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Let $p_0=0$, $t\in[0,1]$ and $p_{n+1}(t):=p_{n+1}=p_{n}+\frac{1}{2}(t-p_{n}^2)$ be a sequence, I must show that it is increasing, that is $p_n\leq p_{n+1}$.

If it were to be increasing then by the relation we have that: $$p_n\leq p_{n+1}\iff p_{n}^2\leq t$$

How would I show that $p_n^2\leq t$ I tried induction but I get stuck.

Any help appreciated.

Sebastian Cor
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1 Answers1

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Here is one approach to your problem. As in your OP, define \begin{align} p_0(t)\equiv0\qquad p_{n+1}(t)=\tfrac12(t+2p_n(t)-(p_n(t))^2). \end{align}

An simple algebraic manipulation yields \begin{align} 2(p_{n+1}(t)-\sqrt{t})&=(2-p_n(t))p_n(t)-(2-\sqrt{t})\sqrt{t}\tag{1}\label{one}\\ 2(p_{n+1}(t)-p_n(t))&=t-(p_n(t))^2\tag{2}\label{two} \end{align} The function $\varphi(x)=(2-x)x$ is monotone increasing on $[0,1]$ ($\varphi'(x)=2(1-x)>0$ for all $0\leq x<1$). Hence, if $0\leq p_n(t)\leq \sqrt{t}$, then \eqref{one} we would imply $0\leq p_{n+1}(t)\leq\sqrt{t}$. Since $0=p_0(t)\leq\sqrt{t}$, then $0\leq p_1(t)\leq\sqrt{t}$. Continuing by induction, we have that
$$\begin{align} 0\leq p_n(t)\leq\sqrt{t},\qquad 0\leq t\leq 1, \quad n\in\mathbb{Z}_+\tag{3}\label{three}\end{align}$$

From \eqref{three} and \eqref{two} it follows that
$$0\leq p_n(t)\leq p_{n+1}(t)\leq\sqrt{t}$$ This shows that $p_n(t)$ converges for any $t$ as $n\rightarrow\infty$. Again, from \eqref{two} we obtain that $\lim_np_n(t)=\sqrt{t}$. Furthermore, convergence is uniform by Dini's theorem.

Mittens
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  • How does $(1)$ imply $p_{n+1}(t)\leq \sqrt(t)$? – Sebastian Cor Sep 04 '21 at 13:51
  • The right-hand side of (1) is $\varphi(p_n)-\varphi(\sqrt{t})$ (valid if $0\leq p_n(t)\leq 1$, which holds by inductive assumption: $0\leq p_n(t)\leq\sqrt{t}\leq 1$) and $\varphi$ is monotone increasing in $[0,1]$. – Mittens Sep 04 '21 at 14:16
  • Such a nice and short answer! – Sebastian Cor Sep 04 '21 at 14:23
  • @SebastianCor: In fact, changing $t$ got $t^2$ in the definition of $p_n$ and extending the domain to $|t|\leq1$, (substitute then any appearance of $\sqrt{t}$ by $|t|$, would give you a slightly more interesting result: $p_n(t)\xrightarrow{n\rightarrow\infty}|t|$ uniformly. Then you can use this to get your result. – Mittens Sep 04 '21 at 14:26