Past exam of the options I course of the University of Oxford / 2006 / P (Numerical Analysis)

The Lagrange interpolation polynomial constructs existence explicitly:Distinct nodes make every denominator nonzero, and ${\ell }_j(x_k)={\delta }_{jk}$, so $p(x_k)=f_k$ and $p$ has degree at most $n$. If two such polynomials existed, their difference would have degree at most $n$ but at least $n+1$ distinct roots. The root bound for a nonzero polynomial forces that difference to vanish. Thus the interpolating polynomial exists and is unique.

The polynomial $p(x)=x^2-2x$ already has degree at most three and gives the specified data at every node. Uniqueness of polynomial interpolation therefore gives

Let $f(x)=x^4-6x^3+11x^2-5x$. The node polynomial isIt vanishes at all four nodes, so subtracting it leaves the data unchanged. Since $f(x)-\omega (x)=x$, uniqueness of polynomial interpolation gives

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.