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I have to prove a lemma:

If $\varphi: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ is monotone increasing and $\varphi(t) < t, \ \forall t \in \mathbb{R}_+$, then $\varphi^n(t) \rightarrow 0$, $(n \rightarrow \infty)$.

To be clear, $\varphi^2(t) = \varphi(\varphi(t))$.

It's just not clear to me how to proceed.

Edit: as stated by Karolis Juodelė, for the lemma to be true, $\varphi$ must be continuous.

Egor N
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It's not true. Take $$\varphi(t) = \begin{cases} \frac t 2 & t \leq 1 \\ \frac 1 2 + \frac t 2 & t > 1 \end{cases}$$ Note that although $\lim_{t \to 1} \varphi(t) = 1$ but $\varphi(1) = \frac 1 2$ and $\varphi(t) < t$ holds. Note also that $\lim \varphi^n(t) = 1$ when $t > 1$. For it to be $0$, $\varphi$ should be continuous.

Assume that $\varphi$ in fact is continuous. Note that the sequence $x_n = \varphi(x_{n-1}) = \varphi^n(t)$ is decreasing and bounded, thus converges to $x$. Now, because $\varphi$ is continuous, $\lim x_n = \lim \varphi(x_{n-1}) = \varphi(\lim x_{n-1}) = \varphi (x) = x$. By definition of $\varphi$, this implies $x = 0$

Karolis Juodelė
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  • That's an excellent remark. Thank you. Could you provide a proof assuming $\varphi$ is continuous? Should I start a knew question or edit this one? – Egor N Apr 12 '14 at 11:05
  • Wouldn't $\varphi^n(t)$ eventually reach $\varphi(t)=1$?After reaching it,function would also reach $\varphi(t)=0$ – kingW3 Apr 12 '14 at 11:12
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    @kingW3, no. It will come arbitrarily close though. Observe that $\varphi(1+x) = 1+\frac x 2$ so $\varphi^n(2) = 1+\frac 1 {2^n} > 1$. – Karolis Juodelė Apr 12 '14 at 11:22
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    @EgorN, proof added. – Karolis Juodelė Apr 12 '14 at 11:33
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    The end of your proof is a proof by contradiction, as it does not hold that $\varphi(x)=x$ anywhere. Also, $\varphi$ is defined only on the positive reals. – mathse Apr 12 '14 at 12:48