I have read this relation: if $y\in C(0,T;L^2(0,L))\subset L^2((0,T)\times (0,L))$ then $y_t\in L^2(0,L;H^{-1}(0,T))$
can anyone explain how to get this?
I know $y(t)(x)\in L^2(0,L)$, but why $y_t(x)(t)\in H^{-1}(0,T)$?
I found it is straightforward:
$y\in C(0,T;L^2(0,L))\subset L^2((0,T)\times (0,L))=L^2(0,L;L^2(0,T))$, hence $y(x)(t)\in L^2(0,T)$, then $y_t(x)(t)\in H^{-1}(0,T)$
We have $y \in C(0,T;L^2(0,L))\subset L^2(0,T;L^2(0,L))$ then by the definition of weak time derivatives we have $y_t \in L^2(0,T;L^2(0,L))$. So by the definition of the Bochner-norm and using Fubini-Tonelli (which works since $y$ is time-continuous) we get
$$ \begin{align} ||y_t||^2_{L^2(0,T;L^2(0,L))}&=\int_0^T ||y_t(t)||_{L^2(0,L)}^2 dt \\ &= \int_0^T \int_0^L y_t^2(t)(x) \ dx \ dt \\ &= \int_0^L \int_0^T y_t^2(x)(t) \ dt \ dx \\ &= \int_0^L ||y_t(x)||^2_{L^2(0,T)} \ dx=||y_t||^2_{L^2(0,L;L^2(0,T))} \end{align}$$
that means $y_t \in L^2(0,L;L^2(0;T))$. And since $H_0^1(0,T) \subset L^2(0,T) \subset H^{-1}(0,T)$ we get $y_t \in L^2(0,L;H^{-1}(0,T))$ as wanted.