Continuity of vector-valued function

Question

Let $X$ and $Z$ be two Banach spaces such that $Z\subset X$ with dense and continuous emmbedding. Let $\phi \in C([0,2],X)$(with respect to the norm of $X$) be a decreasing function such that $range(\phi)\subset Z$. Can we say that $\phi \in C([0,2],Z)$(with respect to the norm of $Z$) ? Thanks.

Answer 1

No, that is not true in general.

First notice that the condition that $t\mapsto\|\phi(t)\|$ does not really pose a restriction: Given $\phi\in C([0,2],X)$ with $\phi(t)\neq0$ for all $t\in[0,2]$ then $\psi(t):=\phi(t)/(\|\phi(t)\|+t)$ would also belong to $C([0,2],X)$ and $\|\psi(t)\|$ would be decreasing (by the compactness of $[0,2]$ the norm $\|\psi(t)\|$ is bounded from below). Then if $\phi\in C([0,2],Z)$ so is $\psi$. Hence, I will not care about this condition in the following example.

Take $X:=L^2(]0,2\pi[)$ and $Z:=H^1(]0,2\pi[)$ and $\phi(t)(x):=2+h(t)\sin(\frac{x}{h(t)^2})$ if $t\neq2$ and $\phi(2,x)=2$ with e.g. $h(t):=2-t$. Then $\phi(t)$ converges in $X$ to $\phi(2)$ but it diverges in $Z$ (taking $t_n=2-1/n$ should give $\|\phi(t_n)\|\rightarrow\infty$). Still $\phi(t)\in Z$ for all $t$.