Example of a function $u\in L^\infty(0,T,H^1)$ such that $u_t\notin L^\infty(0,T,H^1)$

Question

Could someone give me an example of a function $u\in L^\infty(0,T,H^1)$ such that $u_t$ exists (in the distributional sense), $u_t\in L^\infty(0,T,L^2)$ and $u_t\notin L^\infty(0,T,H^1)$?

Thanks.

EDIT (to add context). Let $f_0\in H^1$. If $u(t)=f_0$ (constant function with respect to $t$) then $u\in L^\infty(0,T,H^1)$ and $u_t\in L^\infty(0,T,H^1)$. If $u(t)=\int_0^t f_0\;dt$, the same occurs. These are simple examples in which I've thought. Essentially, my question is: How to derive $u$ with respect to $t$ affects the regularity of $u(t)$ with respect to $x$? I think that examples can help me to understand it.

Answer 1

Let $X$ be a Banach space and $u\in L^1(0,T,X)$. We say that $u'\in L^1(0,T,X)$ the weak derivative of $u$ if $$\int_0^T u(t)\phi'(t)dt=-\int_0^T u'(t)\phi(t)dt,\forall\ \phi\in C_0^\infty(0,T).$$

According to this definition, the problem is not that $u'(t)$ does not belong to $H^1$, it does belong to $H^1$. The question is, if it belong to $L^\infty$, i.e. $$u\in L^\infty (0,T,H^1),\ u'\in L^\infty(0,T,L^2)\ \mbox{implies}\ u'\in L^\infty(0,T,H^1)? $$

Note that $u'\in L^\infty(0,T,L^2)$ is equivalently to $$\sup_{t\in (0,T)} \|u'(t)\|_2<\infty,$$

while $u'\in L^\infty(0,T,H^1)$ is equivalently to $$\sup_{t\in (0,T)} \|u'(t)\|_{1,2}<\infty.$$

These are two different things. Now I think you can construct a example which satisfies your requirements .