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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.

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!

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!

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.

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.

Thank you!

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.

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!

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 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.

Thank you!

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.

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.

Thank you!