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Let $\{T(t),t\ge 0\}$ be a $C_0$ semigroup on a Hilbert space $X$, does that exist a larger Hilbert space $Y$ such that $X\subset Y$, and $T(t)$ extend to a $C_0$ group $T'(t)$ (so $t<0$ make sense now) on $Y$?

Edit: Here "extend" means that $(T(t)x, y)=(T'(t)x, y)$ for all $x, y\in X$, which is weaker than its usual meaning. If the semigroup $T(t)$ is a contraction, then the conclusion is true, see Theorem 8.1 (p.29) in "Harmonic analysis of operators on Hilbert spaces" written by Sz. Nagy, C. Foias, H. Bercovici and L. Kérchy.

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  • $\begingroup$ For something like the heat semigroup on $L^2(\mathbb{R})$, it's difficult to imagine what that would look like. $\endgroup$ Oct 11, 2015 at 15:47
  • $\begingroup$ More specifically, $e^{\Delta}$, say, has an unbounded inverse on $X=L^2$, and making the Hilbert space larger cannot fix this. $\endgroup$ Oct 11, 2015 at 17:08
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    $\begingroup$ There are well known examples of semigroups for which T(t) is not injective for sufficiently large t. Clearly such a semigroup cannot be extended to a group on any space. $\endgroup$ Oct 11, 2015 at 21:05
  • $\begingroup$ @NateEldredge , I also feel hard to believe at the beginning. But if the semigroup is a contraction (e.g. the heat semigroup), then there do exist such an "extension", see Thm 8.1 in the book I mentioned above $\endgroup$
    – Tomas
    Oct 12, 2015 at 12:37
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    $\begingroup$ I cannot read @MichaelRenardy's mind. But for example you can consider for $T$ the nilpotent left-shift on $X=L^2(0,1)$. Then $T(t)=0$ for all $t\geq 1$, and you cannot extend this to a group in the ordinary way. However, in this example the left-shift on $Y=L^2(\mathbb{R})$ is indeed an extension of $T$ in the sense you defined in your edit. $\endgroup$
    – gsa
    Oct 13, 2015 at 13:22

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An operator $A$ is a group generator if and only if both $A$ and $-A$ are semigroup generators. So under your assumptions, if we denote by $\tilde{A}$ the generator of the extension $\{\tilde{T}(t),t\ge 0\}$ of $\{T(t),t\ge 0\}$ to $Y$, then $-\tilde{A}$ would be a semigroup generator, too.

If by "extend" you mean the usual thing (viz, that $X$ is left invariant under $\{\tilde{T}(t),t\ge 0\}$ and the restriction to $X$ of $\{\tilde{T}(t),t\ge 0\}$ is $\{T(t),t\ge 0\}$ again), then $X$ would be left invariant under $\{T(t),t\le 0\}$, too, and the generator of this semigroup would necessarily be the part in $X$ of $-\tilde{A}$. So, the analytic semigroup $\{T(t),t\ge 0\}$ would actually imbed into a group $\{T(t),t\in \mathbb R\}$ on $X$ as well.

Now, an analytic semigroup on a Banach space $X$ can imbed into a group if and only if $X$ is finite dimensional its generator is bounded.

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  • $\begingroup$ I guess I'm misinterpreting something, but your last claim puzzles me: why doesn't the semigroup $e^{it\Delta}$, $t\ge 0$, on $L^2$ embed in the group $e^{it\Delta}$, $t\in\mathbb R$ ? $\endgroup$ Oct 11, 2015 at 18:19
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    $\begingroup$ Yeah, I don't think that can be what you mean. More simply, what about the semigroup $T(t) x = e^{t} x$, whose generator is $I$? That's an analytic semigroup that embeds in a group, no matter what $X$ is. $\endgroup$ Oct 11, 2015 at 19:00
  • $\begingroup$ @NateEldredge: You are right, I have edited. $\endgroup$ Oct 12, 2015 at 7:29
  • $\begingroup$ @ChristianRemling: But $e^{it\Delta}$ is not analytic! $\endgroup$ Oct 12, 2015 at 7:37
  • $\begingroup$ Thanks, here, "extend" is a little bit weaker than the usual meaning, see my edit above. I'm curious to know that if this also implies that the generator of the "smaller" semigroup is part of $-\tilde{A}$. $\endgroup$
    – Tomas
    Oct 13, 2015 at 3:05

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