# Increasing functions property

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!

-
Hi, what have you tried already? Also, is this homework? If yes, please add a homework tag. – Johannes Kloos Apr 3 '12 at 16:10
@Johannes Kloos: Thank you for your comment. This is not a homework. – Nicole Apr 3 '12 at 16:17

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.