are elements of $F[X; Y, \lambda]_c$, then $|f_\mu \pm g_\mu|_\lambda \leq \max\{|f_\mu|_\lambda, |g_\mu|_\lambda\}$, and the quantity $\max\{|f_\mu|_\lambda p^{c\cdot \mu}, |g_\mu|_\lambda p^{c\cdot \mu}\}$ converges to 0 as $|\mu|$ approaches $\infty$. Thus, $f+g$ is an element of $F[X; Y, \lambda]_c$.

Similarly, we see that $$\begin{aligned}
 \left|\sum_{\mu + \nu = \sigma}f_\mu g_\nu\right|_\lambda \leq \max_{\mu + \nu = \sigma}\{|f_\mu|_\lambda\cdot |g_\nu|_\lambda\}\end{aligned}$$ and $\lim_{|\sigma| \rightarrow \infty} \max_{\mu + \nu = \sigma}\{|f_\mu|\cdot |g_\nu|p^{c\cdot \sigma}\} = 0$ as desired. Thus, $fg$ is an element of $F[X; Y, \lambda]_c$.

Now to prove that that this norm is complete, we let $\left(f^{(i)}\right)_{i=1}^\infty = \left(\sum_{\mu}^\infty f_\mu^{(i)} X^\mu\right)_{i=1}^\infty$ be a Cauchy sequence in $F[X; Y, \lambda]_c$. Then we can choose a suitable subsequence of $\left(f^{(i)}\right)_{i=1}^\infty$ (because a Cauchy sequence is convergent if, and only if, it has a convergent subsequence) and assume that $$\begin{aligned}
 | f_\mu^{(j)} - f_\mu^{(i)} |_\lambda p^{c\cdot \mu} \leq \|f^{(j)} - f^{(i)}\|_{\lambda, c} < 1/i \quad \text{ for all } j > i > 0\text{ and all } |\mu|\geq 0.\end{aligned}$$ For all $j$ and $\mu$, $f_\mu^{(j)}$ is an element of $F[[Y]]$ which is complete, so there is an element $f_\mu$ in $F[[Y]]$ such that $f_\mu^{(j)}$ converges to $f_\mu$ as $j$ approaches $\infty$. Define $f =\sum_{\mu} f_\mu X^\mu$. We claim 