Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is based on Lemma 3.3, page 6 in this paper: http://arxiv.org/pdf/1106.0622v4.pdf I changed the notation quite a lot, but it should be a one-to-one correspondence.


$S(x)$ is a compact manifold for each $x \in [0,T]$.

Fix $s \in [0,T]$. Let $t \in [0,T]$. Suppose $f(t,s):H^{-1}(S(s)) \to H^{-1}(S(t))$ (linear functional), with the property that $f(t,t)$ is the identity for any $t$, and suppose the following holds:

$$\frac{1}{1+|s-t|}\lVert u\rVert_{H^{-1}(S(s))} \leq \lVert f(t,s)u \rVert_{H^{-1}(S(t))} \leq \frac{1}{1-|s-t|}\lVert u\rVert_{H^{-1}(S(s))}$$

It might be helpful to know the adjoint of $f(s,t)$, written $f(s,t)^*:H^1(S(s)) \to H^1(S(t))$ has the property that $\lVert f(s,t)^*v \rVert_{H^1(S(t))}$ is continuous as a function of $t$.

The task is to show that $\lVert f(t,s)u \rVert_{H^{-1}(S(t))}$ is continuous as a function of $t$.

Clearly we can see that it is continuous at $t=s$. But how about apart from $s$? How to see that it is continuous?

share
    
You mean continuous when $u$ is fixed, right? –  Davide Giraudo Nov 28 '12 at 20:12
    
Do you mean, show that $\|f(t,s)u\|_{H^{-1}(S(t))}$ is continuous in $t$? Because you can't define the $H^{-1}(S(s))$ norm of that quantity, unless you know something about $S(x)$. –  Christopher A. Wong Nov 28 '12 at 20:41
    
@DavideGiraudo Yeah, $u \in H^{-1}(S(s))$ is fixed. –  soup Nov 28 '12 at 21:33
    
@ChristopherA.Wong Sorry, you're right, it should be S(t) in that norm. I'll edit. –  soup Nov 28 '12 at 21:34

1 Answer 1

up vote 4 down vote accepted
+50

The operators in question are defined in "Lemma and Definition 3.2" as pullbacks by diffeomorphisms $\Phi_t^s : \Gamma(s)\to \Gamma(t)$. ($S$ is $\Gamma$ in the paper). Said diffeomorphisms are flow maps of a (smooth, time-dependent) field (Assumption 2.1), which implies the composition rule $\Phi_u^t \circ \Phi_t^s = \Phi_u^s$. The composition rule passes to pullbacks $\phi_t^s : H^1(\Gamma(t)) \to H^1(\Gamma(s)) $ and their adjoints $(\phi_t^s)^*: H^{-1}(\Gamma(s)) \to H^{-1}(\Gamma(t))$.

Hence, the small-time estimate quoted in the original post is enough to obtain the continuity. It goes like this: $$\|(\phi_{t+\Delta t}^s)^* v\| = \|(\phi_{t+\Delta t}^t)^* ((\phi_t^s)^* v)\| \approx \|(\phi_t^s)^* v\| $$ where $\approx $ hides multiplicative constants that tend to $1$ as $\Delta t\to 0$.

The text below was written for an earlier version of the question, which lacked much of necessary context.


This looks like a counterexample... Let all $S(x)$ be the same compact manifold $S$. For $t,s\in [0,T]$ define $$\mu(t,s)=\begin{cases}1 \ &\text{ if } t\in \mathbb Q \\ \max(1,|s-t|) \ &\text{ if } t\notin \mathbb Q\end{cases}$$ This function is continuous in the second variable but not in the first.

Let $f(t,s)$ be the scalar operator that multiplies each function by the constant $\mu(t,s)$, that is, $f(t,s)u=\mu(t,s)u$. The norm bounds on $\|f(t,s)u\|$ hold. Of course, $\|f(t,s)u\|$ is badly discontinuous with respect to $t$ when $|s-t|>1$.

The adjoint operator $f(s,t)^*$ is the multiplication by $\mu(s,t)$. The norm $\|f(s,t)^*u\|$ is continuous with respect to $t$ because the function $t\mapsto \mu(s,t)$ is continuous.

share
    
Thanks. The thing is I got this from a paper so I assumed it was true (though I left out some details in my post which may have been necessary). –  soup Dec 27 '12 at 17:51
    
@soup They may well have been necessary. Do the operators form a (semi-)group, by any chance? My suggestion is to always include a reference when asking a question about something in the literature. Quite often, questions of this kind omit important information, precisely because the OP did not see how it is relevant. –  user53153 Dec 27 '12 at 17:55
    
Sorry about that. There is no semigroup stuff (at least not on the surface). I'll edit my main post. Anyway, unless somebody answers in the affirmative I'm happy to give you the bounty :P –  soup Dec 27 '12 at 17:59
    
@soup The linear operators are pullbacks by time-dependent flow. You'd have a group if it was time-independent. As it is, you still have the composition rule: see the edited version. –  user53153 Dec 27 '12 at 18:35
1  
@soup If, for example, Is this a valid proof? (integral, homeomorphism, calculus) was resolved, you should indicate so (either by editing the question title, or by posting a short answer of your own). Otherwise someone may be wasting their time trying to come up with an answer you no longer need. –  user53153 Dec 27 '12 at 23:43

This site is currently not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .