Adjoint of $\lambda I - T$

Question

Given a selfadjoint (maybe unbounded) operator $T$ on a Hilbert space $H$, I want to calculate the adjoint of $\lambda I - T$ for a $\lambda \in \Bbb C$.

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I am tempted to argue as follows:

According to Lemma 4.26 of https://people.math.ethz.ch/~kowalski/spectral-theory.pdf the adjoint of $(S+T)$ (with $S \in L(H)$) is given by $S^\ast + T^\ast$. The Lemma is also valid for selfadjoint operators $T$, because a selfadjoint operator is closed.

Note that $S := \lambda I$ is a bounded operator on $H$ (it has operatornorm $|\lambda|$). Therefore we can apply the above Lemma: $(\lambda I - T)^\ast = (\lambda I)^\ast - T^\ast$.

Furthermore it holds $(\lambda I)^\ast = \overline{\lambda} I^\ast = \overline{\lambda} I$ (Eq1).

This results in $(\lambda I - T)^\ast = \overline{\lambda} I - T$.

Is this reasoning correct? Especially (Eq.1)?

Answer 1

You can just use the definition. The adjoint of $A$ is defined as the operator $A^*$ such that $\langle Ax,y\rangle = \langle x,A^* y\rangle$. Therefore, for $A=\lambda I - T$,

$$ \langle Ax,y\rangle=\langle (\lambda I-T)x,y\rangle = \lambda \langle x,y\rangle - \langle Tx,y\rangle = \langle x,\overline{\lambda}y\rangle - \langle x,T^*y\rangle = \langle x, (\overline{\lambda}I-T^*)y\rangle = \langle x,A^*y\rangle $$

Answer 2

Yes, your reasoning is correct (since closed operators are closable), so we can indeed include that $$ (\lambda I - T)^* = \overline \lambda I - T^* $$