Show that a set is compact on $C^K[0,1]$ 
Show that the set of the functions $A_M:=\{f ∈ C^{k+1}([0, 1]) : \|f\|_{C^{K+1}} ≤ M\}$ is compact in $C^{k}[0,1]\ \  \forall M \geq 0$.

N.B.:
$$\| f\|_{C^{K+1}}=\|f\|_{C^{0}}+\|f^{(k+1)}\|_{C^{0}}$$
I started by showing that $C^{k}[0,1]$ is complete with the $C^{k}$ norm.
I'd like to use this to show that the set is complete.
Then prove that it is totally bounded and so compact.
But I don't have an idea of how to do it... 
Thanks.
 A: Start with this statement:

If $f_n :[0,1]\to\Bbb R$ bounded wrt $\|\cdot\|_\infty$ with $f_{n}^{(1)}$ being unbounded, then all derivatives (if they exist) $f_n^{(k)}$ are unbounded in $n$ wrt $\|\cdot\|_\infty$.

Let $\|f_n\|_\infty < C$. I will carry out the first few steps so that the induction argument becomes obvious.
If $f_n^{(1)}$ is unbounded pass to a subsequence to assume $\|f_n^{(1)}\|_\infty > n$. This means that $|f_n^{(1)}|$ must take on a value $>n$, however it is continuous and the integral of $f$ cannot be larger than $C$ as such when it reaches $n$ it has only a time $\frac{C}{n/2}$ to get back under $n/2$ ($n$ large enough for this number to be less than $1/2$). This means that the Lipschitz constant of $f^{(1)}$ is larger than $\frac{n^2}{2C}$. As such the derivative $f^{(2)}_n$ is unbounded as well, satisfying $\|f_n^{(2)}\|_\infty > \frac{n^2}{2C}$. Further $|f^{(2)}_n|$ cannot be larger than $\frac{n^2}{4C}$ on intervals of length $n\frac{8C^2}{n^2}+\frac{2C}n=\frac{8C^2+2C}n$ as otherwise $|f_n^{(1)}|$ would be larger than $n$ on an interval of length $\frac{2C}n$, which is not allowed.
You must now continue in this way. You must find that $f^{(k)}$ is both unbounded but cannot be "big" on an interval of length $> F^k(n)$ where $F^k(n)$ becomes small. This implies that the next derivative is unbounded in $n$, as the Lipschitz constant must be big, and that the next derivative cannot be big on any interval of length $F^{k+1}(n)$ which again becomes small, as otherwise $f^{(k)}$ cannot leave its "big derivative regime" quickly.

The above statement puts you in the position to use Azerla-Ascoli. If $\|f_n\|_\infty < C$ and $\|f_n^{(k+1)}\|_\infty < C$ you must find that all derivatves of order $≤k$ must be bounded (otherwise contradicting the above statment). Azerla-Ascoli tells you that any bounded equicontinuous sequence in $C([0,1])$ will admit a convergent subsequence.
If $f_n^{(k)}$ is bounded and its derivative $f_n^{(k+1)}$ is bounded then the $f_n^{(k)}$ are equicontinuous because they share a common Lipschitz constant. Azerla-Ascoli gives you the convergent subsequence for the $k$-th derivative. Continue in this way to find a sub-sequence so that all derivatives $≤k$ converge. Then make use of the following lemma:

If $f_n$ converges uniformly to $f$ and $f_n'$ converges uniformly to $g$ then $f$ is differentiable with $f'=g$.

To find that what you got actually converges in the topology you are looking at.
A: The set $A_M$ is not compact in $C^k[0,1]$. It is pre-compact, i.e., it is totally bounded, but not complete.
For example, for $k=0$, and $M=3$, then
$$
f_n(x)=\sqrt{(x-1/2)^2+1/n^2}\in A_M\subset C^1[0,1],
$$
and $f_n\to |x-1/2|=f$, in the $C[0,1]$ norm. But $f\not\in A_M$.
