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Let $(Y,\left\|\cdot\right\|_Y)$ be a Banach space and $A:D(A)\subset Y \to Y$ a closed operator. Studying dynamical systems of the form \begin{equation} u'=Au \end{equation} quickly leads to the notion of one-parameter semigroups:

$\textbf{Definition}$: A one-parameter semigroup $G$ is a collection $\left\{S(t):Y\to Y\right\}_{t \geq 0}$ of bounded operators which satisfy $S(0)=I$ and $\forall s,t > 0$: $S(s+t)=S(s)S(t)$. A one-parameter semigroup is called strongly continuous if the maps \begin{equation} t \mapsto S(t)u \end{equation} are continuous $\forall u \in Y$.

My problem is the following: what if the nice $C^1$-solutions $u(t)$ ($u(t) \subset D(A)$) of the $(u'=Au)$-system have not for all $t>0$ a uniform bound \begin{equation} \sup_{u(0) \in D(A),u(0)\text{ generates } C^1-solution}\frac{\left\|u(t)\right\|_Y}{\left\|u(0)\right\|_Y} \end{equation} but nontheless a subvectorspace $X \subset Y$ exists with $D(A) \subset X$, together with some norm $\left\|\cdot\right\|_X$ such that $(X,\left\|\cdot\right\|_X)$ is a Banach space of its own and \begin{equation} \sup_{u(0) \in D(A),u(0)\text{ generates } C^1-solution}\frac{\left\|u(t)\right\|_Y}{\left\|u(0)\right\|_X} \end{equation} does turn out to be finite for all $t>0$. Continuing this line of reasoning I end up with something like

$\textbf{Definition}$: Let $(Y,\left\|\right\|_Y)$ and $(X,\left\|\right\|_X)$ be Banach spaces and $X \subset Y$ such that $X$ is dense in $Y$ for the $Y$-norm. A bilateral one-parameter semigroup $G$ is a collection $\left\{S(t):X\to Y\right\}_{t \geq 0}$ of bounded operators which satisfy...

*$S(0)=I_{X \to Y}$

*$\forall u \in X$: $\varphi_u:[0,\infty) \to Y:t \mapsto S(t)u$ is a continuous map.

*$\forall u \in X$ and $f:[0,\infty) \to \mathbb{C}$ a $K_G$-function (see below for definition), we have $\int_0^{\infty} f(s)S(s)u\text{ d}s \in X$ and \begin{equation} S(t)\left(\int_0^{\infty} f(s)S(s)u\text{ d}s\right) = \int_0^{\infty} f(s)S(t+s)u\text{ d}t. \end{equation} $\textbf{Definition}$: The vector space $K_G$ consists of all complex functions with domain $[0,\infty)$ spanned by functions of the form $\chi_{[a,b]}$ or $f$ $C^1$ such that $\int_0^{\infty} \left\|f'(s)S(s)\right\|\text{d}s < \infty$, $\forall t \geq 0$ $\int_0^{\infty} \left\|f(t)S(s+t)\right\|\text{ d}s <\infty$, and $\left\|f(t)S(t)\right\| \to 0$ as $t\to \infty$. I think we could as well replace "spanned by functions of the form $\chi_{[a,b]}$ or $f$ $C^1$ such that..." in the definition of $K_G$ by "functions which are piecewise $C^1$ up to a set of 0 Lebesgue-measure"

Are there any good reasons to reject this line of reasoning and this generalisation from the start? Are my concerns (which seem generic to me) already adressed somewhere in the literature?

EDIT: an example and elaboration is below in an answer.

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    $\begingroup$ An example would be most welcome, where these conditions would not imply $S(t)u\in X$. Also, what if $Y=X$? Does the usual semigroup property $S(t+s)u=S(t)S(s)u$ hold? (The norm should be $||.||_Y$ in the integral somewhere...) $\endgroup$ – Jean Duchon Sep 16 '15 at 15:46
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    $\begingroup$ There are many generalizations in the literature (distribution-semigroups, integrated semigroups, regularized semigroups, bi-continuous semigroups, etc.) As Jean said, you should come up with an example. $\endgroup$ – András Bátkai Sep 16 '15 at 18:09
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    $\begingroup$ You write $S(t)\left(\int_0^{\infty} f(s)S(s)u\text{ d}s\right)$ in one of the definitions. How does it make sense, as $S(t)$ only maps $X$ to $Y$, not $X\to X$ or $Y\to Y$ ? $\endgroup$ – Jean Duchon Sep 22 '15 at 10:01
  • $\begingroup$ Oops, in the edit I removed the additional requirement that $\int f(s) S(s)u\text{d}s \in X$ (seen as a subset of $Y$). $\endgroup$ – Thibaut Demaerel Sep 22 '15 at 16:39
  • $\begingroup$ Fixed it. I think I'll just accept my example below as an answer as no additional imput seems close at hand. $\endgroup$ – Thibaut Demaerel Sep 22 '15 at 16:57
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Upon reflection, I find this approach interesting enough, although the example you gave is too elementary to be sure.

Provided your $K_G$ contains $\mathcal D_0:=C_c^\infty(0,\infty)$, your "bilateral semigroups" are distribution semigroups : linear maps $\varphi\mapsto G(\varphi)$ from $\mathcal D_0$ to $\mathcal L(X)$ satisfying $G(\varphi*\psi)=G(\varphi)G(\psi)$. (With your notation, $G(\varphi)u=\int_0^\infty \varphi(t)S(t)u\ dt).$ Historically, they were introduced (by J.L. Lions, 1960) to address the same concern as you express (more or less): sometimes solutions to a Cauchy problem $u'=Au,\ u(0)=u_0\in X$ (example: Schrödinger equation $\partial_t u=i\Delta u$, $X=L^p,\ p\neq2$) have $u(t)\notin X$ while $\int u(t)\varphi(t)\ dt\in X$.

So, going beyond strongly continuous semigroups makes perfect sense. Doing it by introducing two Banach spaces may be original, I don't know, and could be the right thing to do in some cases.

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  • $\begingroup$ I think my (academic) example is correct, although I've been a bit sloppy at the end where for my argument that eta(t) in Z I ignore the requirement that initial conditions in Z must yield UNIQUE solutions (an edit is upcoming). You're always welcome to pass your favourite reference on which you may or may not have based your answer. $\endgroup$ – Thibaut Demaerel Sep 28 '15 at 18:47
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In fact, even a snake does not have to look beyond its nose to find a nice example (I post it here such that the original question does not get to long, discouraging potential readers). Take $X$ as the complex-valued sequences with the usual vector-space structure and with norm \begin{equation} \left\|x\right\|_X^2=|x_1|^2+|x_2|^2+|x_3|^3... \end{equation} (i.e. $X=l^2(\mathbb{N})$) and $Y$ again those sequences with the more relaxed norm \begin{equation} \left\|x\right\|_Y^2=\frac{1}{2}(|x_1|^2+\frac{1}{2!}|x_2|^2+\frac{1}{3!}|x_3|^2+...) \end{equation} For the role of closed operator $A:D(A)\subset Y \to Y$ we take the infinite diagonal matrix \begin{equation} A_{kl}=\left(k+i\sqrt{2k!-k^2}\right)\delta_{kl}. \end{equation} This seemingly "arbitrary" operator satisfies ($\forall x \in X$) \begin{equation} \left\|x\right\|_X=\left\|Ax\right\|_Y. \end{equation} which establishes that $D(A)=X$ (as a subset of $Y$ of course).

Next, let $Z\subset X$ be the sequences with finite support. $Z$ is dense in $X$ and $Y$ for their respective norms. It is straightforward to show that the sequence \begin{equation} u=(u_1,...,u_n,0,0...) \end{equation} is equal to $u(0)$ in the following solution of $u'=Au$ which is smooth in both the $X$ and $Y$-norm and unique for this initial condition: \begin{equation} u(t)=(u_1\exp(A_{11} t),...,u_n \exp(A_{kk}t),0,0...) (\in Z) \end{equation} It is not difficult to see that there is no uniform bound ($\forall u \in Y$ or $\in X$) to the quantities $\frac{\left\|u(t)\right\|_Y}{\left\|u\right\|_Y}$ or $\frac{\left\|u(t)\right\|_X}{\left\|u\right\|_X}$.

Assume however that $\left\|u\right\|_X=1$ which means in particular $|u_k|\leq 1$ $\forall k >0$. Then \begin{equation} \left\|u(t)\right\|_Y = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k!}|u_k\exp(A_{kk}t)|^2 \leq \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k!}\exp(kt)^2=\frac{1}{2}\exp(\exp(2t)) \end{equation}

So far, I've produced all prerequisites stated in my question above, which I will show to be enough to produce a bilateral semigroup. That is, I've come up with a $u'=Au$-system which satisfies...

Definition Let $(Y,\left\|\right\|_Y)$ be a Banach space, a closed linear operator $A:D(A)\subset Y \to Y$ is called a "Cauchy-operator" if the set $Z\subset D(A)$, where $u\in Z$ i.f.f. a unique solution to $u'(t)=Au(t)$ ($\forall t \geq 0$) exists with initial condition $u$, is dense in $Y$. From now we define semigroup operators $R(t):Z \to D(A)$ through $u(t)=R(t)u$.

We call $A$ a "regular Cauchy-operator" if there exists moreover a vectorspace $X$ with $D(A) \subset X \subset Y$ with an accompanying norm $\left\|\right\|_X$ such that $(X,\left\|\right\|_X)$ is a Banach space and such that $Z$ is also dense in $(X,\left\|\right\|_X)$ and such that $\sup_{u \in Z}\frac{\left\|R(t)u\right\|_Y}{\left\|u\right\|_X} < \infty$ $\forall t >0$. $A$ is called a "smooth Cauchy-operator" if for every $t^* \in [0,\infty)$, there exists $\epsilon>0$ such that \begin{equation} \sup_{u \in Z, t\in [t^*-\epsilon,t^*+\epsilon]}\frac{\left\|R(t)u\right\|_Y}{\left\|u\right\|_X} < \infty \end{equation}

Theorem Let $A:D(A)\subset Y \to Y$ be a Cauchy operator. $Z$ is a vector space and the $R(t):Z \to D(A)$-operators are linear. Their range is contained in $Z \subset D(A)$ and $\forall s \geq 0$. Redefining them as $R(t):Z\subset Y \to Z \subset Y$ we have \begin{equation} R(s+t)u=R(s)R(t)u. \end{equation} proof: exercise.

Theorem If $A:D(A)\subset Y \to Y$ is a regular Cauchy operator. Then the bounded operators $R(t):Z\subset X \to Z \subset Y$ can be continuously extended to bounded operators \begin{equation} S(t):X \to Y. \end{equation} If at some point $t^*\in [0,\infty)$ and for some $\epsilon>0$, we have that $R(t)$ is uniformly bounded on $[t^*-\epsilon,t^*+\epsilon]$, then for all $u\in X$ the map \begin{equation} \varphi_u:[0,\infty)\to Y: t \mapsto S(t)u \end{equation} is continuous at $t^*$. As a consequence these maps are continuous if $A$ is a smooth Cauchy operator. In this case the operators $S(t)$ constitute a bilateral one-parameter semigroup $G=\left\{S(t)\right\}_{t\geq 0}$. Note: I've edited the definition of bilateral semigroup slightly.

proof The first part of the theorem is straightforward and left to the reader. Keep in mind that the continuity of the $\varphi_u$ maps implies uniform continuity: for all compacts $C \subset [0,\infty)$ we have \begin{equation} \sup_{t\in C} \left\|S(t)\right\| \leq 0. \end{equation} For the second part of the theorem, we first have to show that elements of the form $\int_a^b S(s)u\text{d}s=\int_0^{\infty} \chi_{[a,b]}S(s)u\text{d}s$ or $\int_0^{\infty} f(s)S(s)u\text{d}s$ (where $f:[0,\infty)\to \mathbb{C}$ is differentiable) are in $X$. Secondly, we have to show that "$S(t) \int S(s)... = \int S(s+t)...$". (Important note: all the relevant Bochner integrals are of course defined in the Banach space $Y$!) In this proof I'll only treat the second case, while the first is similar.

Take $(u_n)_n \in Z$ such that $\left\|u_n-u\right\|_X \to 0$. By the boundedness of the operators $S(t)$, we have \begin{equation} \left\|\int_0^{\infty} f(s)S(s)u_n\text{d}s - \int_0^{\infty} f(s)S(s)u\text{d}s\right\|_Y \to 0 \end{equation} (exercise) Moreover, we have \begin{equation} \left\|A\int_0^{\infty} f(s)S(s)u_n\text{d}s + f(0)u+\int_0^{\infty} f'(s)S(s)u\text{d}s\right\|_Y = \left\|\int_0^{\infty} f(s)AS(s)u_n\text{d}s + f(0)u+\int_0^{\infty} f'(s)S(s)u\text{d}s\right\|_Y =\left\|\int_0^{\infty} f(s)(S(s)u_n)'\text{d}s + f(0)u+\int_0^{\infty} f'(s)S(s)u\text{d}s\right\|_Y=\left\|-f(0)u_n - \int_0^{\infty} f'(s)S(s)u_n\text{d}s+ f(0)u+\int_0^{\infty} f'(s)S(s)u\text{d}s\right\|_Y \to 0 \end{equation} So the closedness of $A:D(A) \subset Y \to Y$ implies that $v:=\int_0^{\infty} f(s)S(s)u\text{d}s \in D(A)\subset X$ and $Av =-f(0)u-\int_0^{\infty} f'(s)S(s)u\text{d}s$. To finish, one takes the following steps: *$\eta:[0,\infty)\to Y: t \mapsto \int_0^{\infty} f(s)S(s+t)u_n\text{d}s=\int_t^{\infty} f(s-t)S(s)u_n\text{d}s$ is differentiable and $\eta'(t)=A\eta(t)$. Hence $\eta(t) \in Z$ for all $t\geq 0$.

*But then \begin{equation} \nu:[0,\infty)\to Y: h \mapsto S(t-h)\int_0^{\infty} f(s)S(s+h)u_n\text{d}s = S(t-h)\eta(h) \end{equation} is also differentiable and $\nu'(h)=0$. But then \begin{equation} S(t)\int_0^{\infty} f(s)S(s)u_n\text{d}s = \nu(0)=\nu(t)=\int_0^{\infty} f(s)S(s+t)u_n\text{d}s. \end{equation}

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    $\begingroup$ I think it would have been better to wait for one or two upvotes before accepting your own answer. Personally, I don't see the usefulness of this notion, even after the example you gave, a sequence of independent equations that don't require such machinery. $\endgroup$ – Jean Duchon Sep 23 '15 at 12:10
  • $\begingroup$ Personally, I preferred that too. But the mathoverflow-community urged to accept an answer or start a bounty. I decided to do the former. The usefulness of the "Bilateral semigroup" notion doesn't show from this example. But If you start exploring the consequences of the axioms of the definition I use, you see that those semigroups must be nice objects and there is again some correspondence between semigroup and the "resolvents" of its generator through Laplace transformation. $\endgroup$ – Thibaut Demaerel Sep 23 '15 at 16:41

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