If $\| f\|_k = \sum_{r=0}^k \frac{1}{r!}$, proof that $\|fg\|_k \leq \|f\|_k \, \|g\|_k$. Let $X \subset \mathbb C$ be a perfect and compact. Denote the algebra of functions on $X$ with continuous $k$th derivative by $D^{k} (X)$. I'm trying to proof that if $$\| f\|_k = \sum_{r=0}^k \frac{1}{r!}\, |f^{(r)}|_X, \text{ where } |g|_X = \sup_{x\in X} |g(x)|,$$
then $\|f\|_k$ defines a norm in the algebra $D^{k}(X)$.
It's not difficult see that $\|f\|_k = 0 \iff f = 0$, $\|\lambda f \|_k = |\lambda| \, \|f\|_k$ and $\|f+g\|_k \leq ||f||_{k} + ||g||_{k}$. My problem is showing that $\|fg\|_k \leq \|f\|_k \, \|g\|_k$.
Given $f,g \in D^{k} (X)$, we have
$|(fg)^{(r)}|_X = | \sum_{j=0}^r \binom{r}{j} \, f^{(j)}\, g^{(r-j)} |_X \leq \sum_{j=0}^r \binom{r}{j} \, | f^{(j)}\, g^{(r-j)} |_X \leq \sum_{j=0}^r \binom{r}{j} \, | f^{(j)} |_X \, |g^{(r-j)} |_X$
hence
$$\|fg\|_k \leq \sum_{r=0}^k \frac{1}{r!} \, \left [ \sum_{j=0}^r \binom{r}{j} \, | f^{(j)} |_X \, |g^{(r-j)} |_X \right ] \leq \sum_{r=0}^k \sum_{j=0}^r \frac{1}{j! (r-j)!}\, | f^{(j)} |_X \, |g^{(r-j)} |_X$$
I'm stuck in this part. Help?
 A: Change the order of summation. As $j$ ranges from $0$ to $k$, $r$ ranges from $j$ to $k$. Thus
$$\sum_{r = 0}^k \sum_{j = 0}^r \frac{1}{j!(r-j)!} \lvert f^{(j)}\rvert_X \lvert g^{(r-j)}\rvert_X = \sum_{j = 0}^k \sum_{r = j}^k\frac{1}{j!(r-j)!} \lvert f^{(j)}\rvert_X \lvert g^{(r-j)}\rvert_X = \sum_{j = 0}^k \frac{1}{j!}\lvert f^{(j)}\rvert_X \sum_{r = 0}^{k-j} \frac{1}{r!}\lvert g^{(r)}\rvert_X $$
For $0 \le j \le k$, $$\sum_{r = 0}^{k-j} \frac{1}{r!}\lvert g^{(r)}\rvert_X \le \sum_{r = 0}^k \frac{1}{r!}\lvert g^{(r)}\rvert_X = \|g\|_k$$
Hence
$$\sum_{j = 0}^k \frac{1}{j!}\lvert f^{(j)}\rvert_X \sum_{r = 0}^{k-j} \frac{1}{r!}\lvert g^{(r)}\rvert_X \le \sum_{j = 0}^k \frac{1}{j!}\lvert f^{(j)}\rvert_X \|g\|_k = \|f\|_k\, \|g\|_k$$
A: you can continue like this:
$$\sum_{j} \sum_{r} \frac{1}{j!(r-j)!} |f^{(j)}|_{X}|g^{(r-j)}|_{X} = \sum_{j} \frac{1}{j!}|f^{(j)}|_{X} \sum_{r} \frac{1}{(r-j)!} g^{(r-j)}|_{X} = \left \|  f\right \|_{k} \sum_{r} \frac{1}{r!}|g^{(r)}|_{X}|=\left \|  f\right \|_{k}\left \|  g\right \|_{k}$$
 If you got stuck with things like this is usually useful start from the other side of the inequality and meet in the middle.
