Finding $g_t$ such that $(f|g)=f(t)$ 
Let $V$ be the vector space of the polynomials over the reals of degree less than or equal to $3$ with the inner product $(f,g)=\int_{0}^{1}f(t)g(t)\,dt$.  If $t$ is a real number, find the polynomial $g_t$ in $V$ such that $(f|g_t)=f(t)$ for all $f \in V$.

My only question about this is the $g_t$.  My book defines the notation $f_\beta(\alpha)=(\alpha|\beta)$ as follows:  Let $V$ be any inner product space and let $\beta$ be a fixed vector in $V$.  We define a function $f_\beta$ from $V$ into the scalar field by $f_\beta(\alpha)=(\alpha|\beta)$.  In this case I'm having problems interpreting what the $g_t$ should be.  Is this saying that $g_t(?)=(?|t)$ where the $?$'s are the basis vectors - in other words, is $g_t$ a fixed polynomial?
 A: Hint: for $t$ fixed, let be $g_t(x) = a x^3 + b x^2 + c x + d$ (the coefficients will depend of $t$). By the condition, taking $f(x) = x^n$, $n = 0,1,2,3$:
$$
t^n = f(t) = (f,g_t) = \int_0^1(a x^3 + b x^2 + c x + d)x^n\,dx = \cdots
$$
This gives a linear system of 4 equations with 4 unknowns...
A: I would like to change the notation, to make it clearer. This is the strategy for a the polynomial space or degree at most $1$, but works the same way in your case:
Put $f(t) = a_0 + a_1t$
Fix $\xi \in \mathbb{R}$. We are looking for at a polynomial $g_\xi(t)$ s.t. 
$$ \int f(t)g_{\xi}(t) dt = f(\xi) = a_0 + a_1\xi $$
Now put  $g_\xi(t) = b_0 + b_1t$.
Multiplying and integrating term by term gives: 
$$a_0(b_0 + \frac{b_1}{2}) = a_0 $$
$$a_1(\frac{b_1}{2} + \frac{b_1}{3}) = a_1\xi $$
Now solve this system in two unknowns to get: 
$$b_0 = -6\xi +4 $$
$$b_1 = 12\xi - 6 $$
And you can check that with this choice of $b_0, b_1$
$$ \int f(t)g_{\xi}(t) dt = f(\xi) $$
The same calculations applies in your four-dimensional space, but the calculations are a bit longer.
