if, and only if, $f \equiv c_0 \mod (Y)$ where $c_0$ is a unit in $F$. If an element $f$ in $F[X; Y, \lambda]_c$ is invertible in $F[X; Y, \lambda]$, then $f^{-1}$ is an element of $F[X; Y, \lambda]_c$. Further, if $\|f\|_{\lambda, c} \leq 1$, then $\| f^{-1}\|_{\lambda, c} \leq 1$.

If $f(X,Y)$ is invertible, then it is an invertible polynomial modulo $(Y)$. Therefore, $f$ is a non-zero unit modulo $(Y)$.

If $f \equiv c_0 \mod (Y)$ for $c_0$ in $F^\times$, we can write $f = c_0(1 - g(X,Y))$ as an element of $F[X;Y,\lambda]_c$ for some $c>0$. Then $$\begin{aligned}
 f^{-1} &=&  c_0^{-1}\left(1 + \sum_{k=1}^{\infty}g(X,Y)^k\right) \\
&=& c_0^{-1}\left(1 + \sum_{k=1}^\infty\sum_{j=1}^\infty
\sum_{\substack{\mu^{(1)} + \cdots + \mu^{(k)} = \sigma \\
|\sigma| = j} }\prod_{i=1}^k g_{\mu^{(i)}}(Y) X^\sigma\right)\end{aligned}$$ Observe that, $$\begin{aligned}
 \left(\left|\prod_{i=1}^k g_{\mu^{(i)}}(Y) X^\sigma\right|_\lambda\right) p^{c\cdot \sigma} &=& \prod_{i=1}^k \left(\left|g_{\mu^{(i)}}(Y) X^\sigma\right|_\lambda p^{c\cdot\mu^{(i)}}\right)\end{aligned}$$ converges to $0$ as $|\sigma|$ approaches $\infty$ because $g_\mu = -c_0f_\mu$. Suppose $\|f\|_{\lambda, c} \leq 1$, then this product is also less than or equal to one, because each term satisfies this property, so $\|f^{-1}\|_{\lambda, c} = |c_0^{-1}|_\lambda = 1$.

The ring, $(F[X; Y, \lambda]_c, \| \; \|_{\lambda, c})$, is an $F[[Y]]$-Banach algebra.

Suppose that $f = \sum_{\mu} f_\mu(Y)X^\mu$ and $g = \sum_{\nu} g_\nu(Y)X^\nu$, 