The operator $id/dx,$ on $C[0,1],$ is $C^ 1$-scalar

Abstract:

We consider closed, unbounded linear operators $A$, on a Banach space, $X$, with discrete real spectrum, with corresponding eigenvectors, $\{ {x_k}\} _{ - \infty }^\infty$, such that $A{x_k} = {a_k}{x_k},{a_k} \leq {a_{k + 1}}$, for all $k$, and ${\lim _{k \to \pm \infty }}\left | {{a_k}} \right | = \infty$. For arbitrary $n$, we present necessary and sufficient conditions for $A$ to be ${C^n}$-scalar. Letting $F(s)$ be the projection whose range is the ${\text {span}}\{ {x_k}|{a_k}\;{\text {is between }}0{\text { and }}s\}$, null-space is the closure of ${\text {span}}\{ {x_k}|{a_k}\;{\text {is not between }}0{\text { and }}s\}$, we show that $A$ is ${C^1}$-scalar if and only if the series \[ \sum \limits _{{a_k} > 0} {\varphi (F({a_k})x)\left ( {\frac {1}{{{a_k}}} - \frac {1}{{{a_{k + 1}}}}} \right )} \quad {\text {and}}\quad \sum \limits _{{a_k} < 0} {\varphi (F({a_k})x)\left ( {\frac {1}{{{a_{k - 1}}}} - \frac {1}{{{a_k}}}} \right )} \] both converge absolutely, for all $\varphi$ in ${X^ * },x$ in $X$. As a corollary, we get that $id/dx$, on $\{ f{\text { in }}C[0,1]|f(0) = f(1)\}$, is ${C^1}$-scalar. Also, $id/dx$, on ${L^1}[0,1]$, is ${C^1}$-scalar.