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$f$ is a function in $C^{\infty}$ with compact support and $A$ is a self-adjoint operator. If $\mathrm{supp}(f)\cap\sigma(A)=\varnothing$, does this imply that $f(A)=0$?

Does anyone have an explanation why this is true ?

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Yes, the condition implies that $f|_{\sigma(A)}=0$. So you are doing functional calculus with the zero function. And $C^\infty$ plays no role here, the argument works for any continuous function.

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