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In a physics paper I am currently reading, the following statement seems to be used: Let $A$ be a (not necessarily bounded) positive self-adjoint operator on $L^2(a,b)$ with $C_0^{\infty}(a,b)\subseteq \mathrm{dom}(A)$. Then $\Phi(t)=\cos(A^{1/2}t)$ is well-defined for $t\in\mathbb{R}$.

Is this statment true? And if yes, how can I prove it?

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Is it $\phi(t)=\cos(A^{1/2}t)$? –  Mhenni Benghorbal Mar 18 '13 at 13:54
    
Yes, that's better. –  Scipio Mar 18 '13 at 14:02
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I don't have time to give a full answer right now, but you should look into the Borel functional calculus. Not only is $\Phi(t)$ defined for all $t$, it extends to a bounded operator on $L^2(a,b)$ since $\cos$ is bounded on $\mathbb{R}$. A reference I particularly like for material on operator algebras and functional analysis is Kadison and Ringrose's "Fundamentals of the Theory of Operator Algebras." Volume I is the appropriate one for this material. –  J. Loreaux Mar 18 '13 at 14:33
    
What you are doing is nothing but an extension of the idea of functions of matrices to bounded linear operators. Read under Cauchy Integral. –  Mhenni Benghorbal Mar 18 '13 at 22:57
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You should look at the Spectral theorem for unbounded self-adjoint operators. See Michael E. Taylor's Partial Differential Equations Vol II page 97, equation (1.30) and Theorem 1.7. This is the Borel functional calculus mentioned above. Once you do square root and exponential, you can get to cosine. Also see the discussion in the second paragraph of page 98, which gives the boundedness of the operator.

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