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I am looking at the proof of the following version of the Hille-Yosida theorem given in Ethier and Kurtz' Markov Process.

Some Definitions.

Here, $A$ is a (multivalued) linear operator on $L$, i.e. a subset $A$ of $L \times L$ with domain $D(A)=\{f:(f,g)\in A \;\text{for some}\;g\}$ and range $R(A)=\{g: (f,g)\in A \; \text{for some}\;f\}$. $A\subset L\times L$ is said to be linear if $A$ is a subspace of $L\times L$. If $A$ is linear, then $A$ is said to be single-valued if $(0,g)\in A$ implies $g=0$, in this case, $A$ is a graph of a linear operator on $L$, also denoted by $A$, so we write $Af=g$ if $(f,g)\in A$. If $A\subset L\times L$ is linear, then $A$ is said to be dissipative if $\Vert \lambda f-g\Vert \ge \lambda \Vert f\Vert $ for all $(f,g)\in A$ and $\lambda >0$; the closure $\bar{A}$ of $A$ is just the closure in $L\times L$ of the subspace $A$. Finally, we define $\lambda - A=\{(f,\lambda f-g):(f,g)\in A\}$ for each $\lambda >0$.

Lemma preceding the theorem.

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Below is the Theorem. enter image description here

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Now in the below part of the proof, I don't understand why $\Vert \lambda(\lambda - \bar{A})^{-1}f-f\Vert = \Vert(\lambda - \bar{A})^{-1}g\Vert$ which is above (4.2). Also, why is the domain of $(\lambda - \bar{A})^{-1}$ given as $R(\lambda - A_0)$? Shouldn't it be $R(\lambda-\bar{A})$ by definition? Finally, I don't understand why the range is $D(A_0)$ either. I would greatly appreciate if anyone could explain these to me.

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First note that $R(\lambda - A_0) \subset R (\lambda - \bar A)$. This follows from $A_0 \subset \bar A$.

With this in mind, you can define the restriction of $(\lambda - \bar A)^{-1}: R (\lambda - \bar A) \to D(\bar A)$ to the function $(\lambda - \bar A)^{-1}: R (\lambda - A_0) \to D(A_0)$. This is not the perfect notation choice, since it can lead to confusion but when the authors write $(\lambda - \bar A)^{-1}: R (\lambda - A_0) \to D(A_0)$ they mean the restriction of $(\lambda - \bar A)^{-1}$ to the set $R (\lambda - A_0)$. Note further that if $ h = (\lambda - A_0)f$ then $(\lambda - A_0)^{-1}h = f \in D(A_0)$ so the function above is onto $D(A_0)$.

Now take $(f,g) \in \bar A$. The condition $R(\lambda - A) \supset \bar {D(A)}$ implies that $f, g \in R (\lambda - \bar A)$.

this means that there is $h$ such that $(\lambda - \bar A)h = f$ and $\varphi$ such that $(\lambda - \bar A)\varphi = g$.

therefore $(\lambda - \bar A)^{-1}f = h$ : $$(\lambda - \bar A) (\lambda h - f) = \lambda(\lambda - \bar A)h - (\lambda - \bar A)f = \lambda f - (\lambda f - \bar A f) = \bar A f = g $$ which implies that $$(\lambda - \bar A)^{-1} g = (\lambda h - f) = \lambda (\lambda - \bar A)^{-1}f - f $$

There is one last comment I would like to add (even though you didn't ask about it) and that is why does this implies (4.2).

from the relation

$$ \|\lambda (\lambda - \bar A)^{-1}f - f\| \leq \lambda^{-1}\|g \|$$

we conclude that for $f \in D(\bar A)$ $$\lim_{\lambda \to \infty} \lambda (\lambda - \bar A)^{-1}f = f $$

now note that $D(A) \subset D(\bar A) \subset \bar{D(A)}$, this implies that $\bar{D(\bar A)} = \bar{D(A)} $ =. That is what they meant to say when they wrote "Since $ D(\bar A)$ is dense in $D( A)$". In fact they want to write "Since $ D(\bar A)$ is dense in $\bar{D( A)}$".

Now take $\hat f \in \bar {D(A)}$. consider $f_n \to \hat f$ with $f_n \in D(\bar A)$.

by the triangle inequality and using that $\|(\lambda - \bar A)^{-1}\varphi\|\leq \lambda^{-1}\varphi$ we obtain

$$ \| \lambda (\lambda - \bar A)^{-1}f - f\| \leq \| \lambda (\lambda - \bar A)^{-1}(f - f_n)\| + \| \lambda (\lambda - \bar A)^{-1}f_n - f_n\| + \| f_n - f\|\\ \leq 2 \| f_n - f|\ + \| \lambda (\lambda - \bar A)^{-1}f_n - f_n\|$$

taking the limit in $\lambda$ yields:

$$ \lim_{\lambda \to \infty}\| \lambda (\lambda - \bar A)^{-1}f - f\| \leq 2 \| f_n - f\| $$

now, since $f_n\to f$ we can take for arbitrary $\epsilon > 0$, $n $ such that $$ \lim_{\lambda \to \infty}\| \lambda (\lambda - \bar A)^{-1}f - f\| \leq \epsilon $$ This proves (4.2).

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  • $\begingroup$ Thanks for your excellent work. Can you just clear me with a line in the middle? $D(A) \subset D(\bar A) \subset \bar{D(A)}$, this implies that $\bar{D(\bar A)} = \bar{D(A)} $ =. Why do we have $D(\bar{A})\subset \bar{D(A)}$ and how does this imply $\bar{D(\bar A)} = \bar{D(A)} $? $\endgroup$ – takecare Mar 5 '17 at 2:07
  • $\begingroup$ $\bar A$ is an extension of $A$ so $D(A) \subset D(\bar A)$ (the authors made a typo). Now if $A,B$ are sets, and $A \subset B \subset \bar A$ then $\bar B = \bar A$ since $B \supset A$ implies $\bar B \supset \bar A$ (remember, if $B \supset A$ the smallest closed extension containing B must contain A.... this allows the smallest closed extension of $A$ to be smaller than the smallest closed extension of B) $\endgroup$ – Conrado Costa Mar 5 '17 at 7:31
  • $\begingroup$ Can you help me with two additional questions? In the first paragraph of the proof, why is $R(\lambda - A_0)=\bar{D(A)}$ equivalent to $R(\lambda-\bar{A})\supset \bar{D(A)}$, and why does it suffice, to complete the proof, show that $D(A_0)$ is dense in $\bar{D(A)}$ assuming that $R(\lambda-A_0)=\bar{D(A)}$? $\endgroup$ – takecare May 23 '17 at 17:41
  • $\begingroup$ Also, why does the dissipativity of $\bar{A}$ make it possible for us to regard $(\lambda-\bar{A})^{-1}$ as a single-valued bounded linear operator? $\endgroup$ – takecare May 23 '17 at 18:39

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