Permanent problem IDPPL 072
Analysis
Given a_(i)^n∈ [-1,1] for all 1≤ i≤ n<∈fty we define p_(i)^n as the unique polynomial of degree n-1 such that p_(i)^n(a_(i)^n)=1 and p_(i)^n(a_(i')^n)=0 if 1≤ i'≤ n with i≠ i'. We similarly define L^nf(x) = ∑_(1≤ i≤ n)f(a_i^n)p_i^n(x), the unique polynomial of degree n-1 which agrees with f on a_i^n for 1≤ i≤ n (that is, the sequence of Lagrange interpolation polynomials). Is there such a sequence of a_i^n such that for every continuous f:[-1,1]→ ℝ there exists some x∈ [-1,1] where \limsup_(n→ ∈fty) ∑_(1≤ i≤ n)| p_(i)^n(x)|=∈fty and yet L^nf(x) → f(x)? Is there such a sequence such that \limsup_(n→ ∈fty) ∑_(1≤ i≤ n)| p_(i)^n(x)|=∈fty for every x∈ [-1,1] and yet for every continuous f:[-1,1]→ ℝ there exists x∈ [-1,1] with L^nf(x) → f(x)?
Given a_(i)^n∈ [-1,1] for all 1≤ i≤ n<∈fty we define p_(i)^n as the unique polynomial of degree n-1 such that p_(i)^n(a_(i)^n)=1 and p_(i)^n(a_(i')^n)=0 if 1≤ i'≤ n with i≠ i'. We similarly define L^nf(x) = ∑_(1≤ i≤ n)f(a_i^n)p_i^n(x), the unique polynomial of degree n-1 which agrees with f on a_i^n for 1≤ i≤ n (that is, the sequence of Lagrange interpolation polynomials). Is there such a sequence of a_i^n such that for every continuous f:[-1,1]→ ℝ there exists some x∈ [-1,1] where \limsup_(n→ ∈fty) ∑_(1≤ i≤ n)| p_(i)^n(x)|=∈fty and yet L^nf(x) → f(x)? Is there such a sequence such that \limsup_(n→ ∈fty) ∑_(1≤ i≤ n)| p_(i)^n(x)|=∈fty for every x∈ [-1,1] and yet for every continuous f:[-1,1]→ ℝ there exists x∈ [-1,1] with L^nf(x) → f(x)?
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