Compactness of $A:=${$f \in C[0,1], |f|_\infty \le K, |f'|_\infty \le M$}

Question

Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_\infty \le M$. Is it true that $A$ is close and therefore compact?

Answer 1

No it is not compact. For example, $A$ contains

$$ f_n (x) = \sqrt{(x-1/2)^2 + 1/n}.$$

since

$$ |f_n'(x)| = \frac{2|x-1/2|}{\sqrt{(x-1/2)^2 + 1/n}}\le 2. $$

But $f_n$ converges to $|x-1/2|$ which is not differentiable.

Remark: If instead one consider $$A' = \{ f\in C[0,1] : |f|_\infty\le K, \operatorname{Lip}f \le M\},$$

here $$\operatorname{Lip}f := \sup_{x, y\in [0,1], x\neq y} \frac{|f(x) - f(y)|}{|x-y|},$$

then $A\subset A'$ and $A'$ is compact.