Prove boundedness of recurrence relation

Question

For a number sequence $\{y_n\}$ we know that $y_{n+1} = 2y_n-y^2_n$

If: $0<y_0<1$ show that $0<y_n<1$ for all integers $n>0$

I've tried solving the recurrence relation, but I couldn't solve the inhomogenous part. Anyway, I thought that maybe I didn't need to solve it to figure it out and so I though if I showed that when:

$y_0 = 0 \implies y_1 = 2\times0-0^2 = 0$

and $y_0 = 1 \implies y_1 = 2\times1-1^2 = 1$

Is this really enough for proof for the statement?

Answer 1

By the AM-GM inequality, given that $x\in[0,2]$, $$ x(2-x)\leq\left(\frac{x+(2-x)}{2}\right)^2 = 1, $$ and equality holds only if $x=1$. This gives that if $y_0\in (0,1)$, $y_n\in(0,1)$ holds by induction.

Answer 2

$y_{n+1}=2y_{n} - y_{n}^2 = 1-(1-y_{n})^2 \Rightarrow y_{1}= 1-(1-y_{0})^2 \Rightarrow 0< y_{1}<1$ as $ 0<y_{0}<1$ is given. Now assume it is true for $n=k$ that is $ 0 < y_{k} < 1 $ now $y_{k+1} = 1- (1-y_{k})^2 \Rightarrow 0< y_{k+1}<1$. So by mathematical induction we have the result.