$C^n[a,b]$ as a normed algebra

Question

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!!!

Answer 1

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.

Answer 2

$C^n([a,b])$ 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.