Need help with difficult extra credit analysis question

Question

I don't really understand this problem from my analysis class:

Let $BC([0,1],R)$ be the metric space of functions which are bounded, continuous on $[0,1]$, and in $C^\infty$, meaning that all of their derivatives exist and are continuous. We're using the infinity metric on $BC([0,1],R)$. Notice that there's an issue about defining the derivative at $0$ or $1$. We'll take "differentiable" to mean that, for every $f$ in $BC([0,1],R)$, there is some $a<0$ and $b>1$ so that $f$ is infinitely differentiable on $(a,b)$.

Let $F$ be the set of all $f$ in $BC([0,1],R)$ such that the absolute value of the $K$-th derivative of $f(x)$ is less than or equal to $M$ for all $x$ between $0$ and $1$, inclusive, and also such that $0$ is in the $j$-th derivative of $f([0,1])$ for all $j$ between $0$ and $K$, inclusive.

That is, $F$ is the set of all functions whose $K$-th derivative is uniformly bounded by $M$ and whose image includes $0$. Note only the $K$-th derivative is bounded, not the $(K - 1)$-th or anything.

Prove that for any fixed $K$ in the natural numbers and $M > 0$, we have that the closure of $F$ is compact. Here the closure is taken in $BC([0,1],R)$.

I have been advised to use the Arzelà-Ascoli theorem (Take $K$ to be a closed subset of $C[0,1]$. Then the following are equivalent: $K$ is compact; $K$ is uniformly bounded and equicontinuous).

So basically we need to prove that the closure of $F$ is uniformly bounded and equicontinuous. Can I prove that $F$ is uniformly bounded by saying that the $K$-th derivative is obviously bounded by $M$ for $x$ between $0$ and $1$, so the $(K - 1)$-th derivative must be bounded because it only changes by a finite amount (only achieves at maximum a velocity of $M$) and so must be bounded as well, and then extending this logic all the way to the original $f$?

I also don't really know how to go about proving that the function is equicontinuous. Any help is greatly appreciated!

Answer 1

Hint: So we have, if I understand your question correctly $$ F = \{f \in C^\infty([0,1]) \mid 0 \in f^{(j)}([0,1]), 0\le j \le K, \|f^{(K)}\|_\infty \le M \}$$ Start, as you suggest with proving that indeed all derivatives up to the $K$-th of the $f\in F$ are uniformly bounded. Argue by induction on $j$, that there is an uniform bound for the $f^{(K-j)}$, for $j=0$ that is given, in the step use, that for some $x \in[0,1]$ you have $f^{(K-j-1)}(x) = 0$ and hence the fundamental theorem gives you, as you say $\|f^{(K-j-1)}\|_\infty \le \|f^{(K-j)}\|_\infty$.

For equicontinuity, use that we have by the above, that the derivatives of the $f\in F$ are uniformly bounded. Hence, by the mean value theorem, we have $$ |f(x) - f(y)| = |f'(\xi)|\cdot |x-y| \le M \cdot |x-y| $$ for $x,y \in [0,1]$, $f\in F$ and some uniform(!) $M$. Why does this imply equicontinuity of $F$ (and hence of $\bar F$)?