(January 2006 #7a) Let \(f\) be continuous on \([0,1]\) and \(f(0)=f(1)=0\). Show that there is a sequence of polynomials \(\{P_n\}\) such that \(x(1-x)P_n(x)\) converges to \(f\) uniformly.
Solution
Goal: Given \(f \in C[0,1]\) with \(f(0) = f(1) = 0\), find polynomials \(P_n\) with \(x(1-x)P_n(x) \to f(x)\) uniformly on \([0,1]\).
1. The Bernstein polynomials \(B_N f\) converge to \(f\) uniformly on \([0,1]\).Proof: the Weierstrass approximation theorem via Bernstein polynomials: \(B_Nf(x) = \sum_{k=0}^N f(k/N)\binom{N}{k}x^k(1-x)^{N-k} \to f(x)\) uniformly.
2. \(B_Nf\) vanishes at both endpoints: \(B_Nf(0) = f(0) = 0\) and \(B_Nf(1) = f(1) = 0\).Proof: in the sum, at \(x = 0\) only the \(k = 0\) term survives and equals \(f(0)\); at \(x = 1\) only the \(k = N\) term survives and equals \(f(1)\).Both are \(0\) by hypothesis.
3. Every polynomial \(p\) with \(p(0) = p(1) = 0\) is divisible by \(x(1-x)\): \(p(x) = x(1-x)q(x)\) for a polynomial \(q\).Proof: \(p(0) = 0\) means the constant term of \(p\) is \(0\), so \(p(x) = x\tilde p(x)\); \(p(1) = 0\) means \(\tilde p(1) = 0\), so \(\tilde p(x) = (1-x)q(x)\).Since \(x\) and \(1-x\) are coprime, \(p = x(1-x)q\).
4. Apply 3 to \(p = B_Nf\): there are polynomials \(q_N\) with \(B_Nf(x) = x(1-x)q_N(x)\) for all \(x\).Proof: 2 gives the two endpoint zeros; 3 then gives the factorization.
5. Q.E.D. Proof: by 1, \(x(1-x)q_N(x) = B_Nf(x) \to f(x)\) uniformly; take \(P_N := q_N\).