bounding derivative of a sequence

Question

I've been banging my head against the wall on this one ... define a sequence of polynomials $q_n$ by $q_0 = 0$ and $$q_{n+1} = q_n + .5(t^2 - q_n^2).$$ If $q_n \leq t$ on $[0,1]$ then $$.5(t^2 - q_n^2) = .5(t - q_n)(t+q_n) \leq t - q_n$$ so that $q_{n+1} \leq t$, and also $$t - q_{n+1} = (t-q_n)(1 - .5(t + q_n)) \leq (t-q_n)(1 - .5t),$$ so that inductively $t - q_n \leq (1 - .5t)^n$ on $[0,1]$. Thus $q_n$ increases pointwise to $t$ on $[0,1]$.

Are the derivatives $q_n'$ uniformly bounded on $[0,1]$? If $r_n = q_n'$ then the recurrence relation is $$r_{n+1} = t + (1 - q_n)r_n$$ but I don't see how this helps.

Answer 1

Your last recurrence relation indeed helps. That is:

1) Away from zero, once you consider $t>\epsilon$ and $n\ge 1$, you have $q_n$ bounded away from zero by a positive constant $c=\min_{[\epsilon,1]} q_1(t)$, and hence $r_n$ cannot jump over the fixed point of the map $r\mapsto t+(1-c) r$, which is $r=\frac{t}{c}$.

2) Now, you have to handle the neighborhood of zero. If you had the limit value $t$ or at least $\alpha t$ (with $\alpha$ a constant) instead of $q_n$ in your recurrence relation, the fixed point would be $\frac{t}{\alpha t}=\frac{1}{\alpha}$, and it would be done.

3) Now you handle different $t$'s separately. For each individual $t$, for the first $1/t$ iterations there is nothing to worry about: anyway you have $r_n<nt<1$.

4) On the other hand, it is not so difficult to see that at the $n=1/t$'s iteration you have $q_n>\alpha t$. Indeed, on every step $j$ you add $\frac{1}{2}(t^2-q_j^2)$. Either at some $j<n$ you already have $q_j>\frac{1}{2}t$, and then you are done. Or it does not happen, and then $q_n\ge n\cdot \frac{1}{2}(t^2-\frac{1}{4}t^2) = n\cdot \frac{3}{8}t^2 \sim \frac{3}{8}t$. So you have $\alpha=\frac{3}{8}$, and the proof is complete.