I would like to illustrate my question with an example:


It is well-known that $\Delta$ is the generator of a strongly continuous semigroup $(T(t))$ on $L^2(\mathbb R^n),$ i.e. the heat-semigroup.


It is also known that if one has a strongly continuous semigroup and the domain of the generator is the entire Banach space that this is equivalent to the semigroup being uniformly continuous. 


Because the domain of $\Delta$ is the Sobolev space $W^{2,2}$ the heat-semigroup is therefore not uniformly continuous.


Now, consider $X$ a closed subspace of $L^2(\mathbb R^d).$ Are there sufficient and necessary conditions on $X$ such that $T(t):X \rightarrow L^2(\mathbb R^d)$ is uniformly continuous, i.e. $\lim\_{t \downarrow 0}\sup\_{x \in X; \left\lVert x \right\rVert=1} \left\lVert (T(t)-\operatorname{id})x \right\rVert=0?$


This is true if $X$ is finite-dimensional and false if $X=L^2(\mathbb R^d)$ but what happens for the spaces "in between"? I should add that it must also hold on infinite-dimensional spaces on which $\Delta$ is bounded. Examples of such spaces can be constructed using the spectral measure.
More precisely, let $E$ be the spectral measure of the Laplacian then every subspace $X=E([-n,0])L^2(\mathbb R^d)$ does the job.


I would prefer to get some understanding of this situation in the general case and not only for the heat semigroup, but if there is a nice characterization for the heat semigroup then this would qualify as an answer.


If there are any further questions, please let me know. Thank you! 



Uniform continuity of heat semigroup
**Setting.** Throughout, let $E$ be a complex Banach space and denote the space of bounded linear operators on $E$ by $\mathcal{L}(E)$. Let $X \subseteq E$ be a closed subspace and let $\mathcal{T} = (T(t))\_{t \ge 0}$ be a $C\_0$-semigroup on $E$ with generator $A: E \supseteq D(A) \to E$.


As already mentioned in the comments, the following *sufficient* condition for uniform continuity on $X$ holds:


**Proposition 1.** Assume that at least one of the following two assumptions is fulfilled:


(a) We have $X \subseteq D(A)$.


(b) $X$ is finite dimensional.


Then the semigroup $\mathcal{T}$ is uniformly continuous on $X$ at all times, i.e. the mapping $[0,\infty) \ni t \mapsto T(t)|\_X \in \mathcal{L}(X,E)$ is continuous with respect to the operator norm at each $t \in [0,\infty)$.


*Proof.* The sufficiency of (b) is a simple consequence of the strong continuity of $\mathcal{T}$, so assume that (a) is fulfilled. Then for each $x \in X$ the set 
\begin{align\*}
 \{\frac{T(t)x-x}{t}: \; t \in (0,1]\}
\end{align\*}
bounded in $E$. As $X$ is a Banach space, we conclude from the uniform boundedness principle that the set
\begin{align\*}
 \{\frac{T(t)|\_X - I|\_X}{t}: \; t \in (0,1]\}
\end{align\*}
is bounded in $\mathcal{L}(X,E)$ (here, $I$ denotes the identity operator on $E$). Hence, $T(t)|\_X$ converges to $I|\_X$ with respect to the operator norm as $t \to \infty$. This proves continuity at $t = 0$, and the continuity at other times can be shown by exactly the same argument.


The following result gives a concrete *characterization* of uniform continuity on $X$ in the important special case where the semigroup $\mathcal{T}$ is analytic and compact.


**Theorem 2.** Assume that $\mathcal{T}$ is analytic and that the generator $A$ has compact resolvent (for analytic semigroups this is equivalent to $T(t)$ being a compact operator on $E$ for each $t > 0$). Then the following assertions are equivalent:


(i) The semigroup $\mathcal{T}$ is uniformly continuous on $X$ at each time $t \in [0,\infty)$.


(ii) The semigroup $\mathcal{T}$ is uniformly continuous on $X$ at the time $t = 0$.


(iii) $X$ is finite dimensional.


**Remark 3.** Theorem 2 cannot be applied to the heat semigroup on $\mathbb{R}^n$ since this semigroup does not have compact resolvent. However, the theorem can e.g. be applied to the heat semigroup on bounded domains in $\mathbb{R}^n$ (with, say, Dirichlet boundary conditions - or also with Neumann boundary conditions if the boundary of the domain is sufficiently smooth).


*Proof of Theorem 2.* "(iii) $\Rightarrow$ (i)" This is a special case of Proposition 1.


"(i) $\Rightarrow$ (ii)" Obvious.


"(ii) $\Rightarrow$ (iii)" By (ii) there exists a time $t\_0 > 0$ such that $\|T(t\_0)|\_X - I|\_X\| \le 1/2$ (where $I$ denotes the identity operator on $E$). Hence, we have
\begin{align\*}
 \|T(t\_0)x\| \ge \|x\| - \|x - T(t\_0)x\| \ge \|x\| - 1/2\|x\| = 1/2\|x\|
\end{align\*}
for each $x \in X$. Thus, the operator $T(t\_0)|\_X: X \to E$ is bounded below. Since $X$ is closed and thus a Banach space we conclude that the range $Y := T(t\_0)X$ of $T(t\_0)|\_X$ is also closed in $E$ and that $T(t\_0)|\_X$ is an isomorphism between the Banach spaces $X$ and $Y$. Hence, we only need to show that $Y$ is finite dimensional.


As $\mathcal{T}$ is analytic, the range of $T(t\_0)$ is contained in $D(A)$, so $Y$ is a subspace of $D(A)$ and closed in $E$. As $A$ has compact resolvent, the embedding of $D(A)$ (endowed with the graph norm) into $E$ is compact. Hence, the finite dimensionality of $Y$ is a consequence of the following general lemma.


**Lemma 4.** Let $E,F$ be Banach spaces such that $F$ is compactly embedded into $E$. Assume that $Y$ is a closed subspace of $E$ which is, in addition, contained in $F$. Then $Y$ is finite dimensional.


*Proof.* Since $F$ embedes continuously into $E$, the space $Y$ is also closed in $F$. Thus, both norms $\|\cdot\|\_E$ and $\|\cdot\|\_F$ are equivalent on $Y$, and the unit ball with respect to the second norm on $Y$ is compact with respect to the first norm (and thus also with respect to the second, equivalent norm). Hence, $Y$ is finite dimensional.


**Remark 5.** The proof of Theorem 2 actually shows that we can replace analyticity of $\mathcal{T}$ with the weaker assumption that $\mathcal{T}$ be *immediately differentiable*, meaning that the orbit of each vector in $E$ is differential at each time $t > 0$.


