Let H, V be two Hilbert spaces.
Let $ \textbf{V} \hookrightarrow \textbf{H} $ and let $\textbf{V}$ dense in $\textbf{H}$.
Then, the Hilbert triplet theorem holds, and we can write
$ \textbf{V} \hookrightarrow \textbf{H} \hookrightarrow \textbf{V'} $, where $ \textbf{V'}$ is the dual of $ \textbf{V}$.
My question is the following: since Riesz lemma and since H and V Hilbert spaces, I can identify H with H' (where H' is the dual of H), but why I can't apply the Riesz lemma to V and identify it to its dual V'?
I mean.. That would be absurd - having supposed that the norm of H and V are not necessarily equivalent. Why I can't apply Riesz lemma to V?
Let me denote the embeddings $V\hookrightarrow H$ and $H\hookrightarrow V'$ by $i$ and $i'$. Then $i(V)$ and $i'(H)$ are dense in $H$ and $V'$. So the Hilbert triple method yields a mapping $ii': V \to V'$ with dense range. In many examples $i$ and thus $i'$ are compact, so $ii'$ is not continuously invertible.
By Riesz theorem you get another mapping $I: V \to V'$ which is an isomorphism.
However, $ii' \ne I$, which can be easily seen in the compact case.
Note that both methods give different constructions to get from $V$ to $V'$: The Hilbert triple uses the inner product in $H$, the Riesz theorem the inner product in $V$.
So you still can apply Riesz theorem. It is however not advised to do so, as it bears the danger to silently make use of the faulty identification $I=ii'$.