Let \(V\) be the vector space of the polynomials over \(\RR\) of degree less than or equal to 3, with the inner product \[ (f \mid g) = \int_0^1 f(t)g(t)\,dt. \] If \(t\) is a real number, find the polynomial \(g_t\) in \(V\) such that \((f \mid g_t) = f(t)\) for all \(f\) in \(V\).
Solution
1. Let \(g_t(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3\).
write \(g_t\) in the monomial basis.
2. The condition \((f \mid g_t) = f(t)\) for all \(f \in V\) means \(g_t\) is the reproducing kernel, i.e. \(\int_0^1 f(x) g_t(x)\,dx = f(t)\) for all \(f\).
definition.
3. Taking \(f = 1, x, x^2, x^3\) gives the system:\[\int_0^1 g_t(x)\,dx = 1,\ \int_0^1 x g_t(x)\,dx = t,\ \int_0^1 x^2 g_t(x)\,dx = t^2,\ \int_0^1 x^3 g_t(x)\,dx = t^3.\]
2 with \(f = x^k\).
4. \(\int_0^1 x^k\,dx = \frac{1}{k+1}\), so the system is:\[a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4} = 1,\]\[\frac{a_0}{2} + \frac{a_1}{3} + \frac{a_2}{4} + \frac{a_3}{5} = t,\]\[\frac{a_0}{3} + \frac{a_1}{4} + \frac{a_2}{5} + \frac{a_3}{6} = t^2,\]\[\frac{a_0}{4} + \frac{a_1}{5} + \frac{a_2}{6} + \frac{a_3}{7} = t^3.\]
3, integrating term by term.
5. This is a linear system in \(a_0, a_1, a_2, a_3\) (the Hilbert matrix), which has a unique solution (the Hilbert matrix is invertible).
4.
6. Hence \(g_t\) exists and is unique, given by solving the system in
4.
5.
7. Q.E.D.
6.