Show that norm of a functional is continuous 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? 
 A: 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.  
