Let p(x) be a real polynomial of degree n with leading coefficient 1 and all roots real. Let R be the reals and f : [a, b] → R be an n times differentiable function with at least n + 1 distinct zeros. Show that p(D) f(x) has at least one zero on [a, b], where D denotes d/dx.
Solution
The basic tool is Rolle's theorem which tells us that there is a zero of f '(x) between any two zeros of f(x). But we would like a zero of (D - α)f between any two zeros of f. For that, notice that the zeros of f(x) are also zeros of e-α x f(x). So if f(x) has n + 1 zeros, then the derivative of e-α x f(x) has at least n zeros. But the dervative of e-α x f(x) is e-α x times (D - α) f(x). e-α x is never zero, so (D - α) f(x) has at least n zeros.
That (more than) proves the result for p(x) of degree 1. We now use induction on n, the degree of p(x). Suppose it is true for n. Let p(x) be a real polynomial with real roots and degree n + 1. Then p(x) = q(x) (x - α) for some α and q(x) of degree n. Let f be an n+1 times differentiable real function on [a, b] with at least n + 2 distinct zeros. Then we proved above that (D - α) f(x) has at least n + 1 distinct zeros. It is also at least n times differentiable. So, by the inductive hypothesis, q(D) (D - α) f(x) has at least one zero, which proves the result is true for n + 1.
© John Scholes
jscholes@kalva.demon.co.uk
4 Dec 1999