Gelfand triples and embedding of vector-valued spaces Suppose $u\in L^2(0,T; V)$ and $u_t\in L^2(0,T;V')$, where $V\subset H\subset V'$ is a Gelfand (or Hilbert) triple (the embeddings are dense and continuous and the spaces are all Hilbert). For instance take $H^1_0\subset L^2\subset H^{-1}$. Then it is well-known that $u\in C([0,T];H)$ (see for instance Theorem 6.41 in Hunter's notes: https://www.math.ucdavis.edu/~hunter/pdes/ch6A.pdf)
What goes wrong if we take $V=H$ above (so that instead $V\subset V\subset V'$), and conclude that in fact $u\in C([0,T];V)$? I believe this is incorrect and that the issue stems from how we identify dual spaces, say $H\equiv H'$ and not $V\equiv V'$, but it's unclear exactly how things go wrong. 
 A: Because in the triplet $V \subset V \subset V'$ you no longer have $u_t \in L^2(0,T; V')$.
In fact, denote by $I$ the embedding from $V$ to $H$. Then, the statement $u_t \in L^2(0,T;V')$ really means $(I^* u)_t \in L^2(0,T;V')$. By changing the middle space in the triplet, you change $I$ and hence, $u_t \in L^2(0,T;V')$ is no longer valid.
A: One/another problem is that in $V\to H\to V^*$, the map $H\to V^*$ is (ignoring complex conjugation for this discussion) the adjoint of the map $V\to H$, using self-duality (ignoring complex conjugation) of $H$. That is, the diagram really is $V\to H\approx H^*\to V^*$.
Also, the Riesz-Frechet theorem that gives an isom (ignoring conjugation...) $H\approx H^*$ does not behave as one might have thought! Namely, a continuous Hilbert-space map $V\to W$ and adjoint $W^*\to V^*$ does not make a commuting square with the Riesz-Frechet isomorphisms $V\approx V^*$ and $W\approx W^*$.
(Small wonder then that the Gelfand-Hilbert inclusion $V\to V^*$ is not the Riesz-Fischer map...)
