The non-degenerate properties of duality pairing?

Question

As I'm learning dual space, I am confused of the idea of non-degenerate. On the book it says:

Define the duality pairing as: \begin{align} \langle \cdot,\cdot \rangle : V^* \times V &\longrightarrow \mathbb R, \\ (f,{\bf v}) & \longmapsto f({\bf v}). \end{align} Then, the following properties hold: \begin{align} {\rm (i)} &\quad \forall {\rm v} \in V \ \ \, \, \, \langle f,{\bf v}\rangle = 0 \quad {\rm implies} \quad f=0 \\ {\rm (ii)} &\quad \forall f \in V^* \ \ \langle f,{\bf v}\rangle = 0 \quad {\rm implies} \quad {\bf v}={\bf0} \end{align}

Here is the proof that's on the book:

Proof. $\rm (i)$ It follows trivially from the definition of duality pairing. To prove $\rm (ii)$ assume the contrary, that there exists ${\bf v}\ne{\bf 0}$ such that $\langle f,{\bf v}\rangle={\bf 0}$ for every $f \in V^\ast$. Consider the direct sum $$V=\Bbb R{\bf v}\oplus W$$ where $W$ is a complement of one-dimensional subspace $\Bbb R{\bf v}$. Setting $f({\bf v})=1$ and $f=0$ on $W$, we extend $f$ by linearity $$f(\alpha {\bf v}+{\bf w})=\alpha f({\bf v})+f({\bf w})=\alpha$$ meaning that $f$ is a well-defined linear functional on $V$. Obviously, $f({\bf v})\ne0$, which contradicts the assumption. $\blacksquare$

What I'm confused in this proof is:

1. What does $\Bbb R{\bf v}$ stands for?

2. why he could set $f({\bf v}) = 1$ when\begin{align} \langle f,{\bf v}\rangle = 0 \end{align} (which means $f({\bf v}) = 0$?)
   Alternatively, you could write your own proof if you think this isn't a good way to prove it. I just don't understand why this property must be the case.

3. What's the importance of these two properties? (What will happen if these properties don't hold?) Probably I will understand this once I know how to prove $\rm (ii)$.

Answer 1

$1.$ Typically $\Bbb Rv=\{\lambda v\,\mid\,\lambda\in\Bbb R\}$. In general, if $x$ is an element of a structure $X$ which has a notion of multiplication and $Y\subseteq X$ we write $Yx=\{\lambda x\,\mid\,\lambda\in Y\}$. Another example of this form would be $5\Bbb Z$.

$2.$ While the author attempts to do a proof by contradiction he actually shows the contrapositive (this is a common mistake) which is the cause of the confusion. To put it different, he proves that for every ${\bf v}\ne{\bf 0}$ you can construct a linear functional $f\in V^\ast$ which does not vanish on all of $V$. This is the contrapositive of the implication in $(ii)$ thus proving $(ii)$.

$3.$ The first property is intuitively clear: a functional (i.e. a function) is uniquely determined by its values on all possible arguments. In particular, if the functional vanishes for all values it is already the trivial functional. For the second property: given a non-zero vector one can construct a non-trivial functional, or conversely the only vector which gives rise to the trivial functional (via the given dualitiy pairing) is the zero vector. This can be used in the following way:

For a fixed $v\in V$ defining $\phi_v\colon V^\ast\to\Bbb R,f\mapsto f(v)$ gives an element of $V^{\ast\ast}$. Using this we can define $\varphi\colon V\to V^{\ast\ast},\,\varphi(v)=\phi_v$ which is a linear map, whose injectivity follows from property $(ii)$ (this establishes an isomorphism $V\cong V^{\ast\ast}$ if $\dim V<\infty$).

As the above shows (and the given proof already suggests) property $(ii)$ is related to the injectivity of certain maps involving the dual space.