 Solution

For any $q\in {\Pi }_n$, the difference $p-q$ lies in ${\Pi }_n$ and is orthogonal to $f-p$. The Pythagorean theorem for inner product spaces givesThus the orthogonality condition proves minimality, and equality forces $q=p$.

Write the requested linear polynomial as $p=a+bx$. Orthogonality to one and $x$ givesOdd integrals vanish, leaving $2/5-2a/3=0$ and $-2b/5=0$. ConsequentlyThe weight's single zero does not destroy positive definiteness on continuous functions: a nonzero continuous function is nonzero on an interval containing points where $x^2>0$. The preceding projection argument therefore applies to this inner product.