# The part of an operator as an analytic generator

Let the operator $A$ be the generator of an analytic semigroup on a Banach space $X$. $Y$ is another Banach space embedded in $X$. $A_Y$, the part of $A$ in $Y$ is defined as the operator with domain

$$D(A_Y) := \{ y \in D(A) \cap Y: Ay \in Y \}$$ and $$A_Y \ y := Ay$$

Then it seems to me that $A_Y$ is the generator of an analytic semigroup on $Y$. I didn't find a proof, so I'm asking if someone can give a reference or counterexample if it is not true.

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What is the analytic generator? More details could help. Also, FYI: in order to get brace brackets to show up, you need to precede them by two backslashes rather than the usual one. – MTS Feb 25 '12 at 1:16
By analytic generator is meant that $A$ generates analytic semigroup of operators on $X$. Now what I feel is true but I am not completely sure is that the part $A_Y$ of $A$ also generates analytic semigroup in $Y$. – Martin Feb 25 '12 at 7:54
Sorry I need to add one correction. $Y$ is actually not a subspace of $X$, but another Banach space which is continuously embedded in $X$. I suppose this makes things more interesting. – Martin Feb 25 '12 at 9:04
What's the "analytic semigroup in "Y"?? If $(T_t)_{t\geq 0}$ is your semigroup, is this the assumption that each $T_t$ restricts to an operator on $Y$? Or something else? – Matthew Daws Feb 25 '12 at 9:06
Please edit your original question, rather than leaving "corrections" as comments... – Matthew Daws Feb 25 '12 at 9:06

This is wrong. Let us assume that $Y$ is a closed subspace of $X$ to clearify the problem. As Matthew Daws already said, you have to assume that the semigroup $(T(t))$ generated by $A$ leaves $Y$ invariant: suppose that $A_Y$ indeed generates a (strongly continuous) semigroup $(S(t))$ on $Y$. Then for example the Yosida-approximation shows that $T_Y(t) = S(t)$ for all $t$.
Thank you Stephan. The motivation for this question comes from Cauchy problems where the initial value is often taken to be in some intermediate space of the domain of the operator $A$ and the space X. This operator is usually analytic generator so I was wondering if this il also the case for its part. I didn't quite understand if it is never a generator or it isn't in general. Can you explain it once more? – Martin Feb 25 '12 at 17:38
$A_Y$ can be a generator, but is not in general. If $Y$ is a closed subspace as in your original question, the argument above shows that the invariance of $(T(t))$ is necessary. It is also sufficient: $(T(z))$ can be restricted to an analytic semigroup on $Y$ with generator $A_Y$. So the problem is not interesting in this case. However, if $Y$ is only continuously embedded into $Y$, your question is interesting and has a positive answer in natural situations: for example if $X$ is a Hilbert space and $A$ is induced by a form, see "Analysis of Heat Equations" by Ouhabaz if you're interested. – Stephan Fackler Feb 25 '12 at 19:45
@ Andras Batkai: Also the condition $X$ to be a Hilbert space that Stephan states for this thing to work is restrictive. Do you treat this or similar kind of problems in your book "Semigroups for Delay Equations"? – Martin Feb 25 '12 at 23:15
Let me add a few words: in fact, IF a subspace $Y$ (any subspace! it need not be closed!) is invariant under the semigroup AND the restriction of the semigroup to $Y$ is strongly continuous there, THEN it is known that the part of the generator in $Y$ is the generator of the restriction semigroup. If you need a reference, take a look at the beginning of Engel-Nagel's book, but in fact the proof is quite easy: one direction is simply definition, other one you want to apply the representation of the resolvent as a Laplace transform of the semigroup. – Delio Mugnolo Nov 23 '12 at 0:43