**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 $t > 0$.