TITLE: Counterexample to Riesz representation for Hilbert modules
QUESTION [8 upvotes]: For a Hilbert space $H$, the Riesz representation theorem states that $H$ is isomorphic to its dual $H^*$ via $x \mapsto \langle x, -\rangle$.
It is often stated in the literature that this does not work in full generality for a Hilbert module over a $C^*$ algebra. For example, attempts to define the adjoint of a morphism between Hilbert modules run into the lack  of such a representation theorem. To avoid this, we insist that a morphism between Hilbert modules admit an adjoint.
Is there a simple counterexample to Riesz representability for Hilbert modules?
To be more precise, take $E$ an Hilbert module over $A$ and $\phi : E \to A$ being $A$ linear. The question is on the existence of an $x$ in $E$ such that $\phi = \langle x, - \rangle$.

REPLY [5 votes]: Take $A= \mathcal{C}([0,1])$ and $H$ the ideal of $A$ of functions that vanish at $0$.
$H$ is a Hilbert $A$ module (as any ideal, with the natural multiplication of $A$ and the scalar product $(x,y)=x^*y$ of $xy^*$ depending on if you are talking of right or left modules) the inclusion of $H$ into $A$ is a continuous $A$ linear map and it has no adjoint.
Indeed an adjoint would be a map $p$ from $A$ to $H$ such that for all $y$ in $H$ and $x$ in $A$, $y^* p(x) = y^*x$ hence $p(1)$ would be an element of $H$ which is a unit for the ideal $H$, that does not exists.