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If a nonzero continuous real function on an interval is orthogonal to every polynomial of degree below $n$, it has at least $n$ distinct interior zeros. If it had only finitely many zeros and fewer than $n$ sign changes, multiply it by the product of linear factors at its sign-changing zeros. The product has constant sign and is nonzero on an open interval, so its integral cannot vanish. Orthogonality to this degree-below-$n$ polynomial contradicts that conclusion. Infinitely many interior zeros already satisfy the assertion. Orthogonal polynomials of respective degrees $0,\dots ,n-1$ form a basis of the needed polynomial space.

An $(n+1)$-node quadrature exact on degree $n$ is exact through degree $2n+1$ precisely when its node polynomial $p={\prod }_i(x-x_i)$ is orthogonal to every degree-$n$ polynomial for the integration weight. Necessity applies the rule to $pr$. Sufficiency divides an arbitrary degree-$2n+1$ polynomial as $pq+r$, with both $q,r$ of degree at most $n$; the integral and quadrature of $pq$ vanish while both agree on $r$.

The orthogonal projection $P_F$ onto a closed subspace $F$ sends $x$ to the component in the decomposition $x=P_{F}x+(I-P_F)x$ with $(I-P_F)x\in F^{\perp }$. It is a contraction: $\Vert P_{F}x\Vert \le \Vert x\Vert$.

Near a non-characteristic plane $x=0$, use the small cap ${\omega }_{\epsilon }=\{x>0,x+y^2<\epsilon \}$. Its curved face is non-characteristic by continuity of the principal symbol. A polynomial coordinate change flattens that face. Uniform real analytic coefficient bounds and the uniform Cauchy radius for polynomial forcing solve $P^{\ast }v=g$ throughout a common cap, with zero boundary jet on the curved face. The complete boundary-jet condition for formal adjoints then gives ${\int }_{{\omega }_{\epsilon }}ug=0$ for every polynomial $g$. Uniform approximation of $\bar{u}$ gives $\int |u{|}^2=0$. Repeating on the other side of the plane proves two-sided local uniqueness.

Parameter continuation changes $\lambda$ by a small amount and solves again for $u$, using the previous solution as the initial guess for Newton method. If $F_u$ is an invertible matrix, the implicit function theorem gives a local solution branch $u(\lambda )$, withLocal convergence of Newton method also needs regularity and a sufficiently small step: the cited result assumes a Lipschitz continuous derivative and an invertible $F_u$ at the seed.^[ref]

For the cubic polynomial,The derivative becomes unbounded at either continuation fold, and the scalar Newton method divides by $F_x=0$ there. Just beyond the upper continuation fold, no nearby real solution exists at the increased $\lambda$, although a distant negative-$x$ solution still exists. Converging to that solution would jump away from the local solution branch. Decreasing the parameter step cannot make $x(\lambda )$ a valid smooth local description at the bend.