I am a first year student, studying linear algebra. In the lecture we briefly discussed double dual spaces and I am not sure if I understood it correctly: we take a function f that is an element of the dual space and we evaluate the function on a vector from V and its value is an element of the double dual space?
Thank you very much in advance.
Nope, you're missing a bit here. Long story short: the elements of the double dual space are functions that take a function $f$ from the dual space, and evaluate the function on a vector from $V$. That is, an element of the double dual space is a function of the form $f \mapsto f(v)$. It's nice to think of this weird function as just being that vector $v$ from $V$.
Long story long: for simplicity, I'll talk about vector spaces over $\Bbb R$, but the same applies over arbitrary fields.
The first thing to understand is that the set $\mathcal L(U,V)$ of linear transformations between two vector spaces $U$ and $V$ forms a vector space. For example, if $U = \Bbb R^n$ and $V = \Bbb R^m$, then $\mathcal L(U,V)$ is canonically identified with the space $\Bbb R^{m \times n}$ of $m \times n$ matrices. In general, $\dim (\mathcal L(U,V)) = \dim(U) \cdot \dim(V)$. Dimension is important because any vector spaces of the same (finite) dimension are isomorphic.
Now, for any vector space $V$, $V^* = \mathcal L(V,\Bbb R)$ is the dual space of $V$. The elements of $V^*$ are called linear functionals; they are linear transformations that take vectors and produce numbers. Notably, $\dim(V^*) = \dim(V) \cdot \dim(\Bbb R) = \dim(V) \cdot 1 = \dim (V)$. So, any (finite dimensional) space is isomorphic to its dual space.
The double dual space is the dual of the dual. That is, $V^{**} = \mathcal L(V^*, \Bbb R) = \mathcal L(\mathcal L(V,\Bbb R),\Bbb R)$. The elements of this space are linear transformations that take linear functionals and produce numbers. If that seems weird and unintuitive, that's fine: it should.
Just like $V^*$, $V^{**}$ is isomorphic to $V$, since $\dim(V^{**}) = \dim(V^*)\cdot 1 = \dim(V)$. However, it turns out that $V^{**}$ is canonically isomorphic to $V$. That is (for our purposes), it is isomorphic in a "really nice way". In particular, there is a really nice invertible linear map that takes us from $V$ to $V^{**}$, and it's so slick that we can think of $V$ and $V^{**}$ as being "essentially the same space".
Let's describe that map $\alpha:V \to V^{**}$. For any vector $v \in V$, we want an element $\alpha(v) = \alpha_v \in V^{**}$, which is to say that $\alpha_v$ takes in functionals $f \in V^*$, and produces a number. So, we define $$ \alpha_v(f) = f(v) $$ In other words, the question of "is $V$ canonically isomorphic to $V^{**}$?" can be roughly translated as "is there a natural way to use $v$ to make an element $f \in V^*$ into a number?" Our answer is, "yes: plug $v$ into $f$". For any vector $v \in V$, $\alpha_v$ is the element of $V^{**}$ that tells you to plug in $v$.
Yes, $\psi_x : E' \to \Bbb{R}$ defined by $\psi_x(f) = f(x)$ is an element of $E''$.
That even give us a canonical injection from $E$ to $E''$ ( $x\mapsto \psi_x$)
But note that often not all elements of $E''$ can be written as such (there's not always a bijection between $E$ and $E''$)
