Question about dual of dual of Hilbert space Let $H$ be a real Hilbert space and let $H'$ be the set of continuous linear functionals on $H$.
Then, I know by the Riesz Theorem that for every $L(\cdot) \in H'$, there exists a unique $u\in H$ so that $(u,\cdot)_H = L(\cdot)$, where $(\cdot,\cdot)_H$ denotes the inner product on H. So, I can write $H' = \{(u,\cdot)_H: u\in H\}$ 
Now, my question is if I consider $(H')'$, the set of continuous linear functionals on $H'$, how does an element in $(H')'$ act on $H'$?
That is, if $f(\cdot) \in (H')'$, what is $f$ explicitly and also what is $f((u,\cdot)_H)$??
 A: The Riesz theorem gives the existence of an isometric isomorphism $\phi: \mathbb{H} \to \mathbb{H}^*$ such that
$f(x) = \langle \phi^{-1}(f), x \rangle $ for $f \in \mathbb{H}^*$. This
induces an inner product
$\langle f, g \rangle_{*} = \langle \phi^{-1}(f),\phi^{-1}(g)  \rangle$.
Since $\mathbb{H}^*$ is also a Hilbert space, the Riesz theorem gives the existence of an isometric isomorphism $\eta: \mathbb{H}^* \to \mathbb{H}^{**}$ such that
$J(f) = \langle \eta^{-1}(J), f \rangle_{{*}} $ for all $J\in \mathbb{H}^{**}$.
Then we have $J(f) = \langle \eta^{-1}(J), f \rangle_{{*}} = \langle f, \eta^{-1}(J) \rangle_{{*}} = \langle \phi^{-1}(f), \phi^{-1}(\eta^{-1}(J)) \rangle = f((\eta \circ \phi)^{-1}(J))$.
If we let $j= \phi^{-1}(\eta^{-1}(J) \in \mathbb{H}$, we can identify $J$ with $j$ by $J(f) = f(j)$.
A: Given $f \in H'$ it's not really proper to write $f = \langle \cdot, u\rangle$ and say that this is a closed form solution for $f$ (where $u \in H$ is the unique element guaranteed by the Riesz theorem).  With that being said, the relationship between $H''$ and $H'$ is the same as that between $H'$ and $H$.  That is, given $F \in H''$ there is a unique $f \in H'$ where $F(g) = \langle g, f\rangle_{H'}$ for every $g \in H'$.  There really isn't a need to refer back to the elements in $H$ if you're talking about elements in $H''$ (although we do have the fact that $H$ is isometrically embedded in $H''$).
