Most Popular Thermodynamics Hipolito Sta Maria Solution Manual ➡️

Most Popular Thermodynamics Hipolito Sta Maria Solution Manual ➡️

Most Popular Thermodynamics Hipolito Sta Maria Solution Manual

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$\mathcal{D}(A)$ is dense in $X$, and $\mathcal{D}(A^{ -1})$ is dense in $\mathcal{D}(A)$

Let $X$ be a complex Hilbert space and $A$ a densely defined closed symmetric operator with non-empty resolvent set. The second isomorphic result claims that if $D(A)$ is dense in $X$, then $\mathcal{D}(A)$ is dense in $X$, and $\mathcal{D}(A^{ -1})$ is dense in $\mathcal{D}(A)$.
However, I cannot give a proof of this fact.
Consider $A:H \rightarrow H$ a densely defined closed symmetric operator and $\{x_n\} \subset D(A)$ such that $x_n \rightarrow x \in X$.
Then, $$Ax_n-Ax=x-x_n \in X^{\perp} \Rightarrow \|x_n-x\|=\|Ax_n-Ax\| \Rightarrow \|x-x_n\| \leq \|Ax_n-Ax\| \leq \|Ax_n-Ax\|=\|x_n-x\|$$
And since $\{x_n\}$ is a sequence of elements in $D(A)$ hence, $\forall \lambda \in \mathbb{C}$ such that $\lambda$ is not an eigenvalue of $A$, \|(\lambda I-A)x\|=\lim_{n \rightarrow \infty} \|\lambda x_n-Ax_n\|