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Currently, I am learning the functional calculus for unbounded operators. Let $T$ be a closed and densely defined normal operator on some Hilbert space. Then for any Borel measurable function $f$ defined on the spectrum of $T$ we can define $f(T).$ Now there are two ways of defining $T^2.$ 1. Just define $T^2$ as the usual composition of the operator $T$. Call this $\tilde{T}^2.$ 2. By functional calculus we can also define $T^2$ as $f(T),$ $f(z)=z^2.$ Call this $T^2$

My question-- is $T^2=\tilde{T}^2.$ I can see that the closure of $\tilde{T}^2=T^2.$ can someone help me to understand what is going on here?

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In the functional calculus, $$ x\in\mathcal{D}(T) \iff \int_{\sigma}|\lambda|^2d\|E(\lambda)x\|^2 < \infty,$$ and $$ Tx = \int_{\sigma}\lambda dE(\lambda)x,\;\; x\in\mathcal{D}(T). $$ This is a fundamental result that is always proved as part of the Spectral Theorem for unbounded normal operators.

This result can be extended by induction to show that, for $n=1,2,3,\cdots$, $$ x\in\mathcal{D}(T^n)\iff \int_{\sigma}|\lambda|^{2n}d\|E(\lambda)x\|^2 < \infty $$ and $$ T^nx = \int_{\sigma}\lambda^n dE(\lambda)x,\;\; x\in\mathcal{D}(T^n). $$

A proof can be obtained by applying the projection $E_r = E\{\lambda : |\lambda| \le r\}$, and using an argument where $r\uparrow\infty$. Start by showing that $E_r x \in\mathcal{D}(T^n)$ for all $n=1,2,3,\cdots$ and $T^n E_r x =\int_{|\lambda| \le r}\lambda^n dE(\lambda)x$.

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