 Solution

For any admissible $f$, part (a) rewrites the constraints asHence $f^{(k)}-s$ is orthogonal to every $M_i$ and therefore to their span, which contains $s$. The Pythagorean theorem in an inner-product space givesChoose any $k$-fold antiderivative $\sigma$ of $s$. The identity from part (a) and $(M_i,s)=k!{\gamma }_i$ show that $\sigma$ satisfies all prescribed divided differences, and equality holds in the norm bound. Thereforecharacterizes the minimizers. They are unique up to addition of an arbitrary polynomial in ${\mathcal{P}}_{k-1}$, which changes neither the order-$k$ divided differences nor the $k$th derivative.