Uniform continuity of heat semigroup 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! 
 A: 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 $t > 0$.
A: Using Fourier transformation, your question is about the existence of a subspace $X$ of $L^2(\mathbb R^d)$ such that
$$
\lim_{t\rightarrow 0_+}\left\{\sup_{v\in X, \Vert v\Vert=1}\int(1-e^{-t\vert \xi\vert^2})\vert v(\xi)\vert^2 d\xi\right\}=0.
$$
Of course, as you noted, taking $X=L^2(\mathbb R^d)$ does not work since
$
\Vert 1-e^{-t\vert \xi\vert^2}\Vert_{L^\infty(\mathbb R^d)}=1.
$
On the other hand, one crude way of doing this would be assume that the regularity of $u$ is slightly  better, e.g. $u\in H^s$, $s \in(0,1]$. We get then for $\lambda>0, v=\hat u$,
\begin{multline*}
\int(1-e^{-t\vert \xi\vert^2})\vert v(\xi)\vert^2 d\xi
\\\le 
\int_{\vert \xi\vert\le \lambda}(1-e^{-t\vert \xi\vert^2})\vert v(\xi)\vert^2 
d\xi
+t^s\int_{\vert \xi\vert> \lambda}
\underbrace{(t\vert \xi\vert^2)^{-s}(1-e^{-t\vert \xi\vert^2})}_{\le C_s}\vert v(\xi)\vert^2 \vert\xi\vert^{2s}
d\xi
\\
\le(1-e^{-t\lambda^2})\Vert u\Vert^2_{L^2}+C_st^s\Vert u\Vert^2_{H^s},
\end{multline*}
providing for $s\in (0,1]$,
$
\lim_{t\rightarrow 0_+}\Bigl(\sup_{\Vert u\Vert_{H^s}=1} 
\Vert u-e^{t\Delta}u\Vert_{L^2}\Bigr)=0.
$
Some variations could be made by choosing $\lambda$ dependent of $t$ and $\Vert u\Vert_{H^s}.$
