Past exam of the options I course of the University of Oxford / 2006 / P / P2 / Solution

The form is symmetric and bilinear by the corresponding properties of finite real sums. Its squared norm is $⟨p,p⟩={\sum }_{i}p(x_i{)}^2\ge 0$. Equality means $p$ vanishes at all $n+1$ distinct nodes. A nonzero polynomial of degree at most $n$ has at most $n$ roots, so equality forces $p=0$. Thus this is an inner product on ${\Pi }_n$.

On ${\Pi }_{n+1}$ it is only positive semidefinite: the nonzero node polynomial $\omega (x)={\prod }_{i=0}^n(x-x_i)$ satisfies $⟨\omega ,\omega ⟩=0$. Hence it is not an inner product on ${\Pi }_{n+1}$. The finite-evaluation inner product on polynomials works exactly when the evaluation map is injective on the chosen space.

Finally, if $q\ne 0$, positivity applied to $p-tq$ and the choice $t=⟨p,q⟩/⟨q,q⟩$ giveRearranging proves Cauchy-Schwarz inequality in its required form:If $q=0$ both sides of the asserted inequality are zero. Equality holds exactly when $p,q$ are linearly dependent, since in the nonzero case the minimizing $p-tq$ must have zero norm.