 Solution

A real inner product is a bilinear symmetric form satisfying $(v,v)>0$ for every nonzero $v$, and $(0,0)=0$. Its norm is $\Vert v\Vert =\sqrt{(v,v)}$. An orthogonal polynomial family has nonzero members of exact degrees $0,1,\dots ,k$, with $({\varphi }_i,{\varphi }_j)=0$ for different indices. Exact degree is part of this definition; merely belonging to the indicated polynomial vector spaces would not suffice.

Distinct degrees make ${\varphi }_0,\dots ,{\varphi }_{\ell -1}$ linearly independent. There are $\ell$ of them in the $\ell$-dimensional space ${\Pi }_{\ell -1}$, so they form a basis. Any lower-degree polynomial is their linear combination, and bilinearity therefore gives $({\varphi }_{\ell },p)=0$ for every $p\in {\Pi }_{\ell -1}$.

Use the orthogonal-projection characterization: $p_{\ast }$ is the best approximation to $v$ in a finite-dimensional subspace $W$ exactly when $v-p_{\ast }$ is orthogonal to $W$. Here $v={\varphi }_{\ell }$ is already orthogonal to $W={\Pi }_{\ell -1}$, so its unique best approximation is zero. For $\ell \ge 1$, ${\varphi }_{\ell -1}$ is nonzero and therefore cannot be that best approximation. Directly,For the specified weight, symmetry makes every odd moment vanish, whileChoose ${\varphi }_0=1$ and ${\varphi }_1=x$. A monic quadratic orthogonal to $x$ has no linear term by symmetry, and orthogonality to 1 requires $\int |x|(x^2-c)dx=1/2-c=0$. Thus a valid answer isThe weight vanishes only at one point, so a nonzero polynomial has positive squared norm; the form is indeed positive definite on the polynomial vector space. Nonzero rescalings of individual family members give equally valid unnormalized answers.