Which one of the following is true:

1.If a function real valued function $f$ satisfies $|f(x)-f(y)|\leq |x-y|^{\sqrt2}$ for all $x,y\in \mathbb R$ is $f$ a constant?

2.If $f$ is differentiable and its derivative is bounded then there exists $\epsilon_0>0$ such that $0<\epsilon\le\epsilon_0$,the function $g(x)=x+\epsilon f(x)$ is injective.

I think 1 is not true as $f(x)=x$ is a counter example

for 2 let $g(x_1)=g(x_2)\implies x_1+\epsilon f(x_1)=x_2+\epsilon f(x_2)$

$\implies \frac{1}{\epsilon}=\dfrac{f(x_1)-f(x_2)}{x_1-x_2}$

which is no longer bounded when $\epsilon$ is very small contradiction

Hence $x_1=x_2$

Is my solution correct?Hope someone helps


2 Answers 2


Your counterexample to 1 isn't a good counterexample. $|x-y| \le |x-y|^{\sqrt 2}$ is only true if $|x-y| \ge 1$. But $$\frac{|f(x) - f(y)|}{|x-y|} \le |x-y|^{\sqrt 2 - 1}$$ whenever $x \not= y$. What happens as $y \to x$?

Your contradiction to 2 is on the right track but incomplete. You need to show $\epsilon_0$ exists. If $|f'(x)| \le M$ for all $x$, the triangle inequality leads to $$|g'(x)| = |1 + \epsilon f'(x)| \ge 1 - \epsilon M.$$ Thus $0 < \epsilon < \frac{1}{M}$ implies $|g'(x)| > 0$ for all $x$, which implies in turn that $g$ is injective.

  • 1
    $\begingroup$ for the first case $f^{'}(x)\rightarrow 0$ and hence f is constant right $\endgroup$
    – Learnmore
    Jan 28, 2015 at 12:04
  • $\begingroup$ The expression on the left converges to $0$ as $y \to x$. That implies $f$ is differentiable at $x$ and thus $f'(x) = 0$. Stating $f'(x) \to 0$ isn't quite right. $\endgroup$
    – Umberto P.
    Jan 28, 2015 at 12:09

Regarding 1.: I think your counter example does not hold for $|x-y| < 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.