In this paper we study the question whether $A^{-1}$ is the infinitesimal generator of a bounded C₀-semigroup if A generates a bounded C₀-semigroup. If the semigroup generated by A is analytic and sectorially bounded, then the same holds for the semigroup generated by $A^{-1}$. However, we construct a contraction semigroup with growth bound minus infinity for which $A^{-1}$ does not generate a bounded semigroup. Using this example we construct an infinitesimal generator of a bounded semigroup for which its inverse does not generate a semigroup. Hence we show that the question posed by deLaubenfels in [13] must be answered negatively. All these examples are on Banach spaces. On a Hilbert space the question whether the inverse of a generator of a bounded semigroup also generates a bounded semigroup still remains open.
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In this paper equivalent conditions for exact observability of diagonal systems with a one-dimensional output operator are given. One of these equivalent conditions is the conjecture of Russell and Weiss (1994). The other conditions are given in terms of the eigenvalues and the Fourier coefficients of the system data.
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In this paper we show that from an estimate of the form $sup_{t≥0} ||C(t) - cos(at)I|| < 1$, we can conclude that C(t) equals cos(at)I. Here $(C(t))_{t≥0}$ is a strongly continuous cosine family on a Banach space.
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We show that the growth rates of solutions of the abstract differential equations ẋ(t) = Ax(t), $ẋ(t) = A^{-1} x(t)$, and the difference equation $x_{d}(n+1) = (A+I)(A-I)^{-1} x_{d}(n)$ are closely related. Assuming that A generates an exponentially stable semigroup, we show that on a general Banach space the lowest growth rate of the semigroup $(e^{A^{-1}t})_{t≥0}$ is O(∜t), and for $((A+I)(A-I)^{-1})ⁿ$ it is O(∜n). The similarity in growth holds for all Banach spaces. In particular, for Hilbert spaces the best estimates are O(log(t)) and O(log(n)), respectively. Furthermore, we give conditions on A such that the growth rate of $((A+I)(A-I)^{-1})ⁿ$ is O(1), i.e., the operator is power bounded.
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