# 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!

• If $X\subset D(A)$, then the semigroup restricted to $X$ is uniformly continuous; this follows by applying the uniform boundedness principle to the restriction of the differential quotient $(T(t) - I)/t$ to $X$. This observation includes your spectral measure construction as a special case, but it does not include the case where $X$ is finite dimensional. – Jochen Glueck May 23 '18 at 16:08
• @JochenGlueck thank you, that's right. However, I introduced these two examples only to illustrate what is possible and that it is not clear to me what will happen for arbitrary spaces. – Sascha May 23 '18 at 17:14
• Well, if we would like to be bold, we could conjecture that every subspace $X$ with the desired property is a direct sum of a subspace of $D(A)$ (which is, in addition, closed in $X$) and a finite-dimensional space. Unfortunately though, experience thaught me better not to be bold... – Jochen Glueck May 23 '18 at 17:36
• Does Theorem 2 in my answer below address your intention of getting a general understanding of the situation, or are you more interested in semigroups whose generator does not have compact resolvent (as, e.g., the heat semigroup on the entire space $\mathbb{R}^n$ which is your motivating example). – Jochen Glueck May 29 '18 at 16:09

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 $t > 0$.

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 $\Vert u\Vert_{H^s}.$

• But actually the space $H^s$ is not a closed subspace of $L^2$; it is another Banach space which is continuously embedded in $L^2$, so this is a somewhat different setting, isn't it? – Jochen Glueck May 30 '18 at 14:55
• @Jochen Glueck Yes, this is correct. However, I have strong doubts on the existence of a non-trivial $X$, so I mention what I qualified as a crude answer. Note that you can be logarithmically close to $L^2$, playing a bit with the second integral in the series of inequalities. Also you may bluntly assume some condition of support for the Fourier transform of $u$, providing a closed space of very regular functions (in that case, you are done with the first integral, with a proper choice of $\lambda$). – Bazin May 30 '18 at 20:02
• Yes, restricting the support of the Fourier transform of $u$ to a bounded set is one possibility to obtain a non-trivial example of closed $X$. (It is actually the same example as mentioned by the OP who suggests to apply the spectral measure of the Laplacian to a bounded set). I agree with your assessment that there might not be many more examples of closed spaces $X$ which fulfil the uniform continuity condition. Still, I have no idea how to prove this. – Jochen Glueck May 30 '18 at 20:57