Foucault pendulum vector diagrams: Difference between revisions

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{{Unreferenced|date=December 2009}}
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In [[mathematics]], in the area of [[functional analysis]] and [[operator theory]], the '''Volterra operator''', named after [[Vito Volterra]], represents the operation of [[indefinite integration]], viewed as a [[bounded linear operator]] on the space ''L''<sup>2</sup>(0,1) of complex-valued [[square integrable function]]s on the interval (0,1). It is the operator corresponding to the [[Volterra integral equation]]s.
 
==Definition==
The Volterra operator, ''V'', may be defined for a function ''f''&nbsp;∈&nbsp;''L''<sup>2</sup>(0,1) and a value ''t''&nbsp;∈&nbsp;(0,1), as
 
:<math>V(f)(t) = \int_0^t{f(s)\, ds}.</math>
 
==Properties==
*''V'' is a bounded linear operator between Hilbert spaces, with [[Hermitian adjoint]]
 
::<math>V^*(f)(t) = \int_t^1{f(s)\, ds}.</math>
*''V'' is a [[Hilbert-Schmidt operator]], hence in particular is [[compact operator|compact]].<ref name='stackexchange_indefinite-integrators'>{{cite web|title=Spectrum of Indefinite Integral Operators (From stackexchange.com)|url=http://math.stackexchange.com/questions/151425/spectrum-of-indefinite-integral-operators}}</ref>
*''V'' has no [[eigenvalue]]s and therefore, by the [[spectral theory of compact operators]], its [[spectrum (functional analysis)|spectrum]] σ(''V'') = {0}.<ref name='stackexchange_indefinite-integrators' />
*''V'' is a [[quasinilpotent operator]] (that is, the [[spectral radius]], ''ρ''(''V''), is zero), but it is not [[nilpotent]].
*The [[operator norm]] of ''V'' is exactly ||''V''|| = <sup>2</sup>⁄<sub>π</sub>.<ref name='stackexchange_indefinite-integrators' />
 
==References==
{{Reflist}}
 
{{DEFAULTSORT:Volterra Operator}}
[[Category:Operator theory]]

Latest revision as of 20:23, 21 February 2014

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