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


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 '15 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 '15 at 12:09

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


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