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Choose a best uniform approximation $p_n\in {\mathcal{P}}_n$. It exists because this vector subspace is a finite-dimensional vector space and a minimizing sequence $q_j$ is bounded: $\Vert q_j{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }+\Vert f-q_j{\Vert }_{\infty }$. Finite-dimensional equivalence of norms then supplies a convergent subsequence of coefficients. If $E_n(f)>0$, the Chebyshev alternation theorem gives $n+2$ alternating error extrema. The intermediate value theorem supplies a zero $x_{n,k}$ of $f-p_n$ in each of the $n+1$ intervening open intervals. These interpolation nodes are distinct, and uniqueness of polynomial interpolation givesThe convergence follows from the Weierstrass approximation theorem. If $E_n(f)=0$, choose any $n+1$ distinct nodes: the polynomial interpolant is still $f=p_n$. This constructs interpolation nodes from best uniform approximation; the nodes depend on $f$, and no universal node system for all continuous functions is asserted.

For each $n$, choose the best uniform approximation $p_n^{\ast }$. If its error is nonzero, take its $n+2$ alternating extrema. The intermediate value theorem supplies a point $y_j$ strictly between the $j$th and $(j+1)$st extrema such that $f(y_j)=p_n^{\ast }(y_j)$, for $0\le j\le n$. These $n+1$ points are distinct. The polynomial interpolation problem at them has a unique solution of degree at most $n$, so its solution is exactly $p_n^{\ast }$. This constructs the interpolation nodes from best uniform approximation.

If $E_n(f)=0$, then $f$ itself belongs to the polynomial space, and any $n+1$ distinct nodes give the same conclusion. In either case, take $x_{n,j}=y_j$ or the arbitrary nodes just described. ThenThe Weierstrass approximation theorem gives $E_n(f)\to 0$; alternatively, the bound derived in part 4(b), combined with uniform continuity, gives this convergence directly. ThereforeThe nodes may depend on $f$ and on $n$. The conclusion does not assert a universal sequence of interpolation grids working for every continuous target.