Is there an embedding $L^2(0,T;H^{-1}(U)) \hookrightarrow L^1(0,T;H^1_0(U))$? In the book by Evans, he defines the weak derivative $\mathbf{u}'\in L^1(0,T;X)$  for a function $\mathbf{u}\in L^1(0,T;X)$ as the function satisfying
$$\int_0^T \phi'(t) \mathbf u (t) dt = -\int_0^T \phi (t) \mathbf u '(t) dt$$
if it exists. Now, a few pages later he writes

THEOREM 3 (More calculus). Suppose $\mathbf u \in L^2(0,T;H^1_0(U))$, with $\mathbf u' \in L^2(0,T;H^{-1}(U))$. $\ldots$

Now, we can embed $H^1_0 (U) \hookrightarrow L^2(U) \hookrightarrow H^{-1}(U).$ However, if we assume that the weak derivative lies in $L^2(0,T;H^{-1}(U))$, this means that there should be some sort of embedding into the space $L^1(0,T; H^1_0(U))$, where the weak derivative initially is defined in. While there is an embedding
$$L^2(0,T;H^{-1}(U)) \hookrightarrow L^1(0,T;H^{-1}(U)) $$
by treating $L^2$-functions as $L^1$-functions and
$$L^1(0,T;H^{1}_0(U)) \hookrightarrow L^1(0,T;H^{-1}(U)) $$
by treating $H^1_0(U)$ elements as $H^{-1}(U)$, I fail to see if any embedding
$$L^2(0,T;H^{-1}(U)) \hookrightarrow L^1(0,T;H^1_0(U))$$
exists. If not, how should $\mathbf u'\in L^2(0,T;H^{-1}(U))$ be understood?
Remark: I have seen the posts here and here which concern very similar questions. However, I don't think that my question is answered there.
 A: Ok so I think I have an idea what my misunderstanding might have been. This is not supposed to be an answer unless people agree with it.
Considering PhoemueX's comment, it might be wrong to assume that there is such an embedding after all. The problem is how the theorem is to be understood.
We assume that we are given a function $\mathbf u\in L^2(0,T;H^1_0(U))$ such that a weak derivative exists. This derivative $\mathbf u'$ lies in the space $L^1(0,T;H^1_0(U))$. We make the assumption that $\mathbf u'$ also lies in the space $L^2(0,T;H^1_0(U))$ which can be interpreted as a subspace of $L^2(0,T;H^{-1}(U))$ because we can embed $H^1_0(U)$ in $H^{-1}(U)$. This means that we don't assume that our function $\mathbf u'$ lies in the space $L^2(0,T;H^{-1}(U))\setminus L^2(0,T;H^1_0(U))$, we just assume $\mathbf u'\in L^2(0,T;H^1_0(U))\subset L^1(0,T;H^1_0(U))$ which automatically tells us that $\mathbf u'\in L^2(0,T;H^{-1}(U))$.
The outputs of the function $\mathbf u'$ are still in $H^1_0(U)$, we just "ignore" that fact there. I am still not quite sure why one would want to ignore that fact if more information is given, but someone might be able to answer that.
