# Show that $\{f_1,f_2\}$ is a basis for $V^*$, and find the basis for the dual and find the basis for $V^{**}$ that is $\{f_1,f_2\}$

Let $V=P_1(\mathbb{R})$ and $p \in V$. Define $f_1(p)=\int_0^1 p(t) dt$ and $f_2(p)=\int_0^2 p(t) dt$. Show that $\{f_1,f_2\}$ is a basis for $V^*$, and find the basis for the dual. Next find the basis for $V^{**}$ that is dual to $\{f_1,f_2\}$.

I have found the basis for the dual, which is $\{2-2x,x-1/2\}$. Next I have to show that the $\{f_1,f_2\}$ is linearly independent and spans $V^*$. I'm getting confused in trying to argue that this set is in fact a basis. So far I have: to show that it spans, I have: $\forall f \in V^* \exists\alpha, \beta \in F$ s.t. $f=\alpha f_1(p)+\beta f_2(p)$.
Then for the set to be linearly independent, suppose that $\exists \alpha \neq 0, \beta \neq 0$ s.t. $\alpha f_1(p)+\beta f_2(p)$, then proceed to show a contradiction, so then we have that $\{f_1,f_2\}$ is a basis of $V^*$.

I'm unsure how to go about finding a basis for $V^{**}$, and would very much appreciate any help in finding the basis. Thanks in advance.

1. To verify that $\{f_1, f_2\}$ is a basis for $V^{\ast}$, it suffices to show it is linearly independent, as you know that $\dim(V) = \dim(V^{\ast})$ when $V$ is finite-dimensional. You're on the right track: set $\alpha f_1 + \beta f_2 = 0$. This would mean that, for any $p \in P_1(\mathbb{R})$, $$(\alpha f_1 + \beta f_2)(p) = 0$$ Since $p(t) = a_0 + a_1 t$, can you show that the above can only equal zero for all possible $p(t)$ if $\alpha = \beta = 0$?
2. One of the most interesting features about the double-dual, $V^{\ast \ast}$, is that elements of $V$ can be regarded as elements of $V^{\ast \ast}$ in a very natural way: if $v \in V$, then we can construct the corresponding map $\mathrm{ev}_v \in V^{**}$, called the evaluation map at $v$, such that for any $f \in V^*$, $\mathrm{ev}_v(f) = f(v)$; aka it merely acts on a dual vector $f$ by evaluating $f$ at the particular point $v$. This may help you to find a basis for $V^{**}$.