The neutral derivative shift has a bounded Neumann inverse when $\mathrm{\mid }a\mathrm{\mid }<1$

If the mesh array length is $M=0$, [F1] gives a space with exactly one vector $0$, of norm $0$, and a unique self-map. This map is both the zero operator and the identity $I$, and $S$ is this same operator. Thus $J=\lceil 0\mathrm{/}m\rceil =0$ and $S^J=S^0=I=0$. The map $I-aS$ is also the unique map, whose composite with itself is the identity; hence it is its own inverse. The finite sum over $0\le j<J$ is the zero operator. Every term of the infinite series is also zero, including the $j=0$ term because $S^0=I=0$ as an operator on this space. Finally the operator-norm supremum is over the nonempty unit ball $\{0\}$ and equals $0$. This proves all stated assertions for $M=0$, without an empty real maximum. For the remaining proof use either the segment case $K\ge 1$ or a mesh array with $M\ge 1$.

By [F1], every output coordinate of either shift is zero or an input coordinate. Taking the maximum of the component suprema therefore gives $\mathrm{\parallel }Sv\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }$, so $\mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1$. After $K$ segment shifts every component is zero; after $J=\lceil M\mathrm{/}m\rceil$ mesh shifts every array index has moved below zero and is zero. Thus $S^J=0$ in both cases.

Put $R={\sum }_{j=0}^{J-1}{a}^j{S}^j$. Multiplying the finite sum on either side by $I-aS$ cancels adjacent terms and leaves $I-a^J{S}^J=I$, so $R$ is a two-sided inverse. Since $S^J=0$, this finite sum equals the displayed infinite series.

Iterating the contraction bound in step 1.2 gives $\mathrm{\parallel }{S}^{j}v\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }$ for every $j$, hence $\mathrm{\parallel }{S}^j{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1$. Thus $\mathrm{\parallel }R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le {\sum }_{j=0}^{J-1}\mathrm{\mid }a{\mathrm{\mid }}^j$. Multiplying this finite geometric sum by the positive number $1-\mathrm{\mid }a\mathrm{\mid }$ gives $1-\mathrm{\mid }a{\mathrm{\mid }}^J\le 1$, hence $\mathrm{\parallel }R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le (1-\mathrm{\mid }a\mathrm{\mid }{)}^{-1}$.