It is given that $ |f_{n} ' (x) | \le \frac {1}{x^{\frac {1}{3}}} \forall 0 \lt x \le 1$ , where {$f_{n}$} is a sequence of real valued $C^{1}$ function on $[0,1]$ and each {$f_{n}$} has a zero in $[0,1]$ . Now to prove that the sequence has a uniformly convergent subsequence; somehow I have to try with Ascoli Arzela Theorem.

What I have done: .... To go with Ascoli-Arzela, I have to show pointwise boundedness & equicontinuity... Now, by the condition that: each {$f_{n}$} has a zero in $[0,1]$ , pointwise boundedness is shown. But,

1) I could not show Equicontinuity,

2) & one doubt is: since {$f_{n}$} is a sequence of real valued $C^{1}$ function on $[0,1]$ derivative should be bounded.. isn't it??


Hint: I guess you are using the following argument to show that $f_n$ are uniformly bounded:

Let $x_n\in [0,1]$ such that $f_n(x_n) = 0$, then for all $x\in [0,1]$, $$ |f_n(x)| = |f_n(x) - f_n(x_n)| = \bigg|\int^x_{x_n} f_n'(t) dt \bigg| \leq \int^x_{x_n} t^{-1/3} dt \leq \frac{3}{2} \big| x^{1-1/2} - x_n^{1-1/2}\big| \leq 3.$$

What happens if you use a general $y$ instead of $x_n$?

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  • $\begingroup$ Exactly!!... in this way i got the bound to be $\frac {3}{2}$ .... Since: $\int_{x_{n}}^{x} t^{-1/s} dt \le \int_{0}^{1} t^{-1/s} dt$ ... & so on ... But as you suggest if I use a general $y$ in that case I am getting: $|f_{n}(x) - f_{n}(y)| = |\int_{y}^{x} f'_{n}(t)dt|\le \int_{y}^{x} t^{-1/3} dt = \frac{3}{2} |x^{2/3}-y^{2/3}|$ $\endgroup$ – user92360 Nov 20 '13 at 6:43
  • $\begingroup$ Ok, so $s=3$? (I should wear my glass). If you have your inequality, you are close to showing that $f_n$ is equicontinuous. The RHS is independent of $n$ and is small when $|x-y| $ is small. $\endgroup$ – user99914 Nov 20 '13 at 6:55
  • $\begingroup$ But for $0 \le x,y \le 1$ $|x^{2/3} - y^{2/3}| is NOT \lt |x-y|$... Isn't it??? That's where I'm stucked... Right?? Please check it... $\endgroup$ – user92360 Nov 20 '13 at 7:00
  • $\begingroup$ But the function $x\mapsto x^{2/3}$ is uniformly continuous on $[0,1]$. So for all $\epsilon >0$ there is $\delta > 0$ such that if $|x-y| <\delta$, then.......... $\endgroup$ – user99914 Nov 20 '13 at 7:04
  • $\begingroup$ Oh hellll!!!! Sorrrry to forget that................. Sorrry it was easy... I missed it!!!! Thanks!!! $\endgroup$ – user92360 Nov 20 '13 at 7:05

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