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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?

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    @ChristopherA.Wong Sorry, you're right, it should be S(t) in that norm. I'll edit.2012-11-28

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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.

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    @soup If, for example, [Is this a valid proof? (inte$g$ral, $h$omeomorphism, calculus)](http://math.stacke$x$c$h$ange.com/questions/256680/is-t$h$is-a-valid-proof-integral-$h$omeomorphism-calculus) was resolved, you should indicate so (either by editing the question title, or by posting a short answer o$f$ your own). Otherwise someone may be wasting their time trying to come up with an answer you no longer need.2012-12-27