8
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jochen Glueck May 23 '18 at 16:08
  • $\begingroup$ @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. $\endgroup$ – Sascha May 23 '18 at 17:14
  • $\begingroup$ 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... $\endgroup$ – Jochen Glueck May 23 '18 at 17:36
  • $\begingroup$ 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). $\endgroup$ – Jochen Glueck May 29 '18 at 16:09
1
$\begingroup$

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 $t > 0$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Jochen Glueck May 30 '18 at 14:55
  • $\begingroup$ @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 $\lambda$). $\endgroup$ – Bazin May 30 '18 at 20:02
  • $\begingroup$ 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. $\endgroup$ – Jochen Glueck May 30 '18 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.