choose $c>0,$ such that $\frac{1}{c} \le k$, the previous equation stands and I have not reached a contradition (so I have done nothing!). For proving that $g$ is onto I have no idea. Any help?

1 Answers 1

3

You are on the right track. With $c<1/k$ let $g(x)=x+c\,f(x)$. Then $$ g'(x)=1+c\,f'(x)\ge1-c\,k>0. $$ Then $g$ is strictly increasing, and hence injective. To show that it is surjective prove $\lim_{x\to\pm\infty}g(x)=\pm\infty$ and use the intermediate value theorem.

  • 0
    Isn' t the derivative of $g(x)$ wrongly written? Isn't $g'(x)=1+cf'(x) \ge 1-ck \gt 0$, for $c<\frac{1}{k}$? – 2017-02-20
  • 0
    This also agrees with the result of the contradiction of the previous proof I wrote where it had to be $\frac{1}{c} \le k \Rightarrow c \ge \frac{1}{k}$ in order for $g$ to be non-injective. – 2017-02-20
  • 0
    The only way that I found to prove that: $\lim_{x\to \pm \infty}g(x)=\pm \infty$ is by writting: $g(x)=x(1+\frac{f(x)}{x})$ and by proving that if $f'$ is bounded, then with the help of the Intermediate Value Theorem, $\frac{f(x)}{x} \rightarrow 0$, for $x\to \pm \infty$ and so, $g(x) \approx x$ for such $x$. Is this correct, can you verify? Afterwards, to show surjectiveness I found how to do it, I am OK with that. – 2017-02-20
  • 0
    @JohnZobolas Thakyou for noticing. I have edites the answer. – 2017-02-20
  • 1
    For the limits use $|f(x)|\le c\,k\,|x|+C$ with $c\,k<1$ and some constant $C$. – 2017-02-20