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Ji-Zhang Question on the Power Set of a Quasinilpotent Operator

Douglas and Yang attach to each nonzero vector $x$ of a quasinilpotent operator $T$ a local resolvent-growth exponent $k_x$, giving the power set $\Lambda(T) = \{k_x : x \ne 0\}$. Ji and Zhang asked whether $1$ always belongs to $\Lambda(T)$. It does, for every quasinilpotent operator on every Banach space. Moreover $\Lambda(T) = [0,1]$ for every backward unilateral weighted shift on $\ell^p$ with strictly decreasing, $p'$-summable weights, weakening the hypotheses of Hu and Ji.

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analysisJul 18, 2026Significance 9/100Registry: unreviewed

Ji-Zhang Question on the Power Set of a Quasinilpotent Operator

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Douglas and Yang attach to each nonzero vector $x$ of a quasinilpotent operator $T$ a local resolvent-growth exponent $k_x$, giving the power set $\Lambda(T) = \{k_x : x \ne 0\}$. Ji and Zhang asked whether $1$ always belongs to $\Lambda(T)$. It does, for every quasinilpotent operator on every Banach space. Moreover $\Lambda(T) = [0,1]$ for every backward unilateral weighted shift on $\ell^p$ with strictly decreas…

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Douglas and Yang attach to each nonzero vector $x$ of a quasinilpotent operator $T$ a local resolvent-growth exponent $k_x$, giving the power set $\Lambda(T) = \{k_x : x \ne 0\}$. Ji and Zhang asked whether $1$ always belongs to $\Lambda(T)$. It does, for every quasinilpotent operator on every Banach space. Moreover $\Lambda(T) = [0,1]$ for every backward unilateral weighted shift on $\ell^p$ with strictly decreasing, $p'$-summable weights, weakening the hypotheses of Hu and Ji.

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