Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X, \rho)$ be a metric space and suppose $f: X\rightarrow X$ to be a continuous function. We say that the function $f$ admits $w: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ as a modulus of continuity if $$\rho(f(x),f(y))\leqslant w(\rho(x,y))\;\;\; (x,y\in X).$$ Clearly, when $f$ is a contraction then $w(t)=Lt$ ($t\in\mathbb{R}^+$), where $0<L<1$ and the series $$ \sum_{n=1}^{\infty} w^n(t)$$ is uniformly convergent in $(0,1)$.

I am looking for an example of a function $f$ on noncompact (but at least separable and complete) metric space which is not contraction but admits the modulus of continuity $w$ such that the series $ \sum_{n=1}^{\infty} \varphi(w^n(t))$ is uniformly convergent in some neighborhood of zero, where $\varphi:\mathbb{R}^+\rightarrow\mathbb{R}^+$ is an arbitrary continuous and nondecreasing function.

I'll be very grateful for every hint

share|cite|improve this question
up vote 0 down vote accepted

For example each function $f: (0,1)\rightarrow \mathbb{R}$ of the form: $$f(t)=t-bt^{\alpha+1},$$ where $\alpha \in (0,1)$ and $0<b\leq\frac{1}{1+\alpha}$ are arbitrary constants, satisfies requirments. For details see J.W Thron estimation for iterations of function of the form $v(t)=t-bt^{\alpha+1}$. (answered by Robert Israel).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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