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Suppose $\Delta$, the forward differece operator on some functional space, $z$ complex number, does the equation $(I+\Delta)^z(f(z))= f(z)$ imply that $Re(z)=0$

by using pochhammer symbols, operator $$(I+\Delta)^z =\sum_{k=0}^{\infty}\frac{z^{(k)}}{k!}\Delta^k $$ The equation implies that $$ \sum_{k=1}^{\infty}\frac{z^{(k)}}{k!}\Delta^k(f(z))=0$$ In some sense, this looks like $$ \sum_{k=1}^{\infty}\frac{z^k}{k!}a^k=0$$ but the last equation implies $Re(z)=0$.

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    $\begingroup$ Not if $\Delta=0$. $\endgroup$
    – Ben McKay
    Commented Mar 24, 2017 at 22:08
  • $\begingroup$ The complex power of an operator is not define unless the operator is elliptic. But in this case you should be able to write out the corresponding Mellin transform rather explicitly and working with the trace. Pochhammer symbols (and functional calculus) is useful but may not tell you as much information. Personally I do not see why this has to be true. $\endgroup$
    – Bombyx mori
    Commented Mar 25, 2017 at 3:06

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Here is a very dumb answer. Let your operator to be $$ \Delta=(i-1)I, \Delta(f)=if-f $$ Then it is trivial to verify that $$ (1+\Delta)^{4}f=f, 4>0 $$ So for the problem to be interesting, some extra condition is needed.

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  • $\begingroup$ A very nice dumb answer. In my case, the $\Delta$ is the forward difference operator $\Delta f(x) = f(x+1)-f(x)$. thanks $\endgroup$
    – zhizhi Liu
    Commented Mar 25, 2017 at 23:19

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