Vtyjkyuk

Question. Let $f_n$ be a sequence of functions which are continuous over $[0,1]$ and continuously differentiable in $[0,1]$. Assume that $|f_n(x)|le1$ and that $|f_n'(x)|le 1$ for all $x in (0,1)$ and for each positive integer $n$. Then prove that $f_n$ contains a subsequence which converges in $C[0,1]$.

My Solution: Here I try to use the following theorem from Baby Rudin:

Theorem. If $K$ is compact, if $f_n in C(K)$ for $n = 1,2,3,dots$ and if $f_n$ is pointwise bounded and equicontinuous o $K$, then

1. $f_n$ is uniformly bounded on $K$.




2. $f_n$ contains a uniformly convergent subsequence.

Now using MVT and the given conditions I have managed to prove that $f_n$ is equicontinuous on $[0,1]$. Now I have to prove that $f_n$ is pointwise bounded. Now $|f_n(x)|le 1$ for all $x in (0,1)$. So I have to prove that $f_n(0)_n$ and $f_n(1)_n$ are bounded. How can I proceed now to prove these?

Since $f_n$ is continuous in $[0,1]$ (and so is $|f_n|$), then
$$
|f_n(0)| = lim_xto 0 |f_n(x)| leq 1,
$$
and the same holds for $|f_n(1)|$.

This answer was posted before a correction was made by the OP. I don't know if you mis-typed the question but it is false as stated. Example, $f_n(x)=x^n$. However, if the hypothesis includes the condition $|f_n'(x)| leq 1$for all $n$ and $x$ then you argument works: just use $f_n(frac 1 2)-f_n(0)=int_0^1/2 f_n'(t) , dt$ to prove that $f_n(0)$ is bounded. Similarly, for $f_n(1)$.