# Functional to inner product in Hilbert triple

If $V \subset H \subset V^*$ is a Hilbert triple, and $f \in V^*$, I cannot represent $f(v) = (e,v)_V$ because we don't identify $V$ with $V^*$. But is it true that $f(v) = (e,v)_H$ for some $e$?

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Yes, it is true: the Riesz representation theorem applies to the Hilbert space $V$, regardless of its relation to other spaces.
A sensible way to deal with non-identification is to introduce the duality map $J:V\to V^*$, defined by $J(e)=f$ in your notation. This map is an isomorphism of two distinct Hilbert spaces $V$ and $V^*$. This is different from the case of a single (real) Hilbert space, when we think of $J$ as the identity map.
By the way, in complex Hilbert spaces $J$ is conjugate-linear, which is a constant source of mild irritation (for me at least).