How to show that time-dependent norm is continuous (please verify my proof) For each $t \in [0,T]$, let $H_t$ be a Hilbert space. Suppose for each $t$, the operator $T_t:H_0 \to H_t$ is a linear homeomorphism with inverse $T_{-t}:H_t \to H_0$ also linear homeomorphism. Suppose that $$t \mapsto |T_t w|_{H_t}\tag{1}$$ is continuous on $[0,T]$ for all $ w \in H_0.$ Does this imply that the map
$$t \mapsto |T_{-t}^*f|_{H_t^*}$$ is also continuous? Here $T_{-t}^*:H_0^* \to H_t^*$ is the adjoint operator  and $H_t^*$ is the dual space of $H_t.$ I think this is true because one can write the norm as a supremum, use the adjoint identity and the assumption (1). But I just want a check. Thanks.
Edit Here is my proof:
$$|T_{-t}^*f|_{H_t^*} = \sup_{h \in H_t} \frac{\langle T_{-t}^*f, h \rangle_{H_t^*, H_t}}{|h|_{H_t}} = \sup_{h \in H_t} \frac{\langle f, T_{-t}h \rangle_{H_0^*, H_0}}{|h|_{H_t}} = \sup_{g \in H_0} \frac{\langle f, g \rangle_{H_0^*, H_0}}{|T_tg|_{H_t}}$$
with the last equality because $T_t$ is homeomorphism. This is continuous since we assumed $|T_tg|_{H_t}$ is continuous. Is this correct?
 A: It is correct that for every $f $ and $g$ the function 
$$t\mapsto \frac{\langle f, g \rangle_{H_0^*, H_0}}{|T_tg|_{H_t}}$$
is continuous. However, this does not imply that the supremum over all $g$ will be a continuous function of $t$. 
Here is a counterexample. Let $H_t=L^2[0,1]$ for all $t$, and let $T_t$ be the multiplication operator with the symbol
$$m_t(x) = \begin{cases} t \quad & 0\le x<t \\1 \quad & t\le x\le 1 \end{cases}$$
That is, $(T_tf) (x)= m_t(x) f(x)$. Since $m_t$ is real valued, $T_t$ is self-adjoint. Note that $\inf m_t=t>0$ when $t>0$, and $m_0\equiv 1$. Therefore, $T_t$ is invertible, with the inverse given by multiplication by $m_t^{-1}$. 
For any $f\in L^2[0,1]$, the value 
$$\|T_t f\|_2^2 = \int_0^1 m_t^2(x) |f(x)|^2\,dx$$
depends continuously on $t$. Indeed, as $t\to s$, the integrand converges a.e., and is dominated by $|f|^2$.
On the other hand, 
$$\|T_t^{-1} f\|_2^2 = \int_0^1 m_t^{-2}(x) |f(x)|^2\,dx$$ 
is not a continuous function of $t$ for $f(x)=x^{-1/3}$. Indeed,  for $t>0$
$$\|T_{t}^{-1} f\|_2^2 \ge \int_0^t t^{-2} x^{-2/3}\,dx =3 t^{-5/3} $$ 
while $$\|T_{0}^{-1} f\|_2^2 = \|  f\|_2^2 = \int_0^1  x^{-2/3}\,dx = 3$$
