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Given a function $f(x):\mathbb R\to\mathbb R$, which is continuous, bijective, and nondecreasing on $\mathbb R$. Also, there exists a constant $L>0$ such that $0<|f'(x)|\leq L$ for all $x\in\mathbb R$.

I want to show that

$$ f((x, x+a))\subset (f(x), f(x)+La)$$ for all $x\in\mathbb R$, and $a>0$, where $(x,x+a)$ is an open interval in $\mathbb R$.

Any help! Thanks.

Edit: I know from the above conditions that $f$ will be Lipschitz function with Lipschitz constant $L$, i.e., $$|f(x)-f(y)|\leq L|x-y|$$ and if we consider $y=x+a$, then we get $|f(x)-f(x+a)|\leq La$. But How to use this!

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You should also write explicitly that $f$ is differentiable. In that case you know that $$ f(x)<f(y)<f(x+a) $$ for all $y\in(x,x+a)$ so you only should show that $f(x+a)\leq f(x)+La$ - indeed: $$ f(x+a) = f(x)+\int\limits_x^{x+a}f'(y)dy\leq f(x)+\int\limits_x^{x+a}Ldy = f(x)+La $$ as needed.

If you can only use the Lipschitz condition, then from the monotonicity again you obtain: $$ f(x+a) - f(x)\leq L((x+a)-a) = La $$ so $f(x+a)\leq f(x)+La$ and the argumet above applies.

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Thank you, that make sense. I was working on a awrong direction - using lipschitz definition. – Nicole Apr 3 '12 at 16:16
@Nicole: updated the answer – Ilya Apr 3 '12 at 16:24
Thanks again!... – Nicole Apr 3 '12 at 16:45
@Nicole Hi Nicole. Welcome here! It is common that you accept the best answer (select the green V) and upvote the answer. Of course, you could wait with accepting the answer to see if there will come other ones too. – Jonas Teuwen Apr 3 '12 at 17:44

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