$C^n[a,b]$ as a normed algebra I would like to prove that the space of complex valued functions, differentiable $n$ times with continuous derivative, $C^n[a,b]$, with the metric defined by the norm $$\|x\|=\sum_{k=0}^n\frac{1}{k!}\max_{a\leq t\leq b}|x^{(k)}(t)|$$ is a normed algebra, but I cannot verify that $\forall x,y\in C^n[a,b]\quad \|xy\|\leq \|x\|\cdot\|y\|$.
Could anyone help me?
Thank you so much!!!
 A: $C^n([a,b])$ is the space of the functions over $[a,b]$ such that $f^{(n)}$ is continuous (otherwise, we would not be able the take a $\max$ in the definition of the norm, but just a $\sup$). Anyway,
$$\frac{1}{k!}\frac{d^k}{dx^k}(f\cdot g)=\sum_{j=0}^{k}\frac{f^{(j)}}{j!}\cdot\frac{g^{(k-j)}}{(k-j)!}$$
while:
$$\|f\|\cdot\|g\| = \sum_{0\leq a,b\leq n}\frac{\max |f^{(a)}|}{a!}\cdot \frac{\max |g^{(b)}|}{b!}$$
hence
$$ \|f\cdot g\| \leq \|f\|\cdot\|g\| $$
just follows from the triangle inequality.
A: We have that
$$
(fg)^{(k)}(x)=\sum_{j=0}^k\binom{k}{j}f^{(j)}(x)g^{(k-j)}(x),
$$
hence
$$
\lvert (\,fg)^{(k)}(x)\rvert\le\sum_{j=0}^k\binom{k}{j }\lvert\, f^{(j)}(x)\rvert\lvert
g^{(k-j)}(x)\rvert\le \sum_{j=0}^k\binom{k}{j }\max_{x\in[a,b]}\lvert\, f^{(j)}(x)\rvert
\cdot \max_{x\in[a,b]}\lvert\, g^{(k-j)}(x)\rvert,
$$
and thus
$$
\frac{\max_{x\in[a,b]}\lvert (\,fg)^{(k)}(x)\rvert}{k!}\le\sum_{j=0}^k\frac{1}{j!(k-j)!}\max_{x\in[a,b]}\lvert\, f^{(j)}(x)\rvert
\cdot \max_{x\in[a,b]}\lvert\, g^{(k-j)}(x)\rvert,
$$
and summing over $k$ we obtain
$$
\sum_{k=0}^n\frac{\max_{x\in[a,b]}\lvert (\,fg)^{(k)}(x)\rvert}{k!}\le
\sum_{k=0}^n\sum_{j=0}^k\frac{1}{j!(k-j)!}\max_{x\in[a,b]}\lvert\, f^{(j)}(x)\rvert
\cdot \max_{x\in[a,b]}\lvert\, g^{(k-j)}(x)\rvert \\
=\left(\sum_{k=0}^n \frac{\max_{x\in[a,b]}\lvert (\,f)^{(k)}(x)\rvert}{k!}\right)\cdot
\left(\sum_{k=0}^n \frac{\max_{x\in[a,b]}\lvert (\,g)^{(k)}(x)\rvert}{k!}\right),
$$
which is what we are looking for.
