Equivallence of norms over $C^n([a,b])$ 
Denote by $C^n([a,b])$ the $n$ times continuously diffrentiable functions on the interval $[a,b]$. Prove that the norms $\|f\|=\operatorname{max}_{0\leq k\leq n}\operatorname{sup}_{x\in [a,b]}|f^{(k)}|$ and $\|f\|_{\sim}=\sum_{i=0}^{n-1}|f^{(k)}(a)|+\operatorname{max}_{x\in [a,b]}|f^{(n)}(x)|$ are equivallent.

It's easy to show that for every $f\in C^n([a,b])$, $\|f\|_{\sim}\leq n\|f\|$. So I know that $Id:(C^n([a,b]),\|.\|)\to (C^n([a,b]),\|.\|_{\sim})$ is continuous. Now I want to show that $Id:(C^n([a,b]),\|.\|_{\sim})\to (C^n([a,b]),\|.\|)$ is bounded. This is where I got stuck.
Any hint would be appreciated.
 A: Hint:
By Taylor's theorem,
\begin{align}
|f(x)|&=\left|\sum_{k=0}^{n-1}f^{(k)}(a)\frac{(x-a)^k}{k!}+f^{(n)}(\xi)\frac{(x-a)^n}{n!}\right|\\
&\leq \sum_{k=0}^{n-1}|f^{(k)}(a)|\frac{(b-a)^k}{k!}+\sup_{\xi \in [a,b]}|f^{(n)}(\xi)|\frac{(b-a)^n}{n!} \\
&\leq C_0\left(\sum_{k=0}^{n-1}|f^{(k)}(a)|+\sup_{\xi \in [a,b]}|f^{(n)}(\xi)| \right),
\end{align}
where
\begin{align}
C_0&=\max_{0\leq k \leq n}\frac{(b-a)^k}{k!}.
\end{align}
A similar idea holds for each derivative in that such a constant $C_k$ can be obtained for $k=0,1,\dots,n$. Then obviously, $\|f\|\leq\max\{C_0,\dots,C_k\}\|f\|_{\sim}$.
A: Suppose $X$ is a vector space with two norms $\|.\|_1$ and $\|.\|_2$ defined on it. Let $I:(X,\|.\|_1)\to(X,\|.\|_2)$ be the identity map as you defined, which is one-to-one and onto.

*

*completeness: If both $(X,\|.\|_1)$ and $(X,\|.\|_2)$ are Banach spaces,

*boundedness of I: $\exists M>0\hspace{4mm}\forall x\in X\hspace{4mm}\|x\|_2\leq M\|x\|_1$,

then by the Open Mapping Theorem,  $\big(\exists K>0\hspace{4mm}\forall x\in X\hspace{4mm}\|x\|_1\leq K\|x\|_2\big)$, that is $I^{-1}$ is also bounded.
