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Show that $u_0-u_1+u_2-u_3+...=\frac{1}{2}u_0-\frac{1}{4}\Delta u_0+\frac{1}{8}\Delta^2 u_0-\frac{1}{16}\Delta^3 u_0+...$, where $\Delta$ is the forward difference operator.

My attempt:

$(u_0+u_2+u_4+...)-(u_1+u_3+u_5+...)=(u_0+E^2u_0+E^4u_0+...)-(Eu_0+E^3u_0+E^5u_0+...)$

$E$ being shift operator.

  • $1-0+0-0+-\ldots=\frac{1}{2}-\frac{1}{4}+0-0+-\ldots$? – Leonhardt von M Apr 07 '15 at 09:19
  • $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+-\ldots = \frac{2}{3} $ but $\frac{1}{2} -\frac{1}{4}\frac{-1}{2}+\frac{1}{8}\frac{-1}{4}-\frac{1}{16}\frac{-1}{8}+-$ $\ldots =\frac{1}{2}+\frac{1}{8}-\frac{1}{32}+\frac{1}{128}\ldots=\frac{3}{5}$. – Leonhardt von M Apr 07 '15 at 10:24
  • If $u_k=\delta_{0,k}$, then $(Δ^mu)_0=(-1)^m$ and the right side is $\frac12+\frac14+\frac18+...=1$. In the second case $u_k=2^{-k}$ so that $(Δ^mu)_0=(-2)^{-m}$ and the right side is $\frac12(1+\frac14+\frac1{16}+\frac1{64}+...)=\frac23$. – Lutz Lehmann Jan 11 '23 at 14:46

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Let $T$ be the unit translation operator, $(T^ku)_0=u_k$. Then $Δ=T-1$, $T=1+Δ$, $$ \sum_{k=0}^\infty (-T)^k=\frac1{1+T}=\frac1{2+Δ}=\frac12·\frac1{1-(-\frac12Δ)}=\frac12\sum_{k=0}^\infty \left(-\frac12Δ\right)^k. $$ Which is all rather questionable since the norm and spectral radius of the translation operator is $1$ for the standard norms and thus the geometric or Neumann series does not converge. But these transformations may serve as general guideline for a correct proof in a distributional sense.

Considering the modified supremum norm norm $\|u\|_r=\sup_k r^k|u_k|$ for some $r>1$ one finds that in operator norm $\|T\|=r^{-1}<1$ and $\frac12\|Δ\|\le\frac{1+r^{-1}}2<1$, so that in this topology on the sequence space the identities are also correct as converging sequences.

Lutz Lehmann
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