Show that $p(D) f(x)$ has at least one zero on $[a, b]$, where D denotes $\frac{d}{dx}$.

Question

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 $\frac{d}{dx}$.

Source: Putnam 1956

First thing that comes to mind is Rolle's theorem.

Also, is the question below similar?

Let $P(z)$ be a polynomial with real coefficients whose roots can be covered by a disk of radius $R$. Prove that for any real number $k$, the roots of the polynomial $n P(z)−kP'(z)$ can be covered by a disk of radius $R + |k|$, where n is the degree of $P(z)$, and $P'(z)$ is the derivative.

Answer 1

Rolle’s Theorem is sufficient with sufficiently clever application. In particular, notice that $D-c=e^{cx}De^{-cx}$.

Let’s prove it here with induction that when $p$ has degree $n$ and f has $m$ roots then $p(D)f$ has $m-n$ roots. When $n$ is 0, $p$ is 1, so the $m$ roots of $f$ are the roots of $1f$ so it’s true. Otherwise, since $p$ has real roots, we can write $p(x)=q(x)(x-c)$ for some real $c$. Then, $(D-c)f= e^{cx}De^{-cx}f$. We have that $e^{-cx}f$ has the same $m$ roots as $f$, and by Rolles on the interval between each of the roots, $De^{-cx}f$ has $m-1$ roots. Multiplying this by $e^{cx}$ gives that $(D-c)f$ has the same $m-1$ roots. Applying the inductive hypothesis to $q(D)((D-c)f)$ gives that it has $(m-1)-(n-1)=m-n$ roots as desired.

Letting $m=n+1$ gives the desired result.

Answer 2

I don't think Rolle's theorem quite works for the $n = 1$ case. To see this, let $p(D) = D - c$ for some $c \in R$. Then a zero of $p(D)f$ is a point $x_0 \in [a, b]$ where $Df(x_0) = c f(x_0)$, but if $x_1, x_2$ are two distinct zeros of $f$, Rolle's theorem only guarantees an $x_0$ between them such that $Df(x_0) = c f(x_1) = c f(x_2) = 0$; since $f(x_0)$ might not be zero, Rolle's theorem doesn't work.

What does work for $n = 1$ is to consider $(D - c)f(x_1) = Df(x_1)$ and $(D - c)f(x_2) = Df(x_2)$. If either of these is zero, then we're done. Otherwise, we assume that $x_1, x_2$ are consecutive zeros of $f$. You should then be able to show that $(D - c)f(x_1), (D - c)f(x_2)$ have to have opposite signs. Since $(D - c)f$ is the derivative of some function, it follows that it must have a zero between $x_1$ and $x_2$ (Since $(D - c)f$ may not be continuous, the usual Intermediate Value Theorem might not apply).

With the case $n = 1$ done, I think the best thing to do would be to try the next few cases and see if you can't extract a general proof. Also, note the the case $n = 0$ is trivial.