Existence of adjoint linear map between pre Hilbert spaces Given $\mathcal{H}_1, \mathcal{H}_2$ pre-Hilbert spaces (i.e. vector spaces with a scalar product), then it is true that a linear map $f$ between them has a well-defined adjoint?
My guess is that since this property follows from Riesz representation theorem in the Hilbert spaces case and Riesz representation theorem follows from the existence of orthogonal projection onto closed subspaces. Orthogonal projections exist in Hilbert spaces because the infimum of the distance function is actually a minimum because of completeness, in pre-Hilbert spaces this properties is not guaranteed and all the tower of properties need not hold true. It is however true that we can define the adjoint of a linear map? 
 A: There is always an adjoint for a bounded linear map $T : X \to Y$, where $X$ and $Y$ are normed linear spaces (topological linear spaces, even), but the adjoint is not from $Y$ to $X$, but from $Y^*$ to $X^*$, the topological duals of $Y$ and $X$ respectively (i.e. the space of linear functions from the space to the field). It is defined like so:
$$T^* : Y^* \to X^* : g \mapsto g \circ T.$$
It should be clear that $g \circ T$ is linear, continuous, and maps $X$ to the scalar field, putting $g \circ T \in X^*$.
Recall that, if $f \in X^*$ and $x \in X$, then we sometimes denote $f(x)$ by $\langle x, f \rangle$. Using this notation, the adjoint satisfies:
$$\langle Tx, g\rangle = g(Tx) = (g \circ T)x = (T^*g)(x) = \langle x, T^* g\rangle.$$
When $X$ and $Y$ are Hilbert spaces, we can identify $X^*$ and $Y^*$ with $X$ and $Y$ respectively. As such, the implicit definition of adjoints in Hilbert spaces agrees with the normed linear space definition.
Now, let's consider the case where $X$ and $Y$ are Pre-Hilbert. Then $\overline{X}$, the metric completion of $X$, is a Hilbert space in which $X$ is dense, and hence both $X$ and $\overline{X}$ have the same dual $X^*$. But, since $\overline{X}$ is a Hilbert space, we may identify it with its dual $X^*$, meaning that we may identify $X$ linearly and isometrically with a dense subspace of $X^*$.
All of this is to say, the adjoint of $T$, when $X$ and $Y$ are Pre-Hilbert, has a slightly larger domain and codomain. If you're lucky, you might find that you can restrict the domain of $T^*$ back to $Y$, and have it map into $X$ (a really lazy example would be the $0$ map). However, this need not be the case.
For example, let $Z$ be a Hilbert space, and $X \subsetneq Y \subsetneq Z$ are dense subspaces of $Z$. Let $T : X \to Y$ be the inclusion map, i.e. $Tx = x$ for all $x \in X$. If $g \in Y^*$, then $(T^*g)(x) = g(T(x)) = g(x)$, i.e. $T^* g = g|_X$.
We can also look at $T^*$ as an operator on $Z$. We can extend $g \in Y^*$ and $T^* g \in X^*$ uniquely to some functional $f \in Z^*$, since these functionals are uniformly continuous, $X$ and $Y$ are dense in $Z$, and $g$ is already an extension of $T^* g$ to the larger domain $Y \subsetneq Z$. By Riesz Representation Theorem, there exists some unique $u \in Z$ such that $\langle \cdot, u \rangle = f$, so in this sense, we get $T^* u = u$. That is, $T^*$, considered as an operator on $Z$, is the identity operator.
If $g = \langle \cdot, v \rangle$ for some $v \in Y$ (which is to say, $g$ belongs to the embedding of $Y$ into $Y^* \simeq Z$), then $T^* g = \langle \cdot, v \rangle$, considered as a function on $Z$. If $v \in Y \setminus X$, then this functional will not belong to the embedding of $X$ into $Z$. As such, restricting the domain of $T^*$ to $Y$ will not necessarily map cleanly into $X$. So, indeed, the larger domain/codomain may be necessary.
