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Let $H$ be a Hilbert space and $D:H\rightarrow H$ be a densely-defined, unbounded self-adjoint operator, such that $D$ is a bounded operator when viewed as an operator

$$D:\text{dom}(D)\rightarrow H,$$

where $\text{dom}(D)$ is given the graph norm. Then one knows that

$$(D^2+1)^{-1}:H\rightarrow \text{dom}(D^2)$$

is a positive bounded operator with a square root, which we can denote by $(D^2+1)^{-1/2}$. I believe it follows from functional analysis that the range of $(D^2+1)^{-1/2} = \text{dom}(D)$.

Question: Is $(D^2+1)^{-1/2}$ a bounded operator $H\rightarrow \text{dom}(D)$, where $\text{dom}(D)$ is given the graph norm as above?

geometricK
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1 Answers1

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$(D^2+I)^{-1}$ is a bounded positive operator, and $((D^2+I)^{-1})^{1/2}$ is the unique positive square root of that bounded operator, which makes it positive. To see that $(D^2+I)^{-1}$ is bounded, note that, for all $f\in\mathcal{D}(D^2)$, $$ \langle f,f \rangle \le \langle (D^2+I)f,f\rangle \le \|(D^2+I)f\|\|f\| \\ \|f\| \le \|(D^2+I)f\|. $$ Setting $f = (D^2+I)^{-1}g$ for an arbitrary $g$ leads to the conclusion that $(D^2+I)$ is bounded by $1$ in operator norm: $$ \|(D^{2}+I)^{-1}g\| \le \|g\|. $$

Disintegrating By Parts
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  • I'm not sure how this answers the question, which was whether the square root of the inverse is a bounded operator? – geometricK Dec 12 '17 at 09:38
  • @ougoah : It's the square root of a bounded operator. – Disintegrating By Parts Dec 12 '17 at 09:39
  • Hmm ok, is it clear that it's bounded $H\rightarrow\text{dom}(D)$ where $\text{dom}(D)$ has the graph norm? – geometricK Dec 12 '17 at 09:42
  • I wonder if the question is equivalent to asking whether $\text{dom}(D)$ is equal to the domain of the square root of the unbounded operator $D^2+1$, as per your answer in https://math.stackexchange.com/questions/1061661/square-root-of-unbounded-operator? Is it true that $((D^2+1)^{-1})^{1/2}$ is the inverse of $(D^2+1)^{1/2}$? – geometricK Dec 12 '17 at 21:57